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  • Aside from a prototype, what is the next best thing to satisfy a user

    - by user1639998
    Aside from a prototype, what is the next best thing to satisfy a user who really wants to know what an application will be like? Choice 1 A process model Choice 2 An interaction diagram Choice 3 A data-flow diagram Choice 4 A class diagram Choice 5 A class-state diagram 2 is my partner's choice 3 is my choice I stick to my colors also he stick to his. which one is better?

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  • Camping: Return user to recent entries, but keep errors

    - by echoback
    Users can view a specific entry in my webapp with a URL. /entry/8, for example. If an entry doesn't exist, "Entry not found" gets appended to @messages and I render an error page. I'd like to redirect the user to /recent, but I can't figure out a good way to keep the error message around to be displayed. There are other actions that need to take place in the Recent controller, so I can't just duplicate the query and render :posts.

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  • Checking array of censored words against user submitted content

    - by steve-o
    Hello, I have set up an array of censored words and I want to check that a user submitted comment doesn't contain any of these words. What is the most efficient way of doing this? All I've come up with so far is splitting the string into an array of words and checking it against the array of censored words, but I've a feeling there's a neater way of doing this.

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  • Stop a user directly accesing a page

    - by James Jeffery
    I have a page called create.php. It receives post variables and sets up accounts. I don't want that page to be accessible by a user. What's the conventional way of achieving this? I think I remember reading something about including a page with a CONSTANT. If the CONSTANT is not present the page has been accessed directly. I think Wordpress also do it.

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  • Why can a local root turn into any LDAP user?

    - by Daniel Gollás
    I know this has been asked here before, but I am not satisfied with the answers and don't know if it's ok to revive and hijack an older question. We have workstations that authenticate users on an LDAP server. However, the local root user can su into any LDAP user without needing a password. From my perspective this sounds like a huge security problem that I would hope could be avoided at the server level. I can imagine the following scenario where a user can impersonate another and don't know how to prevent it: UserA has limited permissions, but can log into a company workstation using their LDAP password. They can cat /etc/ldap.conf and figure out the LDAP server's address and can ifconfig to check out their own IP address. (This is just an example of how to get the LDAP address, I don't think that is usually a secret and obscurity is not hard to overcome) UserA takes out their own personal laptop, configures authentication and network interfaces to match the company workstation and plugs in the network cable from the workstation to their laptop, boots and logs in as local root (it's his laptop, so he has local root) As root, they su into any other user on LDAP that may or may not have more permissions (without needing a password!), but at the very least, they can impersonate that user without any problem. The other answers on here say that this is normal UNIX behavior, but it sounds really insecure. Can the impersonated user act as that user on an NFS mount for example? (the laptop even has the same IP address). I know they won't be able to act as root on a remote machine, but they can still be any other user they want! There must be a way to prevent this on the LDAP server level right? Or maybe at the NFS server level? Is there some part of the process that I'm missing that actually prevents this? Thanks!!

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  • asp.net membership create user error

    - by nitroxn
    Hi, I encountered an issue while creating users using the asp.net user membership. The membership provider configuration is as follows- The error generated by ASP.net configuration application is as follows- An error was encountered. Please return to the previous page and try again. The following message may help in diagnosing the problem: Exception has been thrown by the target of an invocation. at System.RuntimeMethodHandle._InvokeMethodFast(IRuntimeMethodInfo method, Object target, Object[] arguments, SignatureStruct& sig, MethodAttributes methodAttributes, RuntimeType typeOwner) at System.RuntimeMethodHandle.InvokeMethodFast(IRuntimeMethodInfo method, Object target, Object[] arguments, Signature sig, MethodAttributes methodAttributes, RuntimeType typeOwner) at System.Reflection.RuntimeMethodInfo.Invoke(Object obj, BindingFlags invokeAttr, Binder binder, Object[] parameters, CultureInfo culture, Boolean skipVisibilityChecks) at System.Reflection.RuntimeMethodInfo.Invoke(Object obj, BindingFlags invokeAttr, Binder binder, Object[] parameters, CultureInfo culture) at System.Web.Administration.WebAdminMembershipProvider.CallWebAdminMembershipProviderHelperMethodOutParams(String methodName, Object[] parameters, Type[] paramTypes) at System.Web.Administration.WebAdminMembershipProvider.CreateUser(String username, String password, String email, String passwordQuestion, String passwordAnswer, Boolean isApproved, Object providerUserKey, MembershipCreateStatus& status) at System.Web.UI.WebControls.CreateUserWizard.AttemptCreateUser() at System.Web.UI.WebControls.CreateUserWizard.OnNextButtonClick(WizardNavigationEventArgs e) at System.Web.UI.WebControls.Wizard.OnBubbleEvent(Object source, EventArgs e) at System.Web.UI.WebControls.CreateUserWizard.OnBubbleEvent(Object source, EventArgs e) at System.Web.UI.WebControls.Wizard.WizardChildTable.OnBubbleEvent(Object source, EventArgs args) at System.Web.UI.Control.RaiseBubbleEvent(Object source, EventArgs args) at System.Web.UI.WebControls.Button.OnCommand(CommandEventArgs e) at System.Web.UI.WebControls.Button.RaisePostBackEvent(String eventArgument) at System.Web.UI.WebControls.Button.System.Web.UI.IPostBackEventHandler.RaisePostBackEvent(String eventArgument) at System.Web.UI.Page.RaisePostBackEvent(IPostBackEventHandler sourceControl, String eventArgument) at System.Web.UI.Page.RaisePostBackEvent(NameValueCollection postData) at System.Web.UI.Page.ProcessRequestMain(Boolean includeStagesBeforeAsyncPoint, Boolean includeStagesAfterAsyncPoint)

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  • not able to register sip user on red5server, using red5phone

