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  • I want to input JSONObject from jquery.post() to a a simple JS chart(bar chart)?

    - by ann-stack
    HI I am very new to both json and js charts. In the example of bar chart, they are giving a hard coded array like this, var myData = new Array(['U.S.A.', 69.5], ['Canada', 2.8], ['Japan & SE.Asia', 5.6] ); var myChart = new JSChart('graph', 'bar'); myChart.setDataArray(myData); Instead of that I want to use the response of $.post() method which is in json. Here is the piece of code. var myData=[]; $.post("JSONServlet", function(data) { $.each(data.Userdetails, function(i, data) { myData[i] = []; myData[i]['text'] = data['firstname']; myData[i]['id'] = data['ssn']; alert("first name " +myData[i]['text']+ " salary " +myData[i]['id']); // I am getting correct data here, but how to assign this myData to barchart }); }, "json"); is this the logic to use or how else can i get username and salary from the response and pass it to the barchart. Please help. I am stuck with this. thanks in advance.

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  • Use of putty in command line

    - by kij
    Hi, I'm trying to use putty in command line from an hudson job. The command is the following one: putty -ssh -2 -P 22 USERNAME@SERVER_ADDR -pw PASS -m command.txt Where 'command.txt' is a shell script to execute in the server through SSH. If i launch this command from the Window command prompt, it works, the shell script is executed on the server machine. If i launch a build of the hudson job configured with this batch command, it doesn't work. The build is running... and running... and running.. without doing anything, and i have to stop it manually. So my question is: Is it possible to launch an external programm (i.e. putty) from an hudson job ? ps: i tried SSH plugin but... not a really good plugin (pre/post build, fail status of the commands launched not caught by hudson, etc.) Thanks in advance for your help. Best regards. kij

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  • Mysql Performance Question - Essentially about normalizing efficiency

    - by freqmode
    Hi there. Just a quick question about database performance. I'll outline my site purpose below as background. I'm creating a dictionary site that saves the words users define to a database. What I'm wondering is whether or not to create a words table for each user or to keep one massive words table. This site will be used for entire schools so the single words table would be massive! The database structure is as follows: A user table with: User_ID PRIMARY KEY Username First Last Password Email Country Research Standings SendInfo Donated JoinedOn LastLogin Logins Correct Attempts Admin Active And one word table with: User_ID PRIMARY KEY Word Vocab Spell Defined DefinedAttempted Spelled SpelledAttempted Sentenced SentencedAttempted So what I'm asking is , performance-wise, should I create a new table for each user when they join the site - each user could have hundreds or thousands of words over time? Or is it better to have one massive table with thousands and thousands of records and filter by User_ID. I don't think I'll perform many table joins. My gut feeling is to create a new table for each user, but I thought I'd ask for expert advice! Thanks in advance.

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  • Java applet loading images from external jars

    - by Mathias
    I have a jar on a server, and users should be able to develop extensions for it. Therefore the jars main class should be extended and some resources should be added to a second user created jar which will be loaded from another server or locally. Now I have problems accessing the resources (images) from the user loaded jars. Heres is the structure: My Server: game.jar containing game.class images.class ... image1.png (...) Local: user.jar containing: user.class extends game userimage.png The extension is loaded via Greasemonkey, it modifies the "archive" attribute to "/home/username/user.jar, game.jar" and the "code" attribute to "user.class". The user should be able to overwrite already defined images. If the image does not exist in game.jar, it is loaded correctly from user.jar. But the images loaded early in the game are always loaded from the game.jar, others seem to be overwritten correctly by the user. Is there a way to make sure they are always loaded in the correct order? This might be because of some caching mechanism. Because Greasemonkey removes the game from the page, changes the archive and code and reinsert it, the game is loaded without a mod for a brief second. In that time, images are loaded as expected from game jar, but those are the ones not being overwritable by the user. But how to avoid it? Another thing: If I overwrite the "run" method in user.class, the game is unable to load any image at all. Not from the user.jar and not from the game.jar. Java doesn't find the image, as the URL object "getClass().getResource(imagename)" returns with null. I tried to overwrite the image.class, but that doesn't fix the problem, unless I overwrite every class from game.class involved into calling the image.class

