Search Results

Search found 22281 results on 892 pages for 'password management'.

Page 463/892 | < Previous Page | 459 460 461 462 463 464 465 466 467 468 469 470  | Next Page >

  • Passing Key-Value pair to a COM method in C#

    - by Sinnerman
    Hello, I have the following problem: I have a project in C# .Net where I use a third party COM component. So the problem is that the above component has a method, which takes a string and a number of key-value pairs as arguments. I already managed to call that method through JavaScript like that: var srcName srcName = top.cadView.addSource( 'Database', { driver : 'Oracle', host : '10.10.1.123', port : 1234, database : 'someSID', user : 'someuser', password : 'somepass' } ) if ( srcName != '' ) { ... } ... and it worked perfectly well. However I have no idea how to do the same using C#. I tried passing the pairs as Dictionary and Struct/Class but it throws me a "Specified cast is not valid." exception. I also tried using Hashtable like that: Hashtable args = new Hashtable(); args.Add("driver", "Oracle"); args.Add("host", "10.10.1.123"); args.Add("port", 1234); args.Add("database", "someSID"); args.Add("user", "someUser"); args.Add("password", "samePass"); String srcName = axCVCanvas.addSource("Database", args); and although it doesn't throw an exception it still won't do the job, writing me in a log file [ Error] [14:38:33.281] Cad::SourceDB::SourceDB(): missing parameter 'driver' Any help will be appreciated, thanks.

    Read the article

  • Can't read from RSOP_RegistryPolicySetting WMI class in root\RSOP namespace

    - by JCCyC
    The class is documented in http://msdn.microsoft.com/en-us/library/aa375050%28VS.85%29.aspx And from this page it seems it's not an abstract class: http://msdn.microsoft.com/en-us/library/aa375084%28VS.85%29.aspx But whenever I run the code below I get an "Invalid Class" exception in ManagementObjectSearcher.Get(). So, does this class exist or not? ManagementScope scope; ConnectionOptions options = new ConnectionOptions(); options.Username = tbUsername.Text; options.Password = tbPassword.Password; options.Authority = String.Format("ntlmdomain:{0}", tbDomain.Text); scope = new ManagementScope(String.Format("\\\\{0}\\root\\RSOP", tbHost.Text), options); scope.Connect(); ManagementObjectSearcher searcher = new ManagementObjectSearcher(scope, new ObjectQuery("SELECT * FROM RSOP_RegistryPolicySetting")); foreach (ManagementObject queryObj in searcher.Get()) { wmiResults.Text += String.Format("id={0}\n", queryObj["id"]); wmiResults.Text += String.Format("precedence={0}\n", queryObj["precedence"]); wmiResults.Text += String.Format("registryKey={0}\n", queryObj["registryKey"]); wmiResults.Text += String.Format("valueType={0}\n", queryObj["valueType"]); } In the first link above, it lists as a requirement something called a "MOF": "Rsopcls.mof". Is this something I should have but have not? How do I obtain it? Is it necessary in the querying machine or the queried machine? Or both? I do have two copies of this file: C:\Windows>dir rsop*.mof /s Volume in drive C has no label. Volume Serial Number is 245C-A6EF Directory of C:\Windows\System32\wbem 02/11/2006 05:22 100.388 rsop.mof 1 File(s) 100.388 bytes Directory of C:\Windows\winsxs\x86_microsoft-windows-grouppolicy-base-mof_31bf3856ad364e35_6.0.6001.18000_none_f2c4356a12313758 19/01/2008 07:03 100.388 rsop.mof 1 File(s) 100.388 bytes Total Files Listed: 2 File(s) 200.776 bytes 0 Dir(s) 6.625.456.128 bytes free

