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  • filling a form with parameter already selected using ajax

    - by kawtousse
    hi every one, My goal untill now is to fill a form with values from a table (html table). it is a kind of refreshing the form. so the user who wants to modify the html table through the form must prefill the form with values wich he already selected. I used the DOM to acces to each row and cell in the table and i used ajax to pass parameter to other jsp. but Iam confused what shall be the next step to fill the form.

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  • Persisting Session Between Different Browser Instances

    - by imran_ku07
        Introduction:          By default inproc session's identifier cookie is saved in browser memory. This cookie is known as non persistent cookie identifier. This simply means that if the user closes his browser then the cookie is immediately removed. On the other hand cookies which stored on the user’s hard drive and can be reused for later visits are called persistent cookies. Persistent cookies are less used than nonpersistent cookies because of security. Simply because nonpersistent cookies makes session hijacking attacks more difficult and more limited. If you are using shared computer then there are lot of chances that your persistent session will be used by other shared members. However this is not always the case, lot of users desired that their session will remain persisted even they open two instances of same browser or when they close and open a new browser. So in this article i will provide a very simple way to persist your session even the browser is closed.   Description:          Let's create a simple ASP.NET Web Application. In this article i will use Web Form but it also works in MVC. Open Default.aspx.cs and add the following code in Page_Load.    protected void Page_Load(object sender, EventArgs e)        {            if (Session["Message"] != null)                Response.Write(Session["Message"].ToString());            Session["Message"] = "Hello, Imran";        }          This page simply shows a message if a session exist previously and set the session.          Now just run the application, you will just see an empty page on first try. After refreshing the page you will see the Message "Hello, Imran". Now just close the browser and reopen it or just open another browser instance, you will get the exactly same behavior when you run your application first time . Why the session is not persisted between browser instances. The simple reason is non persistent session cookie identifier. The session cookie identifier is not shared between browser instances. Now let's make it persistent.          To make your application share session between different browser instances just add the following code in global.asax.    protected void Application_PostMapRequestHandler(object sender, EventArgs e)           {               if (Request.Cookies["ASP.NET_SessionIdTemp"] != null)               {                   if (Request.Cookies["ASP.NET_SessionId"] == null)                       Request.Cookies.Add(new HttpCookie("ASP.NET_SessionId", Request.Cookies["ASP.NET_SessionIdTemp"].Value));                   else                       Request.Cookies["ASP.NET_SessionId"].Value = Request.Cookies["ASP.NET_SessionIdTemp"].Value;               }           }          protected void Application_PostRequestHandlerExecute(object sender, EventArgs e)        {             HttpCookie cookie = new HttpCookie("ASP.NET_SessionIdTemp", Session.SessionID);               cookie.Expires = DateTime.Now.AddMinutes(Session.Timeout);               Response.Cookies.Add(cookie);         }          This code simply state that during Application_PostRequestHandlerExecute(which is executed after HttpHandler) just add a persistent cookie ASP.NET_SessionIdTemp which contains the value of current user SessionID and sets the timeout to current user session timeout.          In Application_PostMapRequestHandler(which is executed just before th session is restored) we just check whether the Request cookie contains ASP.NET_SessionIdTemp. If yes then just add or update ASP.NET_SessionId cookie with ASP.NET_SessionIdTemp. So when a new browser instance is open, then a check will made that if ASP.NET_SessionIdTemp exist then simply add or update ASP.NET_SessionId cookie with ASP.NET_SessionIdTemp.          So run your application again, you will get the last closed browser session(if it is not expired).   Summary:          Persistence session is great way to increase the user usability. But always beware the security before doing this. However there are some cases in which you might need persistence session. In this article i just go through how to do this simply. So hopefully you will again enjoy this simple article too.