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  • MySQL Database is Indexed at Apache Solr, How to access it via URL

    - by Wasim
    data-config.xml <dataConfig> <dataSource encoding="UTF-8" type="JdbcDataSource" driver="com.mysql.jdbc.Driver" url="jdbc:mysql://localhost:3306/somevisits" user="root" password=""/> <document name="somevisits"> <entity name="login" query="select * from login"> <field column="sv_id" name="sv_id" /> <field column="sv_username" name="sv_username" /> </entity> </document> </dataConfig> schema.xml <?xml version="1.0" encoding="UTF-8" ?> <schema name="example" version="1.5"> <fields> <field name="sv_id" type="string" indexed="true" stored="true" required="true" multiValued="false" /> <field name="username" type="string" indexed="true" stored="true" required="true"/> <field name="_version_" type="long" indexed="true" stored="true" multiValued="false"/> <field name="text" type="string" indexed="true" stored="false" multiValued="true"/> </fields> <uniqueKey>sv_id</uniqueKey> <types> <fieldType name="string" class="solr.StrField" sortMissingLast="true" /> <fieldType name="long" class="solr.TrieLongField" precisionStep="0" positionIncrementGap="0"/> </types> </schema> Solr successfully imported mysql database using full http://[localSolr]:8983/solr/#/collection1/dataimport?command=full-import My question is, how to access that mysql imported database now?

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  • Users Hierarchy Logic

    - by user342944
    Hi guys, I am writing a user security module using SQLServer 2008 so threfore need to design a database accordingly. Formally I had Userinfo table with UserID, Username and ParentID to build a recursion and populated tree to represent hierarchy but now I have following criteria which I need to develop. I have now USERS, ADMINISTRATORS and GROUPS. Each node in the user hierarchy is either a user, administrator or group. User Someone who has login access to my application Administrator A user who may also manage all their child user accounts (and their children etc) This may include creating new users and assigning permissions to those users. There is no limit to the number of administrators in user structure. The higher up in the hierarchy that I go administrators have more child accounts to manage which include other child administrators. Group A user account can be designated as a group. This will be an account which is used to group one or more users together so that they can be manage as a unit. But no one can login to my application using a group account. This is how I want to create structure Super Administrator administrator ------------------------------------------------------------- | | | Manager A Manager B Manager C (adminstrator) (administrator) (administrator) | ----------------------------------------- | | | Employee A Employee B Sales Employees (User) (User) (Group) | ------------------------ | | | Emp C Emp D Emp E (User) (User) (User) Now how to build the table structure to achieve this. Do I need to create Users table alongwith Group table or what? Please guide I would really appreciate.

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  • PHP -- automatic SQL injection protection?

    - by ashgromnies
    I took over maintenance of a PHP app recently and I'm not super familiar with PHP but some of the things I've been seeing on the site are making me nervous that it could be vulnerable to a SQL injection attack. For example, see how this code for logging into the administrative section works: $password = md5(HASH_SALT . $_POST['loginPass']); $query = "SELECT * FROM `administrators` WHERE `active`='1' AND `email`='{$_POST['loginEmail']}' AND `password`='{$password}'"; $userInfo = db_fetch_array(db_query($query)); if($userInfo['id']) { $_SESSION['adminLoggedIn'] = true; // user is logged in, other junk happens here, not important The creators of the site made a special db_query method and db_fetch_array method, shown here: function db_query($qstring,$print=0) { return @mysql(DB_NAME,$qstring); } function db_fetch_array($qhandle) { return @mysql_fetch_array($qhandle); } Now, this makes me think I should be able to do some sort of SQL injection attack with an email address like: ' OR 'x'='x' LIMIT 1; and some random password. When I use that on the command line, I get an administrative user back, but when I try it in the application, I get an invalid username/password error, like I should. Could there be some sort of global PHP configuration they have enabled to block these attacks? Where would that be configured? Here is the PHP --version information: # php --version PHP 5.2.12 (cli) (built: Feb 28 2010 15:59:21) Copyright (c) 1997-2009 The PHP Group Zend Engine v2.2.0, Copyright (c) 1998-2009 Zend Technologies with the ionCube PHP Loader v3.3.14, Copyright (c) 2002-2010, by ionCube Ltd., and with Zend Optimizer v3.3.9, Copyright (c) 1998-2009, by Zend Technologies

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  • SQLAlchemy sessions - DetachedInstanceError?