    Read the article

  • Configurating JOOMLA's e-mail notification for new account

    - by Dion
    Dear all... I'm using Joomla 1.5 to create a local site for my office. The site will be accessed locally via intranet, and my PC will be the localhost for the site. I'm using a Login pluggin, so that anyone who wanted to enter the site should create an account. In JOOMLA, all user who created their account for the first time will receive a notification e-mail like : "Hello pras, You have been added as a User to Information Center by an Administrator. This e-mail contains your username and password to log in to http://localhost/yaddayadda/ Username: hadisuryo.prasetio Password: xxxx Please do not respond to this message as it is automatically generated and is for information purposes only." but if the user click the URL in the mail, which is, "localhost/yaddayadda/" they will not be directed to my site, but to their own PC's localhost.... My question is : How can I Modified the e-mail or the site configuration so that the URL will not be "localhost/yaddayadda/" anymore, but will be "(My-IP adress)/yaddayadda" I'm not going to host my site to a web hosting service, just using my PC as a host. I've been trying to trace on each config and .ini files...it seems that i have to do something with the "JURI" function or the "$mosConfig_live_site" on the backlink.php file $mosConfig_absolute_path = JPATH_SITE; $mosConfig_live_site = JURI :: base(); $url_array = explode('/', $_SERVER['REQUEST_URI']); Can anyone give me assistance ? Thank You

    Read the article

  • How do I get ASP.NET login status controls to display a Log In option?

    - by Greg McNulty
    I have the following log in status controls on the top of my master page. It displays the logged in as, manager log in, and Log out options. However, when a user is not logged in, there is nothing displayed there. When the user is NOT logged in, is there a way to display a "Login" text link that takes you to the log in page and then "disappears" once the user is logged in? Any help is appreciated. Thanks! <asp:LoginName ID="LoginName1" runat="server" FormatString="Logged in as {0}" ForeColor="Aqua" /> <asp:LoginView ID="LoginView1" runat="server"> <RoleGroups> <asp:RoleGroup Roles="Managers"> <ContentTemplate> <asp:HyperLink ID="HyperLink1" runat="server" NavigateUrl="~/Management/management.aspx">Manage Site</asp:HyperLink> or <asp:LoginStatus id="LoginStatus1" runat="server" /> </ContentTemplate> </asp:RoleGroup> </RoleGroups> <LoggedInTemplate> (<asp:LoginStatus id="LoginStatus1" runat="server" />) </LoggedInTemplate> </asp:LoginView> ASP.NET 3.5 VWD 2008 C#

    Read the article

  • How can I read the properties of an object that I assign to the Session in ASP.NET MVC?

    - by quakkels
    Hey all, I'm trying my hand at creating a session which stores member information which the application can use to reveal certain navigation and allow access to certain pages and member role specific functionality. I've been able to assign my MemberLoggedIn object to the session in this way: //code excerpt start... MemberLoggedIn loggedIn = new MemberLoggedIn(); if (computedHash == member.Hash) { loggedIn.ID = member.ID; loggedIn.Username = member.Username; loggedIn.Email = member.Email; loggedIn.Superuser = member.Superuser; loggedIn.Active = member.Active; Session["loggedIn"] = loggedIn; } else if (ModelState.IsValid) { ModelState.AddModelError("Password", "Incorrect Username or Password."); } return View(); That works great. I then can send the properties of Session["loggedIn"] to the View in this way: [ChildActionOnly] public ActionResult Login() { if (Session["loggedIn"] != null) ViewData.Model = Session["loggedIn"]; else ViewData.Model = null; return PartialView(); } In the Partial View I can reference the session data by using Model.Username or Model.Superuser. However, it doesn't seem to work that way in the controller or in a custom Action Filter. Is there a way to get the equivalent of Session["loggedIn"].Username?