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  • Ajax call from a form rendered as Ajax response (jQuery + Grails: chaining ajax requests)

    - by bsreekanth
    Hello, I was expecting the below scenario common, but couldn't find much help online. I have a form loaded through Ajax (say, create entity form). It is loaded through a button click (load) event $("#bt-create").click(function(){ $ ('#pid').load('/controller/vehicleModel/create3'); return false; }); the response (a form) is written in to the pid element. The name and id of the form is ajax-form, and the submit event is attached to an ajax post request $(function() { $("#ajax-form").submit(function(){ // do something... var url = "/app/controller/save" $.post(url, $(this).serialize(), function(data) { alert( data ) ; /// alert data from server }); I could make the above ajax operations individually. That is the ajax post operation succeeds if it calls from a static html file. But if I chain the requests (after completing the first), so that it calls from the output form generated by the first request, nothing happens. I could see the post method is called through firebug. Is there a better way to handle above flow? One more interesting thing I noticed. As you could see, I use grails as my platform. If I keep the javascripts in the main.gsp (master layout), the submit event would not register as the breakpoint is not hit in firebug. But, if I define the javascript in the template file (which renders the form above), the breakpoint is hit, but as I explained, the action is not called at the controller. I changes the javascript to the head section but same result. any help greatly appreciated. thanks, Babu.

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  • Zend Form Radio Elements, using images instead of default radio elements.

    - by Davy
    Hello, Ultimately here is my goal. Using Zend_Form I want to turn this idea http://www.sohtanaka.com/web-design/fancy-thumbnail-hover-effect-w-jquery/ into a list of radio buttons. Kind of using this concept. http://theodin.co.uk/tools/tutorials/jqueryTutorial/fancyRadio/ I know there has to be a way to do this but I can't seem to figure anything out! Any ideas? Thanks! -d

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  • error handler isn't called when file is uploaded via ajax using jQuery form plugin

    - by scompt.com
    Here's my test case. If the form is posted, a 500 error response is sent. If not, the form is sent. If the file input tag is commented out, the error handler is called. If the file input tag is present, the error handler isn't called. I think this might have something to do with the fact that jQuery needs to use an iframe to handle the upload and iframes don't seem to respond to the error handler. Edit: If I add iframe: true to the options passed to ajaxSubmit to force the use of an iframe, the non-file-upload case stops working also, so it definitely has to do with the iframe. Edit2: I'm using the jQuery Form Plugin. <?php if($_SERVER['REQUEST_METHOD'] == 'POST') { header('HTTP/1.1 500 Internal Server Error'); die; } else {?> <html><head> <script type='text/javascript' src='http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js?ver=2.9.2'></script> <script type='text/javascript' src='http://github.com/malsup/form/raw/master/jquery.form.js?v2.43'></script> <script type="text/javascript"> jQuery(document).ready(function() { jQuery('a').click(function() {jQuery('form').ajaxSubmit({error: function(){alert('error handler called');}})}); }); </script> </head><body> <form method="POST"> <input type="text" name="mytext" /> <input type="file" name="myfile" /><!-- comment this element out --> <input type="hidden" name="blah" value="blah" /> <a>submit</a> </form> </body></html> <?php } Is there any way to get the error handler to be called in both situations?

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  • cgi.FieldStorage always empty - never returns POSTed form Data

    - by Dan Carlson
    This problem is probably embarrassingly simple. I'm trying to give python a spin. I thought a good way to start doing that would be to create a simple cgi script to process some form data and do some magic. My python script is executed properly by apache using mod_python, and will print out whatever I want it to print out. My only problem is that cgi.FieldStorage() is always empty. I've tried using both POST and GET. Each trial I fill out both form fields. <form action="pythonScript.py" method="POST" name="ARGH"> <input name="TaskName" type="text" /> <input name="TaskNumber" type="text" /> <input type="submit" /> </form> If I change the form to point to a perl script it reports the form data properly. The python page always gives me the same result: number of keys: 0 #!/usr/bin/python import cgi def index(req): pageContent = """<html><head><title>A page from""" pageContent += """Python</title></head><body>""" form = cgi.FieldStorage() keys = form.keys() keys.sort() pageContent += "<br />number of keys: "+str(len(keys)) for key in keys: pageContent += fieldStorage[ key ].value pageContent += """</body></html>""" return pageContent I'm using Python 2.5.2 and Apache/2.2.3. This is what's in my apache conf file (and my script is in /var/www/python): <Directory /var/www/python/> Options FollowSymLinks +ExecCGI Order allow,deny allow from all AddHandler mod_python .py PythonHandler mod_python.publisher </Directory>

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  • How can I get VS2008 winforms designer to render a Form that implements an abstract base class

    - by BeowulfOF
    Hi, I engadged a problem with inherited Controls in WinForms, and need some advice on it. I do use a base class for items in a List (selfmade GUI list made of a panel) and some inherited controls that are for each type of data that could be added to the list. There was no problem with it, but I know found out, that it would be right, to make the base-control an abstract class, since it has methods, that need to be implemented in all inherited controls, called from the code inside the base-control, but must not and can not be implemented in the base class. When I mark the base-control as abstract, the VS2008 Designer refuses to load the window. Is there any way to get the Designer work with the base-control made abstract?