    - by benjaminhkaiser
    I have a function that attempts to take a list of usernames, look each one up in a user table, and then add them to a membership table. If even one username is invalid, I want the entire list to be rolled back, including any users that have already been processed. I thought that using sessions was the best way to do this but I'm running into a DetachedInstanceError: DetachedInstanceError: Instance <Organization at 0x7fc35cb5df90> is not bound to a Session; attribute refresh operation cannot proceed Full stack trace is here. The error seems to trigger when I attempt to access the user (model) object that is returned by the query. From my reading I understand that it has something to do with there being multiple sessions, but none of the suggestions I saw on other threads worked for me. Code is below: def add_members_in_bulk(organization_eid, users): """Add users to an organization in bulk - helper function for add_member()""" """Returns "success" on success and id of first failed student on failure""" session = query_session.get_session() session.begin_nested() users = users.split('\n') for u in users: try: user = user_lookup.by_student_id(u) except ObjectNotFoundError: session.rollback() return u if user: membership.add_user_to_organization( user.entity_id, organization_eid, '', [] ) session.flush() session.commit() return 'success' here's the membership.add_user_to_organization: def add_user_to_organization(user_eid, organization_eid, title, tag_ids): """Add a User to an Organization with the given title""" user = user_lookup.by_eid(user_eid) organization = organization_lookup.by_eid(organization_eid) new_membership = OrganizationMembership( organization_eid=organization.entity_id, user_eid=user.entity_id, title=title) new_membership.tags = [get_tag_by_id(tag_id) for tag_id in tag_ids] crud.add(new_membership) and here is the lookup by ID query: def by_student_id(student_id, include_disabled=False): """Get User by RIN""" try: return get_query_set(include_disabled).filter(User.student_id == student_id).one() except NoResultFound: raise ObjectNotFoundError("User with RIN %s does not exist." % student_id)

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  • Problem in Application_Error in Global.asax

    - by mmtemporary
    my problem is User.Identity.Name or Request.Url.AbsoluteUri in exception handling is empty when exception email to me. this is Application_Code: void Application_Error(object sender, EventArgs e) { Server.Transfer("~/errors/default.aspx"); } and this is default.aspx code: protected void Page_Load(object sender, EventArgs e) { if (Server.GetLastError() == null) return; Exception ex = Server.GetLastError().GetBaseException(); if (ex == null) return; string message = string.Format("User: ", User.Identity.Name); message += Environment.NewLine; message += string.Format("AbsoluteUri: ", Request.Url.AbsoluteUri); message += Environment.NewLine; message += string.Format("Form: ", Request.Form.ToString()); message += Environment.NewLine; message += string.Format("QueryString: ", Request.QueryString.ToString()); message += Environment.NewLine; HttpBrowserCapabilities browser = Request.Browser; string s = "Browser Capabilities:\n" + "Type = " + browser.Type + "\n" + "Name = " + browser.Browser + "\n" + "Version = " + browser.Version + "\n" + "Platform = " + browser.Platform + "\n" + "Is Crawler = " + browser.Crawler + "\n" + "Supports Cookies = " + browser.Cookies + "\n" + "Supports JavaScript = " + browser.EcmaScriptVersion.ToString() + "\n" + "\n"; message += s; message += Environment.NewLine; message += ex.ToString(); Exception lastException = (Exception)Application["LastException"]; if (lastException == null || lastException.Message != ex.Message) { Application.Lock(); Application["LastException"] = ex; Application.UnLock(); SiteHelper.SendEmail(SiteHelper.AdministratorEMail, "Error!!!", message, false); } Server.ClearError(); } but i receive email like this (this is header without full exception content): User: AbsoluteUri: Form: QueryString: Browser Capabilities: Type = IE8 Name = IE Version = 8.0 Platform = WinXP Is Crawler = False Supports Cookies = True Supports JavaScript = 1.2 why username and request url is emty? this problem is exist when i replace transfer with redirect or i don't use both. tanx