    - by sunil221
    I start the red5, and then i start red5phone i try to register sip user , details i provide are username = 999999 password = **** ip = asteriskserverip and i got --- Registering contact -- sip:[email protected]:5072 the right contact could be --- sip :99999@asteriskserverip this is the log: SipUserAgent - listen -> Init... Red5SIP register [SIPUser] register RegisterAgent: Registering contact <sip:[email protected]:5072> (it expires in 3600 secs) RegisterAgent: Registration failure: No response from server. [SIPUser] SIP Registration failure Timeout RegisterAgent: Failed Registration stop try. Red5SIP Client leaving app 1 Red5SIP Client closing client 35C1B495-E084-1651-0C40-559437CAC7E1 Release ports: sip port 5072 audio port 3002 Release port number:5072 Release port number:3002 [SIPUser] close1 [SIPUser] hangup [SIPUser] closeStreams RTMPUser stopStream [SIPUser] unregister RegisterAgent: Unregistering contact <sip:[email protected]:5072> SipUserAgent - hangup -> Init... SipUserAgent - closeMediaApplication -> Init... [SIPUser] provider.halt RegisterAgent: Registration failure: No response from server. [SIPUser] SIP Registration failure Timeout please let me know if i am doing anything wrong. regards Sunil

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  • Ruby on Rails user login form in main layout

    - by Jimmy
    Hey guys I have a simple ror application for some demo stuff. I am running into a problem with trying to move my login form from the users controller and just have it displayed in the main navigation so that a user can easily log in from anywhere. The problem is the form doesn't generate the correct action for the html form. Ruby code: <% form_for(url_for(:action => 'login'), :method => 'post') do |f| %> <li><%= f.text_field("username") %></li> <li><%= f.password_field("password") %></li> <li><%= submit_tag("Login")%></li> <% end %> The problem is depending on the controller I am currently in this generates HTML actions like <form action="/home" method="post">...</form> when it should be generating HTML like so <form action="/login" method="post">...</form> I know I could simply do an HTML form here but I want to keep things as easy to maintain as possible. Any help?

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  • Designing Mobile SMS text advertising system

    - by Ramraj Edagutti
    Currently, I am working on a product where we have an SMS text advertising system, and using this, we setup advertising campaigns for clients, and later these campaigns are sent to the end users. This is very similar to Google Adwords, but targeted to Mobile users via SMS. Just to give an overview of the system Each Campaign is mapped to an advertiser Campaign has start date and end date Campaign has a filter condition(s) or query to select the target user base from our database (to whom we send Campaigns) Target user base can be fixed, for e.g send campaign to 10000 users Target user base can also be dynamic based on query condition, for e.g send campaign to users who are active and from a particular state, district, town etc. (this way user base will be keep changing on daily basis) Campaign can have multiple campaign messages Each campaign message has start date and end date Each campaign message can have multiple message texts for different locales, for e.g English,Hindi,Telugu etc After creating an advertisement campaign, we run daily night job to provision the target user base for that a particular campaign in a separate table, and another daily job runs on morning times and checks provisioned table for campaigns and targeted users and sends the campaign to users via SMS. Problem is, current UI for creating advertising campaigns is designed in a very technical manner, I mean, normal user or business owner or clients can not use the UI to create a campaign. Below are reasons why the UI is very technical in nature Filter condition(s) or query input filed, takes user ids or mobile numbers or SQL queries. Most of times or almost every time, we use big SQL queries So we end up storing SQL queries in a database for a campaign, later we use this SQL query to fetch targeted user base. For scheduling these campaigns, we have input filed on UI which takes quartz cron expression(s) ( for e.g. send campaign on "0 0 9 1-10 MAR 2012" ), again very technical in nature Normal user or business owner, can not use the UI for creating campaigns for reasons mentioned above, Currently, we ourself (developers) helping clients to setup/create campaigns. we are trying to re-design the UI to make it more user friendly so that any user can go to UI and create an advertisement campaign by himself. I am thinking of re-designing the current UI similar to Google Adwords interface, especially for selecting target users based on user geography like country, state, city etc. I also need to select users based user subscription(s), which might make system even more complex. And also, for campaign scheduling, I am thinking of using weekdays with hours. For example, I will shows Monday to Sunday on UI, and user can select the from hours, to hours etc. Any better ideas or suggestion on how to design UI in very user friendly manner and what design should be followed on server side code (we write backend code on java/jpa/spring/quartz)? And I am looking for ideas or design patterns on how to build SQL queries (using JPA/Hinernate) programmatically on server side, based on varies conditions like based on country, state, town, village, and user subscriptions.

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  • Can Google Employees See My Saved Google Chrome Passwords?