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  • Why does asp.net mvc form submits itself on button clicks when javascript function error?

    - by melaos
    hi guys, i'm new to the asp.net mvc, and while working on this, i used very basic asp.net mvc stuff like beginform, etc. i used a lot of jquery codes this round for client side validation, ajax data retrieval, and other gui works. and i used a combinations of html inputs buttons, etc and the asp.net mvc type of controls. what i noticed is that whenever i click on a button control, which sometimes are tied to either jquery oclick events, when there's a javascript error, the page will just go on and submit. why is this happening and what am i missing here? my bad for the dumb questions.. thanks

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  • Calling a method within Portlet when submitting form

    - by Roland
    I have a Portlet that contains a form. Now what I want to achieve is the following. 1) A Porlet containing a form is called within a page via <?php $this->widget('form'); ?> 2) The user fills in this form and clicks on submit "The submit button should be an ajax button" 3) When submit has been pressed the form should call a method within the form portlet class and the form should be replaced with a Thank you message. 4) I only want the current view in the portlet replaced with another view. My portlet class looks like this Yii::import('zii.widgets.CPortlet'); class Polls extends CPortlet{ public $usr_id=''; public function init(){ $cs = Yii::app()->clientScript; $cs->registerCoreScript('jquery'); parent::init(); } protected function renderContent(){ $this->render('form'); } public function update(){ $this->render('thankyou'); } } } Any advise, help would be highly appreciated.

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  • HTML Form HIdden Fields added with Javascript not POSTing

    - by dscher
    I have a form where the user can enter a link, click the "add link" button, and that link is then(via jQuery) added to the form as a hidden field. The problem is it's not POSTing when I submit the form. It's really starting to confound me. The thing is that if I hardcode a hidden field into the form, it is posted, but my function isn't working for some reason. The hidden field DOES get added to my form as I can see with Firebug but it's just not being sent with the POST data. Just to note, I'm using an array in Javascript to hold the elements until the form is submitted which also posts them visibly for the user to see what they've added. I'm using [] notation on the "name" field of the element because I want the links to feed into an array in PHP. Here is the link creation which is being appended to my form: function make_hidden_element_tag(item_type, item_content, item_id) { return '<input type="hidden" name="' + item_type + '[]" id="hidden_link_' + item_id + '" value="' + item_content + '"/>'; Does anyone have an idea why this might not be posting. As stated above, any hard-coded tags that are nearly identical to the above works fine, it's just that this tag isn't working. Here is how I'm adding the tag to the form with jQUery: $('#link_td').append( make_hidden_element_tag('links', link, link_array.length - 1)); I'm using the Kohana 3 framework, although I'm not sure that has any bearing on this because it's not really doing anything from the time the HTML is added to the page and the submit button is pressed.

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  • inserting a form to session raises picklingerror - django

    - by shanyu
    I receive an exception when I add a form to the session: PicklingError: Can't pickle <class 'django.utils.functional.__proxy__'>: attribute lookup django.utils.functional.__proxy__ failed The form includes a few simple fields and has some javascript attached to a widget. It might be that Django forms cannot be pickled at all, but the exception seems to point to unicode lazy translation. To test further, I have also tried to insert only the form errors (an errordict) to the session and received the same error. I appreciate some help here, thanks in advance. EDIT: Here's why I insert a form into the session: I have an app that has a form. This form is rendered by a template tag in another app. When posted, if the form is valid, no problem, I do stuff and redirect to "next". However if it is not valid, I want to go back to the posting page to show errors. Recall that the comments app in this case redirects to an intermediate "hey, please fix the errors" page. I am trying to avoid this, and hence redirect back to the posting page with the form and its errors in the session that the template tag will render.

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  • How to add a specific class to an input which has generated a form error?