    Read the article

  • Java : HTTP POST Request

    - by SpunkerBaba
    I have to do a http post request to a web-service for authenticating the user with username and password. The Web-service guy gave me following information to construct HTTP Post request. POST /login/dologin HTTP/1.1 Host: webservice.companyname.com Content-Type: application/x-www-form-urlencoded Content-Length: 48 id=username&num=password&remember=on&output=xml The XML Response that i will be getting is <?xml version="1.0" encoding="ISO-8859-1"?> <login> <message><![CDATA[]]></message> <status><![CDATA[true]]></status> <Rlo><![CDATA[Username]]></Rlo> <Rsc><![CDATA[9L99PK1KGKSkfMbcsxvkF0S0UoldJ0SU]]></Rsc> <Rm><![CDATA[b59031b85bb127661105765722cd3531==AO1YjN5QDM5ITM]]></Rm> <Rl><![CDATA[[email protected]]]></Rl> <uid><![CDATA[3539145]]></uid> <Rmu><![CDATA[f8e8917f7964d4cc7c4c4226f060e3ea]]></Rmu> </login> This is what i am doing HttpPost postRequest = new HttpPost(urlString); How do i construct the rest of the parameters?

    Read the article

  • Saving a record in Authlogic table

    - by denniss
    I am using authlogic to do my authentication. The current model that serves as the authentication model is the user model. I want to add a "belongs to" relationship to user which means that I need a foreign key in the user table. Say the foreign key is called car_id in the user's model. However, for some reason, when I do u = User.find(1) u.car_id = 1 u.save! I get ActiveRecord::RecordInvalid: Validation failed: Password can't be blank My guess is that this has something to do with authlogic. I do not have validation on password on the user's model. This is the migration for the user's table. def self.up create_table :users do |t| t.string :email t.string :first_name t.string :last_name t.string :crypted_password t.string :password_salt t.string :persistence_token t.string :single_access_token t.string :perishable_token t.integer :login_count, :null => false, :default => 0 # optional, see Authlogic::Session::MagicColumns t.integer :failed_login_count, :null => false, :default => 0 # optional, see Authlogic::Session::MagicColumns t.datetime :last_request_at # optional, see Authlogic::Session::MagicColumns t.datetime :current_login_at # optional, see Authlogic::Session::MagicColumns t.datetime :last_login_at # optional, see Authlogic::Session::MagicColumns t.string :current_login_ip # optional, see Authlogic::Session::MagicColumns t.string :last_login_ip # optional, see Authlogic::Session::MagicColumns t.timestamps end end And later I added the car_id column to it. def self.up add_column :users, :user_id, :integer end Is there anyway for me to turn off this validation?

    Read the article

  • sending email with codeigniter

    - by Maru
    I have this MODEL and I get the email which I want to send class Cliente_Model extends CI_Model{ public function getInfo($id){ $this->db->select('*'); $this->db->from('pendientes'); $query = $this->db->get(); if($query->num_rows() > 0) { foreach ($query->result_array() as $row) { return $row['email']; } } else { return FALSE; } } } CONTROLLER $this->load->model('cliente_model', 'client'); $clientInfo = $this->client->getInfo($id); $this->email->from('[email protected]', 'Demo'); $this->email->to($clientInfo); $this->email->subject('Email Test'); $this->email->message('your user is '.$clientInfo.' and your password is '.$clave); $this->email->send(); and I need some help here, I can get the email and it can send it perfectly but in the message I need to send the password also and I don't know how I can get it from the model. thanks in advance!

    Read the article

  • Powershell invoke-command with PSCredential in line

    - by jaffa
    I need to be able to run a command on another server. This script acts as a bootstrap to another script which is run on the actual server. This works great on servers on the same domain, but if I need to run this script on a remote server, I need to specify credentials. The command is kicked off from a Msbuild targets file like so: <Target Name="PreDeployment" Condition="true" BeforeTargets="MSDeployPublish"> <Exec Command="powershell.exe -ExecutionPolicy Bypass invoke-command bootstrapScript.ps1 -computername $(MyServer) -argumentlist param1, param2" /> </Target> However, I need to be able to supply the credentials by creating a new PSCredentials object with a secure password for my deployment script to run on a remote server: <Target Name="PreDeployment" Condition="true" BeforeTargets="MSDeployPublish"> <Exec Command="powershell.exe -ExecutionPolicy Bypass invoke-command bootstrapScript.ps1 -computername $(MyServer) -credential New-Object System.Management.Automation.PSCredential ('admin', (convertto-securestring $(Password) -asplaintext -force)) -argumentlist param1, param2" /> </Target> When I run the build, a dialog pops up with the username set to System.Management.Automation.PSCredential. I need to be able to create the credentials in-line on the executable target. How do I accomplish this?