    - by Jason Fitzpatrick
    Storing your passwords in your web browser seems like a great time saver, but are the passwords secure and inaccessible to others (even employees of the browser company) when squirreled away? Today’s Question & Answer session comes to us courtesy of SuperUser—a subdivision of Stack Exchange, a community-driven grouping of Q&A web sites. The Question SuperUser reader MMA is curious if Google employees have (or could have) access to the passwords he stores in Google Chrome: I understand that we are really tempted to save our passwords in Google Chrome. The likely benefit is two fold, You don’t need to (memorize and) input those long and cryptic passwords. These are available wherever you are once you log in to your Google account. The last point sparked my doubt. Since the password is available anywhere, the storage must in some central location, and this should be at Google. Now, my simple question is, can a Google employee see my passwords? Searching over the Internet revealed several articles/messages. Do you save passwords in Chrome? Maybe you should reconsider: Talks about your passwords being stolen by someone who has access to your computer account. Nothing mentioned about the central storage security and vulnerability. There is even a response from Chrome browser security tech lead about the first issue. Chrome’s insane password security strategy: Mostly along the same line. You can steal password from somebody if you have access to the computer account. How to Steal Passwords Saved in Google Chrome in 5 Simple Steps: Teaches you how to actually perform the act mentioned in the previous two when you have access to somebody else’s account. There are many more (including this one at this site), mostly along the same line, points, counter-points, huge debates. I refrain from mentioning them here, simply carry a search if you want to find them. Coming back to my original query, can a Google employee see my password? Since I can view the password using a simple button, definitely they can be unhashed (decrypted) even if encrypted. This is very different from the passwords saved in Unix-like OS’s where the saved password can never be seen in plain text. They use a one-way encryption algorithm to encrypt your passwords. This encrypted password is then stored in the passwd or shadow file. When you attempt to login, the password you type in is encrypted again and compared with the entry in the file that stores your passwords. If they match, it must be the same password, and you are allowed access. Thus, a superuser can change my password, can block my account, but he can never see my password. So are his concerns well founded or will a little insight dispel his worry? The Answer SuperUser contributor Zeel helps put his mind at ease: Short answer: No* Passwords stored on your local machine can be decrypted by Chrome, as long as your OS user account is logged in. And then you can view those in plain text. At first this seems horrible, but how did you think auto-fill worked? When that password field gets filled in, Chrome must insert the real password into the HTML form element – or else the page wouldn’t work right, and you could not submit the form. And if the connection to the website is not over HTTPS, the plain text is then sent over the internet. In other words, if chrome can’t get the plain text passwords, then they are totally useless. A one way hash is no good, because we need to use them. Now the passwords are in fact encrypted, the only way to get them back to plain text is to have the decryption key. That key is your Google password, or a secondary key you can set up. When you sign into Chrome and sync the Google servers will transmit the encrypted passwords, settings, bookmarks, auto-fill, etc, to your local machine. Here Chrome will decrypt the information and be able to use it. On Google’s end all that info is stored in its encrpyted state, and they do not have the key to decrypt it. Your account password is checked against a hash to log in to Google, and even if you let chrome remember it, that encrypted version is hidden in the same bundle as the other passwords, impossible to access. So an employee could probably grab a dump of the encrypted data, but it wouldn’t do them any good, since they would have no way to use it.* So no, Google employees can not** access your passwords, since they are encrypted on their servers. * However, do not forget that any system that can be accessed by an authorized user can be accessed by an unauthorized user. Some systems are easier to break than other, but none are fail-proof. . . That being said, I think I will trust Google and the millions they spend on security systems, over any other password storage solution. And heck, I’m a wimpy nerd, it would be easier to beat the passwords out of me than break Google’s encryption. ** I am also assuming that there isn’t a person who just happens to work for Google gaining access to your local machine. In that case you are screwed, but employment at Google isn’t actually a factor any more. Moral: Hit Win + L before leaving machine. While we agree with zeel that it’s a pretty safe bet (as long as your computer is not compromised) that your passwords are in fact safe while stored in Chrome, we prefer to encrypt all our logins and passwords in a LastPass vault. Have something to add to the explanation? Sound off in the the comments. Want to read more answers from other tech-savvy Stack Exchange users? Check out the full discussion thread here.     

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  • Can see samba shares but not access them

    - by nitefrog
    For the life of me I cannot figure this one out. I have samba installed and set up on the ubuntu box and on the Win7 box I CAN SEE all the shares I created. I created two users on ubuntu that map to the users in windows. On ubuntu they are both admins, user A & B on Windows User A is admin and user B is poweruser. User A can see both shares and access them, but user B can see everythin, but only access the homes directory, the other directory throws an error. I have two drives in Ubuntu and this is the smb.config file (I am new to samba): [global] workgroup = WORKGROUP server string = %h server (Samba, Ubuntu) wins support = no dns proxy = yes name resolve order = lmhosts host wins bcast log file = /var/log/samba/log.%m max log size = 1000 syslog = 0 panic action = /usr/share/samba/panic-action %d security = user encrypt passwords = true passdb backend = tdbsam obey pam restrictions = yes unix password sync = yes passwd program = /usr/bin/passwd %u passwd chat = *Enter\snew\s*\spassword:* %n\n *Retype\snew\s*\spassword:* %n\n *password\supdated\ssuccessfully* . pam password change = yes map to guest = bad user ; usershare max shares = 100 usershare allow guests = yes And here is the share section: Both user A & B can access this from windows. No problems. [homes] comment = Home Directories browseable = no writable = yes Both User A & B can see this share, but only user A can access it. User B get an error thrown. [stuff] comment = Unixmen File Server path = /media/data/appinstall/ browseable = yes ;writable = no read only = yes hosts allow = The permission for the media/data/appinstall/ is as follows: appInstall properties: share name: stuff Allow others to create and delete files in this folder is cheeked Guest access (for people without a user account) is checked permissions: Owner: user A Folder Access: Create and delete files File Access: --- Group: user A Folder Access: Create and delete files File Access: --- Others Folder Access: Create and delete files File Access: --- I am at a loss and need to get this work. Any ideas? The goal is to have a setup like this. 3 users on window machines. Each user on the data drive will have their own personal folder where they are the ones that can only access, then another folder where 2 of the users will have read only and one user full access. I had this setup before on windows, but after what happened I am NEVER going back to windows, so Unix here I am to stay! I am really stuck. I am running Ubuntu 11. I could reformat again and put on version 10 if that would make life easier. I have been dealing with this since Wed. 3pm. Thanks.