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  • Sorting and Pagination does not work after I build a custom keyword search that is build using relat

    - by Roland
    I recently started to build a custom keyword search using Yii 1.1.x The search works 100%. But as soon as I sort the columns and use pagination in the admin view the search gets lost and all results are shown. So with otherwords it's not filtering so that only the search results show. Somehow it resets it. In my controller my code looks as follows $builder=Messages::model()->getCommandBuilder(); //Table1 Columns $columns1=array('0'=>'id','1'=>'to','2'=>'from','3'=>'message','4'=>'error_code','5'=>'date_send'); //Table 2 Columns $columns2=array('0'=>'username'); //building the Keywords $keywords = explode(' ',$_REQUEST['search']); $count=0; foreach($keywords as $key){ $kw[$count]=$key; ++$count; } $keywords=$kw; $condition1=$builder->createSearchCondition(Messages::model()->tableName(),$columns1,$keywords,$prefix='t.'); $condition2=$builder->createSearchCondition(Users::model()->tableName(),$columns2,$keywords); $condition = substr($condition1,0,-1) . " OR ".substr($condition2,1); $condition = str_replace('AND','OR',$condition); $dataProvider=new CActiveDataProvider('Messages', array( 'pagination'=>array( 'pageSize'=>self::PAGE_SIZE, ), 'criteria'=>array( 'with'=>'users', 'together'=>true, 'joinType'=>'LEFT JOIN', 'condition'=>$condition, ), 'sort'=>$sort, )); $this->render('admin',array( 'dataProvider'=>$dataProvider,'keywords'=>implode(' ',$keywords),'sort'=>$sort )); and my view looks like this $this->widget('zii.widgets.grid.CGridView', array( 'dataProvider'=>$dataProvider, 'columns'=>array( 'id', array( 'name'=>'user_id', 'value'=>'CHtml::encode(Users::model()->getReseller($data->user_id))', 'visible'=>Yii::app()->user->checkAccess('poweradministrator') ), 'to', 'from', 'message', /* 'date_send', */ array( 'name'=>'error_code', 'value'=>'CHtml::encode($data->status($data->error_code))', ), array( 'class'=>'CButtonColumn', 'template'=>'{view} {delete}', ), ), )); I really do not know what do do anymore since I'm terribly lost, any help will be hihsly appreciated

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  • mysql_connect() acces denied for system@localhost(using password NO)

    - by user309381
    class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { $this-connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db(DB_NAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query(/*$sql*/) { $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result; while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database-open_connection(); $database-query(); $database-close_connection(); ?

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  • SSRS 2008 - Sending report as email

    - by Mozy
    Hi all, What I want to do is render a SSRS 2008 report as HTML (actually MHTML) and send that as the body in an email. Has anyone done this? I've almost got it (I think) but I seem to have a encoding problem. The email shows up as what looks like some sort of base64 encoding. Like this: MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C35DB7.4B204430" X-MSSQLRS-ProducerVersion: V10.0.2531.0 This is a multi-part message in MIME format. ------=_NextPart_01C35DB7.4B204430 Content-ID: Content-Disposition: inline; filename="FollowUpNotification" Content-Type: text/html; name="FollowUpNotification"; charset="utf-8" Content-Transfer-Encoding: base64 PCFET0NUWVBFIEhUTUwgUFVCTElDICItLy9XM0MvL0RURCBIVE1MIDQuMDEgVHJhbnNp... Any ideas on what I'm missing? Here is a code snippet: myMail.BodyEncoding = System.Text.Encoding.UTF8; myMail.IsBodyHtml = true; WebClient client = new WebClient(); System.Net.NetworkCredential repAuthenticationInfo = new System.Net.NetworkCredential(@"username", @"password"); client.Credentials = repAuthenticationInfo; client.Encoding = System.Text.Encoding.UTF8; string messageBody = client.DownloadString( "http://<<reportserver>>&rs%3aFormat=MHTML&Parameter=" + Parameter); myMail.Body = messageBody;