    Read the article

  • Http Digest Authentication, Handle different browser char-sets...

    - by user160561
    Hi all, I tried to use the Http Authentication Digest Scheme with my php (apache module) based website. In general it works fine, but when it comes to verification of the username / hash against my user database i run into a problem. Of course i do not want to store the user´s password in my database, so i tend to store the A1 hashvalue (which is md5($username . ':' . $realm . ':' . $password)) in my db. This is just how the browser does it too to create the hashes to send back. The Problem: I am not able to detect if the browser does this in ISO-8859-1 fallback (like firefox, IE) or UTF-8 (Opera) or whatever. I have chosen to do the calculation in UTF-8 and store this md5 hash. Which leads to non-authentication in Firefox and IE browsers. How do you solve this problem? Just do not use this auth-scheme? Or Store a md5 Hash for each charset? Force users to Opera? (Terms of A1 refer to the http://php.net/manual/en/features.http-auth.php example.) (for digest access authentication read the according wikipedia entry)

    Read the article

  • Problem in encoding and decoding the string in Iphone sdk

    - by monish
    HI Guys, Here I am having a problem In encoding/decoding the strings. Actually I had a string which I am encoding it using the base64.which was working fine. And now I need to decode the string that was encoded before and want to print it. I code I written as: I imported the base64.h and base64.m files into my application which contains the methods as: + (NSData *) dataWithBase64EncodedString:(NSString *) string; - (id) initWithBase64EncodedString:(NSString *) string; - (NSString *) base64EncodingWithLineLength:(unsigned int) lineLength; And the code in my view controller where I encode the String is: - (id)init { if (self = [super init]) { // Custom initialization userName = @"Sekhar"; password = @"Bethalam"; } return self; } -(void)reloadView { NSString *authStr = [NSString stringWithFormat:@"%@:%@",userName,password]; NSData *authData = [authStr dataUsingEncoding:NSASCIIStringEncoding]; NSString *authValue = [NSString stringWithFormat:@"%@", [authData base64EncodingWithLineLength:30]]; NSLog(authValue); //const char *str = authValue; //NSString *decStr = [StringEncryption DecryptString:authValue]; //NSLog(decStr); //NSData *decodeData = [NSData decode:authValue]; //NSString *decStr = [NSString stringWithFormat:@"%@",decodeData]; //NSStr //NSLog(decStr); } -(void)viewWillAppear:(BOOL)animated { [self reloadView]; } and now I want to decode the String that I encoded. But I dont know How to do that.can anyone suggest me with code how to get it. Anyone's help will be much appreciated. Thank you, Monish.

    Read the article

  • how to retrive pK using spring security

    - by aditya
    i implement this method of the UserDetailService interface, public UserDetails loadUserByUsername(final String username) throws UsernameNotFoundException, DataAccessException { final EmailCredential userDetails = persistentEmailCredential .getUniqueEmailCredential(username); if (userDetails == null) { throw new UsernameNotFoundException(username + "is not registered"); } final HashSet<GrantedAuthority> authorities = new HashSet<GrantedAuthority>(); authorities.add(new GrantedAuthorityImpl("ROLE_USER")); for (UserRole role:userDetails.getAccount().getRoles()) { authorities.add(new GrantedAuthorityImpl(role.getRole())); } return new User(userDetails.getEmailAddress(), userDetails .getPassword(), true, true, true, true, authorities); } in the security context i do some thing like this <!-- Login Info --> <form-login default-target-url='/dashboard.htm' login-page="/login.htm" authentication-failure-url="/login.htm?authfailed=true" always-use-default-target='false' /> <logout logout-success-url="/login.htm" invalidate-session="true" /> <remember-me user-service-ref="emailAccountService" key="fuellingsport" /> <session-management> <concurrency-control max-sessions="1" /> </session-management> </http> now i want to pop out the Pk of the logged in user, how can i show it in my jsp pages, any idea thanks in advance