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  • PeopleSoft New Design Solves Navigation Problem

    - by Applications User Experience
    Anna Budovsky, User Experience Principal Designer, Applications User Experience In PeopleSoft we strive to improve User Experience on all levels. Simplifying navigation and streamlining access to the most important pages is always an important goal. No one likes to waste time waiting for pages to load and watching a spinning glass going on and on. Those performance-affecting server trips, page-load waits and just-too-many clicks were complained about for a long time. Something had to be done. A few new designs came in PeopleSoft 9.2 helping users to access their everyday work areas easier and faster. For example, Dashboard and Work Center aggregate most accessed information sections on a single page; Related Information allows users to complete transaction-related-research without interrupting a transaction and Secure Search gets users to a specific page directly. Today we’ll talk about the Actions menu. Most PeopleSoft pages are shared between individual products and product lines. It means changing the content on a single page involves Oracle development and quality assurance time for making and testing the changes. In order to streamline the navigation and cut down on accessing PeopleSoft pages one-page-at-a-time, we introduced a new menu design. The new menu allows accessing shared pages without the Oracle development team making any local changes, and it works as an additional one-click-path to specific high-traffic actionable pages. Let’s look at how many steps it took to Change Salary for an employee in HCM 9.1 before: Figure 1. BEFORE: The 6 steps a user would take to Change Salary in PeopleSoft HCM 9.1 In PeopleSoft 9.1 it took 5 steps + page loading time + additional verification time for making sure a correct employee is selected from the table. In PeopleSoft 9.2 it only takes 2 steps. To complete Ad Hoc Change Salary action, the user can start from the HCM Manager's Dashboard, click the Action menu within a table, choose a menu option, and access a correct employee’s details page to take an action. Figure 2. AFTER: The 2 steps a user would take to Change Salary in PeopleSoft HCM 9.2 The new menu is placed on a row level which ensures the user accesses the correct employee’s details page. The Actions menu separates menu options into hierarchical sections which help to scan and access the correct option quickly. The new menu’s small size and its structure enabled users to access high-traffic pages from any page and from any part of the page. No more spinning hourglass, no more multiple pages upload. The flexible design fits anywhere on a page and provides a fast and reliable path to the correct destination within the product. Now users can: Access any target page no matter how far it is buried from the starting point; Reduce navigation and page-load time; Improve productivity and reduce errors. The new menu design is available and widely used in all PeopleSoft 9.2 product lines.

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  • JavaOne Session Report: “50 Tips in 50 Minutes for GlassFish Fans”

    - by Janice J. Heiss
    At JavaOne 2012 on Monday, Oracle’s Engineer Chris Kasso, and Technology Evangelist Arun Gupta, presented a head-spinning session (CON4701) in which they offered 50 tips for GlassFish fans. Kasso and Gupta alternated back and forth with each presenting 10 tips at a time. An audience of about (appropriately) 50 attentive and appreciative developers was on hand in what has to be one of the most information-packed sessions ever at JavaOne!Aside: I experienced one of the quiet joys of JavaOne when, just before the session began, I spotted Java Champion and JavaOne Rock Star Adam Bien sitting nearby – Adam is someone I have been fortunate to know for many years.GlassFish is a freely available, commercially supported Java EE reference implementation. The session prioritized quantity of tips over depth of information and offered tips that are intended for both seasoned and new users, that are meant to increase the range of functional options available to GlassFish users. The focus was on lesser-known dimensions of GlassFish. Attendees were encouraged to pursue tips that contained new information for them. All 50 tips can be accessed here.Below are several examples of more elaborate tips and a final practical tip on how to get in touch with these folks. Tip #1: Using the login Command * To execute a remote command with asadmin you must provide the admin's user name and password.* The login command allows you to store the login credentials to be reused in subsequent commands.* Can be logged into multiple servers (distinguish by host and port). Example:     % asadmin --host ouch login     Enter admin user name [default: admin]>     Enter admin password>     Login information relevant to admin user name [admin]     for host [ouch] and admin port [4848] stored at     [/Users/ckasso/.asadminpass] successfully.     Make sure that this file remains protected.     Information stored in this file will be used by     asadmin commands to manage the associated domain.     Command login executed successfully.     % asadmin --host ouch list-clusters     c1 not running     Command list-clusters executed successfully.Tip #4: Using the AS_DEBUG Env Variable* Environment variable to control client side debug output* Exposes: command processing info URL used to access the command:                           http://localhost:4848/__asadmin/uptime Raw response from the server Example:   % export AS_DEBUG=true  % asadmin uptime  CLASSPATH= ./../glassfish/modules/admin-cli.jar  Commands: [uptime]  asadmin extension directory: /work/gf-3.1.2/glassfish3/glassfish/lib/asadm      ------- RAW RESPONSE  ---------   Signature-Version: 1.0   message: Up 7 mins 10 secs   milliseconds_value: 430194   keys: milliseconds   milliseconds_name: milliseconds   use-main-children-attribute: false   exit-code: SUCCESS  ------- RAW RESPONSE  ---------Tip #11: Using Password Aliases * Some resources require a password to access (e.g. DB, JMS, etc.).* The resource connector is defined in the domain.xml.Example:Suppose the DB resource you wish to access requires an entry like this in the domain.xml:     <property name="password" value="secretp@ssword"/>But company policies do not allow you to store the password in the clear.* Use password aliases to avoid storing the password in the domain.xml* Create a password alias:     % asadmin create-password-alias DB_pw_alias     Enter the alias password>     Enter the alias password again>     Command create-password-alias executed successfully.* The password is stored in domain's encrypted keystore.* Now update the password value in the domain.xml:     <property name="password" value="${ALIAS=DB_pw_alias}"/>Tip #21: How to Start GlassFish as a Service * Configuring a server to automatically start at boot can be tedious.* Each platform does it differently.* The create-service command makes this easy.   Windows: creates a Windows service Linux: /etc/init.d script Solaris: Service Management Facility (SMF) service * Must execute create-service with admin privileges.* Can be used for the DAS or instances* Try it first with the --dry-run option.* There is a (unsupported) _delete-serverExample:     # asadmin create-service domain1     The Service was created successfully. Here are the details:     Name of the service:application/GlassFish/domain1     Type of the service:Domain     Configuration location of the service:/work/gf-3.1.2.2/glassfish3/glassfish/domains     Manifest file location on the system:/var/svc/manifest/application/GlassFish/domain1_work_gf-3.1.2.2_glassfish3_glassfish_domains/Domain-service-smf.xml.     You have created the service but you need to start it yourself. Here are the most typical Solaris commands of interest:     * /usr/bin/svcs  -a | grep domain1  // status     * /usr/sbin/svcadm enable domain1 // start     * /usr/sbin/svcadm disable domain1 // stop     * /usr/sbin/svccfg delete domain1 // uninstallTip #34: Posting a Command via REST* Use wget/curl to execute commands on the DAS.Example:  Deploying an application   % curl -s -S \       -H 'Accept: application/json' -X POST \       -H 'X-Requested-By: anyvalue' \       -F id=@/path/to/application.war \       -F force=true http://localhost:4848/management/domain/applications/application* Use @ before a file name to tell curl to send the file's contents.* The force option tells GlassFish to force the deployment in case the application is already deployed.* Use wget/curl to execute commands on the DAS.Example:  Deploying an application   % curl -s -S \       -H 'Accept: application/json' -X POST \       -H 'X-Requested-By: anyvalue' \       -F id=@/path/to/application.war \       -F force=true http://localhost:4848/management/domain/applications/application* Use @ before a file name to tell curl to send the file's contents.* The force option tells GlassFish to force the deployment in case the application is already deployed.Tip #46: Upgrading to a Newer Version * Upgrade applications and configuration from an earlier version* Upgrade Tool: Side-by-side upgrade– GUI: asupgrade– CLI: asupgrade --c– What happens ?* Copies older source domain -> target domain directory* asadmin start-domain --upgrade* Update Tool and pkg: In-place upgrade– GUI: updatetool, install all Available Updates– CLI: pkg image-update– Upgrade the domain* asadmin start-domain --upgradeTip #50: How to reach us?* GlassFish Forum: http://www.java.net/forums/glassfish/glassfish* [email protected]* @glassfish* facebook.com/glassfish* youtube.com/GlassFishVideos* blogs.oracle.com/theaquariumArun Gupta acknowledged that their method of presentation was experimental and actively solicited feedback about the session. The best way to reach them is on the GlassFish user forum.In addition, check out Gupta’s new book Java EE 6 Pocket Guide.