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  • Error occurs while validating form input using jQuery in Firebug

    - by Param-Ganak
    I have written a custom validation code in jQuery, which is working fine. I have a login form which has two fields, i.e. userid and password. I have written a custom code for client side validation for these fields. This code is working fine and gives me proper error messages as per the situation. But the problem with this code is that when I enter the invalid data in any or both field and press submit button of form then it displays the proper error message but at the same time when I checked it in Firebug it displays following error message when submit button of the form is clicked validate is not defined function onclick(event) { javascript: return validate(); } (click clientX=473, clientY=273) Here is the JQUERY validation code $(document).ready(function (){ $("#id_login_form").validate({ rules: { userid: { required: true, minlength: 6, maxlength: 20, // basic: true }, password: { required: true, minlength: 6, maxlength: 15, // basic: true } }, messages: { userid: { required: " Please enter the username.", minlength: "User Name should be minimum 6 characters long.", maxlength: "User Name should be maximum 15 characters long.", // basic: "working here" }, password: { required: " Please enter the password.", minlength: "Password should be minimum 6 characters long.", maxlength: "Password should be maximum 15 characters long.", // basic: "working here too.", } }, errorClass: "errortext", errorLabelContainer: "#messagebox" } }); }); /* $.validator.addMethod('username_alphanum', function (value) { return /^(?![0-9]+$)[a-zA-Z 0-9_.]+$/.test(value); }, 'User name should be alphabetic or Alphanumeric and may contain . and _.'); $.validator.addMethod('alphanum', function (value) { return /^(?![a-zA-Z]+$)(?![0-9]+$)[a-zA-Z 0-9]+$/.test(value); }, 'Password should be Alphanumeric.'); $.validator.addMethod('basic', function (value) { return /^[a-zA-Z 0-9_.]+$/.test(value); }, 'working working working'); */ So please tell me where is I am wrong in my jQuery code. Thank You!

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  • sql server 2005 command line install error ADD_LOCAL property already installed

    - by Belliez
    I have a silent installation of SQL Server 2005 that works great when installing SQL Server on a machine that does not have it already installed. I use the following parameters when I perform the installation: #define SQL_SILENT "/passive /qb" #define SQL_USERNAME "username=MyUserName" #define SQL_COMPANYNAME "companyname=MyCompanyName" #define SQL_ADDLOCAL "ADDLOCAL=SQL_Engine" #define SQL_UPGRADE "" #define SQL_DISABLENETWORKPROTOCOLS "disablenetworkprotocols=0" #define SQL_INSTANCENAME "instancename=MYSQLINSTANCE" #define SQL_SQLAUTOSTART "SQLAUTOSTART=1" #define SQL_SECURITYMODE "SECURITYMODE=SQL" #define SQL_SAPWD "SAPWD=StrongPassword" #define SQL_SQLACCOUNT "SQLACCOUNT=""""" #define SQL_SQLPASSWORD "SQLPASSWORD=""""" It installs the instance of SQL Server Express without a problem. However, when I attempt to install SQL Server on a machine that already has another instance with components I get the following error: *"A component that you have specified in the ADD_LOCAL property is already installed. To upgrade the existing component, refer to the template.ini and set the UPGRADE property to the name of the component."* I have also tried using the UPGRADE method as per the error message #define SQL_UPGRADE "UPGRADE=SQL_Engine INSTANCENAME=MYSQLINSTANCE" but get the following error: "SQL Server Setup cannot perform the upgrade because the component is not installed on the computer. To proceed, verify the component to be upgraded in currently installed, and that the component to be upgraded is specified in the ADDLOCAL property." Does anyone have any suggestions? Thank you

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  • Servlet Security question about j_security_check, j_username and j_password