    Read the article

  • C# Uploading files to file server

    - by JustFoo
    Hello All, Currently I have an application that receives an uploaded file from my web application. I now need to transfer that file to a file server which happens to be located on the same network (however this might not always be the case). I was attempting to use the webclient class in C# .NET. string filePath = "C:\\test\\564.flv"; try { WebClient client = new WebClient(); NetworkCredential nc = new NetworkCredential(uName, password); Uri addy = new Uri("\\\\192.168.1.28\\Files\\test.flv"); client.Credentials = nc; byte[] arrReturn = client.UploadFile(addy, filePath); Console.WriteLine(arrReturn.ToString()); } catch (Exception ex) { Console.WriteLine(ex.Message); } The machine located at 192.168.1.28 is a file server and has a share c:\Files. As of right now I am receiving an error of Login failed bad user name or password, but I can open explorer and type in that path login successfully. I can also login using remote desktop, so I know the user account works. Any ideas on this error? Is it possible to transfer a file directly like that? With the webclient class or maybe some other class? Thanks

    Read the article

  • Getting rid of the Expires node (xml) in the WS security header

    - by Nick
    From the snippet below, how do i get rid of the (xml node) <wsu:Expires> tag? I want to either get rid of it or pass it in as a empty element. It is a read only property in objClient.RequestSoapContext.Security.Timestamp.Expires. Any help is appreciated. <wsse:Security soap:mustUnderstand="1"> <wsu:Timestamp wsu:Id="Timestamp-26d09d54-10ef-4141-aa2c-11c75ed8172b"> <wsu:Created>2010-03-08T15:32:16Z</wsu:Created> <wsu:Expires>2010-03-08T15:37:16Z</wsu:Expires> </wsu:Timestamp> <wsse:UsernameToken xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd" wsu:Id="SecurityToken-7c9b80ec-98e9-4e41-af2e-ad37070cbdd3"> <wsse:Username>bubba</wsse:Username> <wsse:Password Type="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-username-token-profile-1.0#PasswordDigest">dsfdfsdfsfs+-dasdf=</wsse:Password> <wsse:Nonce>QQ3C4HUfO2CyGx7HrjzMzg==</wsse:Nonce> <wsu:Created>2010-03-08T15:32:16Z</wsu:Created> </wsse:UsernameToken> </wsse:Security>

    Read the article

  • Network Authentication when running exe from WMI

    - by Andy
    Hi, I have a C# exe that needs to be run using WMI and access a network share. However, when I access the share I get an UnauthorizedAccessException. If I run the exe directly the share is accessible. I am using the same user account in both cases. There are two parts to my application, a GUI client that runs on a local PC and a backend process that runs on a remote PC. When the client needs to connect to the backend it first launches the remote process using WMI (code reproduced below). The remote process does a number of things including accessing a network share using Directory.GetDirectories() and reports back to the client. When the remote process is launched automatically by the client using WMI, it cannot access the network share. However, if I connect to the remote machine using Remote Desktop and manually launch the backend process, access to the network share succeeds. The user specifed in the WMI call and the user logged in for the Remote Desktop session are the same, so the permissions should be the same, shouldn't they? I see in the MSDN entry for Directory.Exists() it states "The Exists method does not perform network authentication. If you query an existing network share without being pre-authenticated, the Exists method will return false." I assume this is related? How can I ensure the user is authenticated correctly in a WMI session? ConnectionOptions opts = new ConnectionOptions(); opts.Username = username; opts.Password = password; ManagementPath path = new ManagementPath(string.Format("\\\\{0}\\root\\cimv2:Win32_Process", remoteHost)); ManagementScope scope = new ManagementScope(path, opts); scope.Connect(); ObjectGetOptions getOpts = new ObjectGetOptions(); using (ManagementClass mngClass = new ManagementClass(scope, path, getOpts)) { ManagementBaseObject inParams = mngClass.GetMethodParameters("Create"); inParams["CommandLine"] = commandLine; ManagementBaseObject outParams = mngClass.InvokeMethod("Create", inParams, null); }

    Read the article

  • How to remove a status message added by the seam security module?