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  • Using C# FindControl to find a user-control in the master page

    - by Jisaak
    So all I want to do is simply find a user control I load based on a drop down selection. I have the user control added but now I'm trying to find the control so I can access a couple properties off of it and I can't find the control for the life of me. I'm actually doing all of this in the master page and there is no code in the default.aspx page itself. Any help would be appreciated. MasterPage.aspx <body> <form id="form1" runat="server"> <div> <asp:ScriptManager runat="server"> </asp:ScriptManager> </div> <asp:UpdatePanel ID="UpdatePanel2" runat="server" UpdateMode="Conditional" ChildrenAsTriggers="false" OnLoad="UpdatePanel2_Load"> <ContentTemplate> <div class="toolbar"> <div class="section"> <asp:DropDownList ID="ddlDesiredPage" runat="server" AutoPostBack="True" EnableViewState="True" OnSelectedIndexChanged="goToSelectedPage"> </asp:DropDownList> &nbsp; <asp:DropDownList ID="ddlDesiredPageSP" runat="server" AutoPostBack="True" EnableViewState="True" OnSelectedIndexChanged="goToSelectedPage"> </asp:DropDownList> <br /> <span class="toolbarText">Select a Page to Edit</span> </div> <div class="options"> <div class="toolbarButton"> <asp:LinkButton ID="lnkSave" CssClass="modal" runat="server" OnClick="lnkSave_Click"><span class="icon" id="saveIcon" title="Save"></span>Save</asp:LinkButton> </div> </div> </div> </ContentTemplate> <Triggers> </Triggers> </asp:UpdatePanel> <div id="contentContainer"> <asp:UpdatePanel ID="UpdatePanel1" runat="server" OnLoad="UpdatePanel1_Load" UpdateMode="Conditional" ChildrenAsTriggers="False"> <ContentTemplate> <asp:ContentPlaceHolder ID="ContentPlaceHolder1" runat="server"> </asp:ContentPlaceHolder> </ContentTemplate> <Triggers> <asp:AsyncPostBackTrigger ControlID="lnkHome" EventName="Click" /> <asp:AsyncPostBackTrigger ControlID="rdoTemplate" EventName="SelectedIndexChanged" /> </Triggers> </asp:UpdatePanel> </div> MasterPage.cs protected void goToSelectedPage(object sender, System.EventArgs e) { temp1 ct = this.Page.Master.LoadControl("temp1.ascx") as temp1; ct.ID = "TestMe"; this.UpdatePanel1.ContentTemplateContainer.Controls.Add(ct); } //This is where I CANNOT SEEM TO FIND THE CONTROL //////////////////////////////////////// protected void lnkSave_Click(object sender, System.EventArgs e) { UpdatePanel teest = this.FindControl("UpdatePanel1") as UpdatePanel; Control test2 = teest.ContentTemplateContainer.FindControl("ctl09") as Control; temp1 test3 = test2.FindControl("TestMe") as temp1; string maybe = test3.Col1TopTitle; } Here I don't understand what it's telling me. for "par" I get "ctl09" and I have no idea how I am supposed to find this control. temp1.ascx.cs protected void Page_Load(object sender, EventArgs e) { string ppp = this.ID; string par = this.Parent.ID; }

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  • ASP.NET: Button in user control not posting back