    - by Nitesh Panchal
    Hello, I used jdbcRealm in my web application and it's working fine. I defined all constraints also in my web.xml. Like all pages of url pattern /Admin/* should be accessed by only admin. I have a login form with uses standard j_security_check, j_username and j_password. Now, when i type Admin/home.jsf it rightly redirects me login.jsf and there when i type the password i am redirected to home.jsf. This works alright but problem comes i directly go to login.jsf and then type password and username. This time it again redirects me to login.jsf. Is there any way through which i can specify which page to go when successful login is there? I need to specify different different pages for different roles. For Admin, it is /Admin/home.jsf for general users it is /General/home.jsf because login form is shared between different type of users. Where do i specify all these things? Secondly, i want to have a remember me checkbox at the end of login form. How do i do this? By default, it is submitted to j_security_check servlet and i have no control over its execution. Please help. This doesn't seem so hard but looks like i am missing something.

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  • Uploading an xml direct to ftp

    - by Joshua Maerten
    i put this direct below a button: XmlDocument doc = new XmlDocument(); XmlElement root = doc.CreateElement("Login"); XmlElement id = doc.CreateElement("id"); id.SetAttribute("userName", usernameTxb.Text); id.SetAttribute("passWord", passwordTxb.Text); XmlElement name = doc.CreateElement("Name"); name.InnerText = nameTxb.Text; XmlElement age = doc.CreateElement("Age"); age.InnerText = ageTxb.Text; XmlElement Country = doc.CreateElement("Country"); Country.InnerText = countryTxb.Text; id.AppendChild(name); id.AppendChild(age); id.AppendChild(Country); root.AppendChild(id); doc.AppendChild(root); // Get the object used to communicate with the server. FtpWebRequest request = (FtpWebRequest)WebRequest.Create("ftp://users.skynet.be"); request.Method = WebRequestMethods.Ftp.UploadFile; request.UsePassive = false; // This example assumes the FTP site uses anonymous logon. request.Credentials = new NetworkCredential("fa490002", "password"); // Copy the contents of the file to the request stream. StreamReader sourceStream = new StreamReader(); byte[] fileContents = Encoding.UTF8.GetBytes(sourceStream.ReadToEnd()); sourceStream.Close(); request.ContentLength = fileContents.Length; Stream requestStream = request.GetRequestStream(); requestStream.Write(fileContents, 0, fileContents.Length); requestStream.Close(); FtpWebResponse response = (FtpWebResponse)request.GetResponse(); response.Close(); MessageBox.Show("Created SuccesFully!"); this.Close(); but i always get an error of the streamreader path, what do i need to place there ? the meening is, creating an account and when i press the button, an xml file is saved to, ftp://users.skynet.be/testxml/ the filename is from usernameTxb.Text + ".xml".

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  • PEAR:DB connection parameters

    - by Markus Ossi
    I just finished my first PHP site and now I have a security-related question. I used PEAR:DB for the database connection and made a separate parameter file for it. How should I hide this parameter file? I found a guide (http://www.kitebird.com/articles/peardb.html) that says: Another way to specify connection parameters is to put them in a separate file that you reference from your main script. ... It also enables you to move the parameter file outside of the web server's document tree, which prevents its contents from being displayed literally if the server becomes misconfigured and starts serving PHP scripts as plain text. I have now put my file in a directory like this /include/db_parameters.inc However, if I go to this URL, the web server shows me the contents of the file including my database username and password. From what I've understood, I should protect this file so, that even though PHP would be served as text, nobody could read this. What does outside of web server's document tree mean here? Put the PHP file out of public_html directory altogether deeper into the server file system? Some CHMOD?