    - by Joshua
    I would like to show a different status message, when a suspended user tries to login. If the user is active we return true from the authenticate method, if not we add a custom StatusMessage message mentioning that the "User X has been suspended". The underlying Identity authentication also fails and adds a StatusMessage. I tried removing the seam generated statusMessage with the following methods, but it doesn't seem to work and shows me 2 different status messages (my custom message, seam generated). What would be the issue here? StatusMessages statusMessages; statusMessages.clear() statusMessages.clearGlobalMessages() statusMessages.clearKeyedMessages(id) EDIT1: public boolean authenticate() { log.info("Authenticating {0}", identity.getCredentials().getUsername()); String username = identity.getCredentials().getUsername(); String password = identity.getCredentials().getPassword(); // return true if the authentication was // successful, false otherwise try { Query query = entityManager.createNamedQuery("user.by.login.id"); query.setParameter("loginId", username); // only active users can log in query.setParameter("status", "ACTIVE"); currentUser = (User)query.getSingleResult(); } catch (PersistenceException ignore) { // Provide a status message for the locked account statusMessages.clearGlobalMessages(); statusMessages.addFromResourceBundle( "login.account.locked", new Object[] { username }); return false; } IdentityManager identityManager = IdentityManager.instance(); if (!identityManager.authenticate(username, "password")) { return false; } else { log.info("Authenticated user {0} successfully", username); } }

    Read the article

  • how to read an arraylist from a txt file in java?

    - by lox
    how to read an arraylist from a txt file in java? my arraylist is the form of: public class Account { String username; String password; } i managed to put some "Accounts" in the a txt file, but now i don't know how to read them. this is how my arraylist look in the txt file: username1 password1 | username2 password2 | etc this is a part of the code i came up with, but it doesn't work. it looks logic to me though... :) . public static void RdAc(String args[]) { ArrayList<Account> peoplelist = new ArrayList<Account>(50); int i,i2,i3; String[] theword = null; try { FileReader fr = new FileReader("myfile.txt"); BufferedReader br = new BufferedReader(fr); String line = ""; while ((line = br.readLine()) != null) { String[] theline = line.split(" | "); for (i = 0; i < theline.length; i++) { theword = theline[i].split(" "); } for(i3=0;i3<theline.length;i3++) { Account people = new Account(); for (i2 = 0; i2 < theword.length; i2++) { people.username = theword[i2]; people.password = theword[i2+1]; peoplelist.add(people); } } } } catch (IOException ex) { System.out.println("Could not read from file"); } }

    Read the article

  • Proper use of HttpRequestInterceptor and CredentialsProvider in doing preemptive authentication with

    - by Preston
    I'm writing an application in Android that consumes some REST services I've created. These web services aren't issuing a standard Apache Basic challenge / response. Instead in the server-side code I'm wanting to interrogate the username and password from the HTTP(S) request and compare it against a database user to make sure they can run that service. I'm using HttpClient to do this and I have the credentials stored on the client after the initial login (at least that's how I see this working). So here is where I'm stuck. Preemptive authenticate under HttpClient requires you to setup an interceptor as a static member. This is the example Apache Components uses. HttpRequestInterceptor preemptiveAuth = new HttpRequestInterceptor() { @Override public void process( final HttpRequest request, final HttpContext context) throws HttpException, IOException { AuthState authState = (AuthState) context.getAttribute(ClientContext.TARGET_AUTH_STATE); CredentialsProvider credsProvider = (CredentialsProvider) context.getAttribute( ClientContext.CREDS_PROVIDER); HttpHost targetHost = (HttpHost) context.getAttribute(ExecutionContext.HTTP_TARGET_HOST); if (authState.getAuthScheme() == null) { AuthScope authScope = new AuthScope(targetHost.getHostName(), targetHost.getPort()); Credentials creds = credsProvider.getCredentials(authScope); if (creds != null) { authState.setAuthScheme(new BasicScheme()); authState.setCredentials(creds); } } } }; So the question would be this. What would the proper use of this be? Would I spin this up as part of the application when the application starts? Pulling the username and password out of memory and then using them to create this CredentialsProvider which is then utilized by the HttpRequestInterceptor? Or is there a way to do this more dynamically?