    - by Ronnie Overby
    I have a simple user control with this code: <%@ Control Language="C#" AutoEventWireup="true" CodeFile="Pager.ascx.cs" Inherits="Pager" %> <table style="width: 100%;"> <tr> <td runat="server" id="PageControls"> <!-- This button has the problem: --> <asp:Button ID="btnPrevPage" runat="server" Text="&larr;" OnClick="btnPrevPage_Click" /> Page <asp:DropDownList runat="server" ID="ddlPage" AutoPostBack="true" OnSelectedIndexChanged="ddlPage_SelectedIndexChanged" /> of <asp:Label ID="lblTotalPages" runat="server" /> <!-- This button has the problem: --> <asp:Button ID="btnNextPage" runat="server" Text="&rarr;" OnClick="btnNextPage_Click" /> </td> <td align="right" runat="server" id="itemsPerPageControls"> <asp:Literal ID="perPageText1" runat="server" /> <asp:DropDownList ID="ddlItemsPerPage" runat="server" AutoPostBack="true" OnSelectedIndexChanged="ddlItemsPerPage_SelectedIndexChanged" /> <asp:Literal ID="perPageText2" runat="server" /> </td> </tr> </table> As you can see, the 2 buttons are wired to click events, which are defined correctly in the code-behind. Now, here is how I include an instance of the control on my page: <uc:Pager ID="Pager1" runat="server" TotalRecords="100" DisplayItemsPerPage="true" ItemsPerPageChoices="10,25,50,100" ItemsPerPageFormatString="Sessions/Page: {0}" PageSize="25" OnPageChanged="PageChanged" OnPageSizeChanged="PageChanged" /> I noticed though, that the 2 buttons in my user control weren't causing a post back when clicked. The drop down list does cause postback, though. Here is the rendered HTML: <table style="width: 100%;"> <tr> <td id="ctl00_MainContent_Pager1_PageControls" align="left"> <!-- No onclick event! Why? --> <input type="submit" name="ctl00$MainContent$Pager1$btnPrevPage" value="?" id="ctl00_MainContent_Pager1_btnPrevPage" /> Page <select name="ctl00$MainContent$Pager1$ddlPage" onchange="javascript:setTimeout('__doPostBack(\'ctl00$MainContent$Pager1$ddlPage\',\'\')', 0)" id="ctl00_MainContent_Pager1_ddlPage"> <option value="1">1</option> <option selected="selected" value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> <option value="6">6</option> </select> of <span id="ctl00_MainContent_Pager1_lblTotalPages">6</span> <!-- No onclick event! Why? --> <input type="submit" name="ctl00$MainContent$Pager1$btnNextPage" value="?" id="ctl00_MainContent_Pager1_btnNextPage" /> </td> <td id="ctl00_MainContent_Pager1_itemsPerPageControls" align="right"> Sessions/Page: <select name="ctl00$MainContent$Pager1$ddlItemsPerPage" onchange="javascript:setTimeout('__doPostBack(\'ctl00$MainContent$Pager1$ddlItemsPerPage\',\'\')', 0)" id="ctl00_MainContent_Pager1_ddlItemsPerPage"> <option value="10">10</option> <option selected="selected" value="25">25</option> <option value="50">50</option> <option value="100">100</option> </select> </td> </tr> </table> And, as you can see, there is no onclick attribute being rendered in the button's input elements. Why not?

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  • initializing structs using user-input information

    - by johnny boy
    I am trying to make a program that works with poker (texas holdem) starting hands; each hand has a value from 1 to 169, and i want to be able to input each card and whether they are suited or not, and have those values correspond to a series of structs. Here is the code so far, i cant seem to get it to work (im a beginning programmer). oh and im using visual studio 2005 by the way #include "stdafx.h" #include <iostream> int main() { using namespace std; struct FirstCard { struct SecondCard { int s; //suited int n; //non-suited }; SecondCard s14; SecondCard s13; SecondCard s12; SecondCard s11; SecondCard s10; SecondCard s9; SecondCard s8; SecondCard s7; SecondCard s6; SecondCard s5; SecondCard s4; SecondCard s3; SecondCard s2; }; FirstCard s14; //ace FirstCard s13; //king FirstCard s12; //queen FirstCard s11; //jack FirstCard s10; FirstCard s9; FirstCard s8; FirstCard s7; FirstCard s6; FirstCard s5; FirstCard s4; FirstCard s3; FirstCard s2; s14.s14.n = 169; // these are the values that each combination s13.s13.n = 168; // will evaluate to, would eventually have s12.s12.n = 167; // hand combinations all the way down to 1 s11.s11.n = 166; s14.s13.s = 165; s14.s13.s = 164; s10.s10.n = 163; //10, 10, nonsuited s14.s13.n = 162; s14.s11.s = 161; s13.s12.s = 160;// king, queen, suited s9.s9.n = 159; s14.s10.s = 158; s14.s12.n = 157; s13.s11.s = 156; s8.s8.n = 155; s12.s11.s = 154; s13.s10.s = 153; s14.s9.s = 152; s14.s11.n = 151; cout << "enter first card: " << endl; cin >> somthing?//no idea what to put here, but this would somehow //read out the user input (a number from 2 to 14) //and assign it to the corresponding struct cout << firstcard.secondcard.suited_or_not << endl; //this would change depending //on what the user inputs system("Pause"); }

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  • get user selection and convert it to a String [Android]