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  • Django: Serving a Download in a Generic View

    - by TheLizardKing
    So I want to serve a couple of mp3s from a folder in /home/username/music. I didn't think this would be such a big deal but I am a bit confused on how to do it using generic views and my own url. urls.py url(r'^song/(?P<song_id>\d+)/download/$', song_download, name='song_download'), The example I am following is found in the generic view section of the Django documentations: http://docs.djangoproject.com/en/dev/topics/generic-views/ (It's all the way at the bottom) I am not 100% sure on how to tailor this to my needs. Here is my views.py def song_download(request, song_id): song = Song.objects.get(id=song_id) response = object_detail( request, object_id = song_id, mimetype = "audio/mpeg", ) response['Content-Disposition'= "attachment; filename=%s - %s.mp3" % (song.artist, song.title) return response I am actually at a loss of how to convey that I want it to spit out my mp3 instead of what it does now which is to output a .mp3 with all of the current pages html contained. Should my template be my mp3? Do I need to setup apache to serve the files or is Django able to retrieve the mp3 from the filesystem(proper permissions of course) and serve that? If it do need to configure Apache how do I tell Django that? Thanks in advanced. These files are all on the HD so I don't need to "generate" anything on the spot and I'd like to prevent revealing the location of these files if at all possible. A simple /song/1234/download would be fantastic.

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  • custom rss feed suddenly does not work anymore

    - by krike
    I just don't understand what's happening, I haven't change anything to the site for a few months but now suddenly the rss feed doesn't work anymore. I create a php file with the following code: header('Content-type: text/xml'); include("config/config.inc.php"); $result = mysqli_query($link, "SELECT * FROM tutorials ORDER BY tutorial_id DESC LIMIT 50"); ?> <rss version="2.0"> <channel> <title>CMS tutorial site</title> <description>Bringing you the best CMS tutorials from the web</description> <link>http://cmstutorials.org</link> <?php while($row = mysqli_fetch_object($result)) { $user = mysqli_fetch_object(mysqli_query($link, "SELECT * FROM user_extra WHERE userid=".$row->user_id."")); ?> <item> <title><?php echo $row->title; ?></title> <author><?php echo $user->username; ?></author> <description><?php echo $row->description; ?></description> <pubDate><?php echo $row->date; ?></pubDate> <link>http://cmstutorials.org/view_tutorial.php?tutorial_id=<?php echo $row->tutorial_id; ?></link> </item> <?php } ?> </channel> </rss> I checked the query by executing it in phpmyadmin and it works, doesn't give any error. When I delete the header content type and the rss tag it will print out each line from the query but the feed won't display anything this is the link of the feed http://cmstutorials.org/rss (or http://cmstutorials.org/rss.php)

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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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  • add a number in while loop

    - by Luke
    if ($num_rows > 0) { while($row=mysql_fetch_assoc($res)) { $fromuser=$row['username']; $comment=$row['comment']; $commentdate=$row['date']; $date=strtotime($commentdate); $final_date=date("g:i a", $date); $final_date2=date("F j Y", $date); ?> <table align="center" width="100%"style='border-top: 1px dotted;'bgcolor="#eeeeee" > <tr><td><?echo "<a href=\"userprofile.php?user=$fromuser\"><b>$fromuser</b></a> commented:\n";?></td></tr> <tr><td><?echo "at $final_date on $final_date2\n";?></td></tr> <tr bgcolor="#ffffff"><td><?echo "$comment\n";?></td></tr> </table><br> <? } } else { echo"There are currently no comments on this user"; } ?> I am looking for a way to add a number to each comment. So 1, 2, 3, 4, etc in DESC order. I can't think how I can do this?

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  • pulling a value from NSMutableDictionary