    Read the article

  • Using Javascript to submit forms.

    - by Razor Storm
    I am using a jQuery function to submit a form when a certain button is pressed, however this seems to have no effect on the form. My code is as follows: HTML: <form id="loginForm" action="" method="POST"> <input class="loginInput" type="hidden" name="action" value="login"> <input id="step1a" class="loginInput" type="text" name="username"> <input id="step2a" class="loginInput" type="password" name="password" style="display:none;"> <input id="step1b" class="loginSubmit" onclick="loginProceed();" type="button" name="submit" value="Proceed" title="Proceed" /> <input id="step2b" class="loginSubmit" onclick="submitlogin();" type="button" value="Validate" title="Validate" style="display:none;" /> Javascript: function submitlogin() { $("form").submit(); } However, when I press the button, absolutely nothing occurs. PS. This function may seem meaningless since I can just use a input type="submit" but I originally intended this to have some more functionality, I stripped the function to its bare bones for testing purposes.

    Read the article

  • SQL query in JSP file pulling variable from VXML file

    - by s1066
    Hi I'm trying to get an SQL query to work within a JSP file. The JSP file is pulled by a VXML file here is my JSP file code: <?xml version="1.0"?> <%@ page import="java.util.*" %> <%@ page import="java.sql.*" %> <% boolean success = true; // Always optimistic String info = ""; String schoolname = request.getParameter("schoolname"); String informationtype = request.getParameter("informationtype"); try { Class.forName("org.postgresql.Driver"); String connectString = "jdbc:postgresql://localhost:5435/N0176359"; String user = "****"; String password = "*****"; Connection conn = DriverManager.getConnection(connectString, user, password); Statement st = conn.createStatement(); ResultSet rsvp = st.executeQuery("SELECT * FROM lincolnshire_school_information_new WHERE school_name=\'"+schoolname+"\'"); rsvp.next(); info = rsvp.getString(2); }catch (ClassNotFoundException e) { success = false; // something went wrong } %> As you can see I'm trying to insert the value of the variable declared as "schooname" into the end of the SQL query. However when I come to run the jsp file it doesn't work and I get an error "ResultSet not positioned properly". When I put a standard query in (without trying to make it value of the variable it works fine) Hope that makes sense, and thank you for any help!

    Read the article

  • Does Java 6 open a default port for JMX remote connections?

    - by Bob Cross
    My specific question has to do with JMX as used in JDK 1.6: if I am running a Java process using JRE 1.6 with com.sun.management.jmxremote in the command line, does Java pick a default port for remote JMX connections? Backstory: I am currently trying to develop a procedure to give to a customer that will enable them to connect to one of our processes via JMX from a remote machine. The goal is to facillitate their remote debugging of a situation occurring on a real-time display console. Because of their service level agreement, they are strongly motivated to capture as much data as possible and, if the situation looks too complicated to fix quickly, to restart the display console and allow it to reconnect to the server-side. I am aware the I could run jconsole on JDK 1.6 processes and jvisualvm on post-JDK 1.6.7 processes given physical access to the console. However, because of the operational requirements and people problems involved, we are strongly motivated to grab the data that we need remotely and get them up and running again. EDIT: I am aware of the command line port property com.sun.management.jmxremote.port=portNum The question that I am trying to answer is, if you do not set that property at the command line, does Java pick another port for remote monitoring? If so, how could you determine what it might be?