    - by Kira
    Hello, I just got a Droid, and after having used it for a while, I felt like I wanted to make a program for it. The program that I am trying to make calculates the actual storage capacity of secondary storage mediums. The user select from a list of units that ranges from KB to YB and the size the entered gets put into a formula depending on the chosen unit. However, there is a bit of a problem with the program. From my testing, I have narrowed it down to the fact that the user's selection is not really being obtained from the spinner. Everything I look up seems to point me to a method quite similar to how it works in J2SE, but it does nothing. How am I actually supposed to get that data? Here is the Java source code for the app: package com.Actual.android; import android.app.Activity; import android.os.Bundle; import android.widget.*; import android.view.*; public class ActualStorageActivity extends Activity { Spinner selection; /* declare variable, in order to control spinner (ComboBox) */ ArrayAdapter adapter; /* declare an array adapter object, in order for spinner to work */ EditText size; /* declare variable to control textfield */ EditText result; /* declare variable to control textfield */ Button calculate; /* declare variable to control button */ Storage capacity = new Storage(); /* import custom class for formulas */ /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); // load content from XML selection = (Spinner)findViewById(R.id.spinner); adapter = ArrayAdapter.createFromResource(this, R.array.choices_array, android.R.layout.simple_spinner_dropdown_item); size = (EditText)findViewById(R.id.size); result = (EditText)findViewById(R.id.result); calculate = (Button)findViewById(R.id.submit); adapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); /* set resource for dropdown */ selection.setAdapter(adapter); // attach adapter to spinner result.setEnabled(false); // make read-only result.setText("usable storage"); } public void calcAction(View view) { String initial = size.getText().toString(); String unit = selection.getSelectedItem().toString(); String end = "Nothing"; double convert = Double.parseDouble(initial); capacity.setStorage(convert); if (unit == "KB") { end = Double.toString(capacity.getKB()); } else if (unit == "MB") { end = Double.toString(capacity.getMB()); } else if (unit == "GB") { end = Double.toString(capacity.getGB()); } else if (unit == "TB") { end = Double.toString(capacity.getTB()); } else if (unit == "PB") { end = Double.toString(capacity.getPB()); } else if (unit == "EB") { end = Double.toString(capacity.getEB()); } else if (unit == "ZB") { end = Double.toString(capacity.getZB()); } else if (unit == "YB") { end = Double.toString(capacity.getYB()); } else; result.setText(end); } }

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  • No route matches [GET] "/user/sign_out"

    - by user3399101
    So, I'm getting the below error when clicking on Sign Out on my drop down menu on the nav: No route matches [GET] "/user/sign_out" However, this only happens when using the sign out on the drop down nav (the hamburger menu for mobile devices) and not when clicking the sign out on the regular nav. See the code below: <div class="container demo-5"> <div class="main clearfix"> <div class="column"> <div id="dl-menu" class="dl-menuwrapper"> <button class="dl-trigger">Open Menu</button> <ul class="dl-menu dl-menu-toggle"> <div id="closebtn" onclick="closebtn()"></div> <% if user_signed_in? %> <li><%= link_to 'FAQ', faq_path %></li> <li><a href="#">Contact Us</a></li> <li><%= link_to 'My Account', account_path %></li> <li><%= link_to 'Sign Out', destroy_user_session_path, method: 'delete' %></li> <--- this is the line <% else %> <li><%= link_to 'FAQ', faq_path %></li> <li><a href="#">Contact Us</a></li> <li><%= link_to 'Sign In', new_user_session_path %></li> <li><%= link_to 'Free Trial', plans_path %></li> <% end %> </ul> </div><!-- /dl-menuwrapper --> </div> </div> </div><!-- /container --> </div> And this is the non-drop down code that works: <div class="signincontainer pull-right"> <div class="navbar-form navbar-right"> <% if user_signed_in? %> <%= link_to 'Sign out', destroy_user_session_path, class: 'btn signin-button', method: :delete %> <div class="btn signin-button usernamefont"><%= link_to current_user.full_name, account_path %></div> <% else %> ....rest of code here Updated error: ActionController::RoutingError (No route matches [GET] "/user/sign_out"): actionpack (4.0.4) lib/action_dispatch/middleware/debug_exceptions.rb:21:in `call' actionpack (4.0.4) lib/action_dispatch/middleware/show_exceptions.rb:30:in `call' railties (4.0.4) lib/rails/rack/logger.rb:38:in `call_app' railties (4.0.4) lib/rails/rack/logger.rb:20:in `block in call' activesupport (4.0.4) lib/active_support/tagged_logging.rb:68:in `block in tagged' activesupport (4.0.4) lib/active_support/tagged_logging.rb:26:in `tagged' activesupport (4.0.4) lib/active_support/tagged_logging.rb:68:in `tagged' railties (4.0.4) lib/rails/rack/logger.rb:20:in `call' quiet_assets (1.0.2) lib/quiet_assets.rb:18:in `call_with_quiet_assets' actionpack (4.0.4) lib/action_dispatch/middleware/request_id.rb:21:in `call' rack (1.5.2) lib/rack/methodoverride.rb:21:in `call' rack (1.5.2) lib/rack/ru

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  • Can't get SSH public key authentication to work

    - by Trey Parkman
    My server is running CentOS 5.3. I'm on a Mac running Leopard. I don't know which is responsible for this: I can log on to my server just fine via password authentication. I've gone through all of the steps for setting up PKA (as described at http://www.centos.org/docs/5/html/Deployment_Guide-en-US/s1-ssh-beyondshell.html), but when I use SSH, it refuses to even attempt publickey verification. Using the command ssh -vvv user@host (where -vvv cranks up verbosity to the maximum level) I get the following relevant output: debug2: key: /Users/me/.ssh/id_dsa (0x123456) debug1: Authentications that can continue: publickey,gssapi-with-mic,password debug3: start over, passed a different list publickey,gssapi-with-mic,password debug3: preferred keyboard-interactive,password debug3: authmethod_lookup password debug3: remaining preferred: ,password debug3: authmethod_is_enabled password debug1: Next authentication method: password followed by a prompt for my password. If I try to force the issue with ssh -vvv -o PreferredAuthentications=publickey user@host I get debug2: key: /Users/me/.ssh/id_dsa (0x123456) debug1: Authentications that can continue: publickey,gssapi-with-mic,password debug3: start over, passed a different list publickey,gssapi-with-mic,password debug3: preferred publickey debug3: authmethod_lookup publickey debug3: No more authentication methods to try. So, even though the server says it accepts the publickey authentication method, and my SSH client insists on it, I'm rebutted. (Note the conspicuous absence of an "Offering public key:" line above.) Any suggestions?