    - by Jared Gross
    I have a dictionary array with a key:@"titleLabel". I am trying to load a pickerView with ONE instance of each @"titleLabel" key so that if there are multiple objects with the same @"titleLabel" only one title will be displayed. I've done some research on this forum and looked at apples docs but haven't been able to put the puzzle together. Below is my code but I am having trouble pulling the values. Right now when I run this code it throws an error Incompatible pointer types sending 'PFObject *' to parameter of type 'NSString' which i understand but am just not sure how to remedy. Cheers! else { // found messages! self.objectsArray = objects; NSMutableDictionary *dict = [[NSMutableDictionary alloc] init]; for(id obj in self.objectsArray){ PFObject *key = [self.objectsArray valueForKey:@"titleLabel"]; if(![dict objectForKey:@"titleLabel"]){ [dict setValue:obj forKey:key]; } } for (id key in dict) { NSLog(@"Objects array is %d", [self.objectsArray count]); NSLog(@"key: %@, value: %@ \n", key, [dict objectForKey:key]); } [self.pickerView reloadComponent:0]; } }];` Here is where I define the PFObject and keys: PFObject *image = [PFObject objectWithClassName:@"Images"]; [image setObject:file forKey:@"file"]; [image setObject:fileType forKey:@"fileType"]; [image setObject:title forKey:@"titleLabel"]; [image setObject:self.recipients forKey:@"recipientIds"]; [image setObject:[[PFUser currentUser] objectId] forKey:@"senderId"]; [image setObject:[[PFUser currentUser] username] forKey:@"senderName"]; [image saveInBackground];

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  • SugarmCRM REST API always returns "null"

    - by TuomasR
    I'm trying to test out the SugarCRM REST API, running latest version of CE (6.0.1). It seems that whenever I do a query, the API returns "null", nothing else. If I omit the parameters, then the API returns the API description (which the documentation says it should). I'm trying to perform a login, passing as parameter the method (login), input_type and response_type (json) and rest_data (JSON encoded parameters). The following code does the query: $api_target = "http://example.com/sugarcrm/service/v2/rest.php"; $parameters = json_encode(array( "user_auth" => array( "user_name" => "admin", "password" => md5("adminpassword"), ), "application_name" => "Test", "name_value_list" => array(), )); $postData = http_build_query(array( "method" => "login", "input_type" => "json", "response_type" => "json", "rest_data" => $parameters )); echo $parameters . "\n"; echo $postData . "\n"; echo file_get_contents($api_target, false, stream_context_create(array( "http" => array( "method" => "POST", "header" => "Content-Type: application/x-www-form-urlencoded\r\n", "content" => $postData ) ))) . "\n"; I've tried different variations of parameters and using username instead of user_name, and all provide the same result, just a response "null" and that's it.

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  • Codeigniter Session Data not available in other pages after login

    - by jswat
    So, I have set up a login page that verifies the user's credentials, and then sets codeigniter session data 'email' and 'is_logged_in' and a few other items. The first page after the login, the data is accessible. After that page, I can no longer access the session data. In fact, if I try reloading that first page, the session data is gone. I have tried storing it in the database, storing it unencrypted (bad idea I know, but it was for troubleshooting), and storing it encrypted. I have autoloaded the session library in config.php. Here's an example of the code I'm using to set the session data: $data = array( 'email' => $this->input->post('username'), 'is_logged_in' => true ); $this->session->set_userdata($data); And to retrieve it, I'm using : $this->session->userdata('email'); Or $this->session->userdata('is_logged_in'); I've done lots of work with PHP and cookies, and sessions before, but this is my first project with Codeigniter and I'm perplexed. Could it have something to do with directory issues? I have the login page and process controlled by a 'login' controller, and then it redirects to a 'site' controller. Thanks for your help, and please let me know if I need to clarify anything.

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  • Problem with impersonating a specific user in WCF service

    - by aJ
    I am having a WCF service hosted in IIS on WindowsServer 2008. This service needs to write to a shared folder present on another machine(Windows XP). The shared folder has write permissions for a particular user say "X" which is present on both the machines .i.e on the server where the service is running as well as the machine where the shared folder is present. The service runs under the NETWORK SERVICE account. For the service to access the shared folder I have added code to impersonate the user "X"in the service so that it gets the permission to write to the shared folder. Since I want to impersonate user "X" only when I run a particular section of code I have used the sample code. Even after the impersonation the service fails to write to the shared folder sometimes. It works sporadically. Whereas if I add tag in the Web.config file it works perfectly fine. <identity impersonate="true" userName="accountname" password="password" /> But the above is not desirable since it impersonates a specific user for all the requests. What I need is to impersonate a specific user only when I run a particular section of code. Also, the impersonation code works absolutely fine when the shared folder is present on another WindowsServer 2008. Could anyone give me ideas on what's going wrong here.

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