    Read the article

  • Ignore case in Python strings

    - by Paul Oyster
    What is the easiest way to compare strings in Python, ignoring case? Of course one can do (str1.lower() <= str2.lower()), etc., but this created two additional temporary strings (with the obvious alloc/g-c overheads). I guess I'm looking for an equivalent to C's stricmp(). [Some more context requested, so I'll demonstrate with a trivial example:] Suppose you want to sort a looong list of strings. You simply do theList.sort(). This is O(n * log(n)) string comparisons and no memory management (since all strings and list elements are some sort of smart pointers). You are happy. Now, you want to do the same, but ignore the case (let's simplify and say all strings are ascii, so locale issues can be ignored). You can do theList.sort(key=lambda s: s.lower()), but then you cause two new allocations per comparison, plus burden the garbage-collector with the duplicated (lowered) strings. Each such memory-management noise is orders-of-magnitude slower than simple string comparison. Now, with an in-place stricmp()-like function, you do: theList.sort(cmp=stricmp) and it is as fast and as memory-friendly as theList.sort(). You are happy again. The problem is any Python-based case-insensitive comparison involves implicit string duplications, so I was expecting to find a C-based comparisons (maybe in module string). Could not find anything like that, hence the question here. (Hope this clarifies the question).

    Read the article

  • A way to specify a different host in an SSH tunnel from the host in use

    - by Tom
    I am trying to setup an SSH tunnel to access Beanstalk (to bypass an annoying proxy server). I can get this to work, but with one caveat: I have to map my Beanstalk host URL (username.svn.beanstalkapp.com) in my hosts file to 127.0.0.1 (and use the ip in place of the domain when setting up the tunnel). The reason (I think) is that I am creating the tunnel using the local SSH instance (on Snow Leopard) and if I use localhost or 127.0.0.1 when talking to Beanstalk, it rejects the authorisation credentials. I believe this is because Beanstalk use the hostname specified in a request to determine which account the username / password combination should be checked against. If localhost is used, I think this information is missing (in some manner which Beanstalk requires) from the requests. At the moment I dig the IP for username.svn.beanstalkapp.com, map username.svn.beanstalkapp.com to 127.0.0.1 in my hosts file, then for the tunnel I use the command: ssh -L 8080:ip:443 -p 22 -l tom -N 127.0.0.1 I can tell Subversion that the repo. is located at: https://username.svn.beanstalkapp.com:8080/repo-name This uses my tunnel and the username and password are accepted. So, my question is if there is an option when setting up the SSH tunnel which would mean I wouldn't have to use my hosts file workaround?

    Read the article

  • how to add data to database from rails console

    - by rails_guy
    I have a User model. >> @u = User.new => #<User id: nil, userid: nil, password: nil, created_at: nil, updated_at: nil, user_first_name: nil, user_last_name: nil, user_status: nil, user_type: nil> I am not able to add data to the Users table from the console. I am doing the following: >> @u.userid="test1" => "test1" >> @u.password="test2" => "test2" >> @u.user_first_name="test3" => "test3" >> @u.user_last_name="test4" => "test4" >> @u.user_status="test5" => "test5" >> @u.user_type="test6" => "test6" >> @u.save NoMethodError: undefined method `userid' for nil:NilClass what am i doing wrong? I just simply want to add one row of data to the app.

    Read the article

  • widget within module in Yii

    - by Karolis
    I'm trying to create a widget within the module and then load that widget from 'outside' of the module. More particularly I'm using user module written by someone else. I don't want to have a separate page for displaying a login form, therefore I tried to make a CPortlet/widget (confusion) displaying the login form. Basically, I've moved the code from LoginController into that widget. Then I try to display the widget on some random page by <?php $this->widget('user.components.LoginForm'); ?> However, I get an error CWebApplication does not have a method named "encrypting". in UserIdentity class in this line: else if(Yii::app()->controller->module->encrypting($this->password)!==$user->password) This happens, because I'm basically trying to execute this code within context of the app and not the module. Thus the "Yii::app()-controller-module" trick doesn't really work as expected. What am I doing wrong:-\ Is there a better way to achieve this. I.e. display that login form in some other page, which is normally displayed by accessing login controller within user module (user/login) or is a widget the right way of doing it? Thanks.

    Read the article

< Previous Page | 459 460 461 462 463 464 465 466 467 468 469 470  | Next Page >