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  • Passwordless SSH not working - keys copied and permissions set

    - by Comcar
    I know this question has been asked, but I'm certain I've done what all the other answers suggest. Machine A: used keygen -t rsa to create id_rsa.pub in ~/.ssh/ copied Machine A's id_rsa.pub to Machine B user's home directory Made the file permissions of id_rsa.pub 600 Machine B added Machine A's pub key to authorised_keys and authorised_keys2: cat ~/id_rsa.pub ~/.ssh/authorised_keys2 made the file permissions of id_rsa.pub 600 I've also ensured both the .ssh directories have the permission 700 on both machine A and B. If I try to login to machine B from machine A, I get asked for the password, not the ssh pass phrase. I've got the root users on both machines to talk to each other using password-less ssh, but I can't get a normal user to do it. Do the user names have to be the same on both sides? Or is there some setting else where I've missed. Machine A is a Ubuntu 10.04 virtual machine running inside VirtualBox on a Windows 7 PC, Machine B is a dedicated Ubuntu 9.10 server UPDATE : I've run ssh with the option -vvv, which provides many many lines of output, but this is the last few commands: debug3: check_host_in_hostfile: filename /home/pete/.ssh/known_hosts debug3: check_host_in_hostfile: match line 1 debug1: Host '192.168.1.19' is known and matches the RSA host key. debug1: Found key in /home/pete/.ssh/known_hosts:1 debug2: bits set: 504/1024 debug1: ssh_rsa_verify: signature correct debug2: kex_derive_keys debug2: set_newkeys: mode 1 debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug3: Wrote 16 bytes for a total of 1015 debug2: set_newkeys: mode 0 debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug3: Wrote 48 bytes for a total of 1063 debug2: service_accept: ssh-userauth debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /home/pete/.ssh/identity ((nil)) debug2: key: /home/pete/.ssh/id_rsa (0x7ffe1baab9d0) debug2: key: /home/pete/.ssh/id_dsa ((nil)) debug3: Wrote 64 bytes for a total of 1127 debug1: Authentications that can continue: publickey,password debug3: start over, passed a different list publickey,password debug3: preferred gssapi-keyex,gssapi-with-mic,gssapi,publickey,keyboard-interactive,password debug3: authmethod_lookup publickey debug3: remaining preferred: keyboard-interactive,password debug3: authmethod_is_enabled publickey debug1: Next authentication method: publickey debug1: Trying private key: /home/pete/.ssh/identity debug3: no such identity: /home/pete/.ssh/identity debug1: Offering public key: /home/pete/.ssh/id_rsa debug3: send_pubkey_test debug2: we sent a publickey packet, wait for reply debug3: Wrote 368 bytes for a total of 1495 debug1: Authentications that can continue: publickey,password debug1: Trying private key: /home/pete/.ssh/id_dsa debug3: no such identity: /home/pete/.ssh/id_dsa debug2: we did not send a packet, disable method debug3: authmethod_lookup password debug3: remaining preferred: ,password debug3: authmethod_is_enabled password debug1: Next authentication method: password

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  • Odd log entries when starting up PotgreSQL

    - by Shadow
    When restarting pgSQL, I get the following log entries: 2010-02-10 16:08:05 EST LOG: received smart shutdown request 2010-02-10 16:08:05 EST LOG: autovacuum launcher shutting down 2010-02-10 16:08:05 EST LOG: shutting down 2010-02-10 16:08:05 EST LOG: database system is shut down 2010-02-10 16:08:07 EST LOG: database system was shut down at 2010-02-10 16:08:05 EST 2010-02-10 16:08:07 EST LOG: autovacuum launcher started 2010-02-10 16:08:07 EST LOG: database system is ready to accept connections 2010-02-10 16:08:07 EST LOG: connection received: host=[local] 2010-02-10 16:08:07 EST LOG: incomplete startup packet 2010-02-10 16:08:07 EST LOG: connection received: host=[local] 2010-02-10 16:08:07 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:08 EST LOG: connection received: host=[local] 2010-02-10 16:08:08 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:08 EST LOG: connection received: host=[local] 2010-02-10 16:08:08 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:09 EST LOG: connection received: host=[local] 2010-02-10 16:08:09 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:09 EST LOG: connection received: host=[local] 2010-02-10 16:08:09 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:10 EST LOG: connection received: host=[local] 2010-02-10 16:08:10 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:10 EST LOG: connection received: host=[local] 2010-02-10 16:08:10 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:11 EST LOG: connection received: host=[local] 2010-02-10 16:08:11 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:11 EST LOG: connection received: host=[local] 2010-02-10 16:08:11 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:12 EST LOG: connection received: host=[local] 2010-02-10 16:08:12 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:12 EST LOG: connection received: host=[local] 2010-02-10 16:08:12 EST FATAL: password authentication failed for user "postgres" 2010-02-10 16:08:12 EST LOG: connection received: host=[local] 2010-02-10 16:08:12 EST LOG: incomplete startup packet My question regarding a potential consequence of this is posted here: http://stackoverflow.com/questions/2238954/mdb2-says-connection-failed-db-logs-say-otherwise , but I didn't realize this was happening when I asked that question, and I figured this [part of the] problem is for SF. Edit: I can connect to the database and manipulate things normally with the psql CLI and the postgres user.

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