    - by Kamil Mroczek
    I want to add a specific class to an input if an error is genereted by the input. For example, if input is empty and has required validator it shouls look like this: <dd id="login-element"> <input type="text" name="login" id="login" value="" class="input-text error" /> <ul class="errors"> <li>Value is required and can't be empty</li> </ul> </dd> class="input-text error" Please tell me how to do that.

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  • How can I adjust the position of a label for a zend form Radio element?

    - by murze
    Hi, with this piece of code $feOnline = New Zend_Form_Element_Radio('online'); $feOnline->setValue($article->online) ->addMultiOptions(array(0=>'offline', 1=>'online')) ->setLabel('Online'); this html is generated <dd id="online-element"> <label for="online-0"> <input type="radio" checked="checked" value="0" id="online-0" name="online">offline </label><br> <label for="online-1"><input type="radio" value="1" id="online-1" name="online">online </label> </dd> However I don't want the input-tag inside the label-tag. No need for the "" either... What decorators must I add to get this output? <dd id="online-element"> <input type="radio" checked="checked" value="0" id="online-0" name="online"><label for="online-0">offline</label> <input type="radio" value="1" id="online-1" name="online"><label for="online-1">online</label> </dd>

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  • Submit a form and get a JSON response with jQuery

    - by Leopd
    I expect this is easy, but I'm not finding a simple explanation anywhere of how to do this. I have a standard HTML form like this: <form name="new_post" action="process_form.json" method=POST> <label>Title:</label> <input id="post_title" name="post.title" type="text" /><br/> <label>Name:</label><br/> <input id="post_name" name="post.name" type="text" /><br/> <label>Content:</label><br/> <textarea cols="40" id="post_content" name="post.content" rows="20"></textarea> <input id="new_post_submit" type="submit" value="Create" /> </form> I'd like to have javascript (using jQuery) submit the form to the form's action (process_form.json), and receive a JSON response from the server. Then I'll have a javascript function that runs in response to the JSON response, like function form_success(json) { alert('Your form submission worked'); // process json response } How do I wire up the form submit button to call my form_success method when done? Also it should override the browser's own navigation, since I don't want to leave the page. Or should I move the button out of the form to do that?

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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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  • Symfony2 Forms: is it possible to bind a form in an "unconventional way"?

    - by DonCallisto
    Imagine this scenario: in our company there is an employee that "play" around graphic,css,html and so on. Our new project will born under symfony2 so we're trying some silly - but "real" - stuff (like authentication from db, submit data from a form and persist it to db and so on..) The problem As far i know, learnt from symfony2 "book" that i found on the site (you can find it here), there is an "automated" way for creating and rendering forms: 1) Build the form up into a controller in this way $form = $this->createFormBuilder($task) ->add('task','text'), ->add('dueDate','date'), ->getForm(); return $this->render('pathToBundle:Controller:templateTwig', array('form'=>$form->createview()); 2) Into templateTwig render the template {{ form_widget(form) }} // or single rows method 3) Into a controller (the same that have a route where you can submit data), take back submitted information if($rquest->getMethod()=='POST'){ $form->bindRequest($request); /* and so on */ } Return to scenario Our graphic employee don't want to access controllers, write php and other stuff like those. So he'll write a twig template with a "unconventional" (from symfony2 point of view, but conventional from HTML point of view) method: /* into twig template */ <form action="{{ path('SestanteUserBundle_homepage') }}" method="post" name="userForm"> <div> USERNAME: <input type="text" name="user_name" value="{{ user.username}}"/> </div> <div> EMAIL: <input type="text" name="user_mail" value="{{ user.email }}"/> </div> <input type="hidden" name="user_id" value="{{ id }}" /> <input type="submit" value="modifica i dati"> </form> Now, if into the controller that handle the submission of data we do something like that public function indexAction(Request $request) { if($request->getMethod() == 'POST'){ // sono arrivato per via di un submit, quindi devo modificare i dati prima di farli vedere a video $defaultData = array('message'=>'ho visto questa cosa in esempio, ma non capisco se posso farne a meno'); $form = $this->createFormBuilder($defaultData) ->add('user_name','text') ->add('user_mail','email') ->add('user_id','integer') ->getForm(); $form->bindRequest($request); //bindo la form ad una request $data = $form->getData(); //mi aspetto un'array chiave=>valore /* .... */ We expected that $data will contain an array with key,value from the submitted form. We found that it isn't true. After googling for a while and try with other "bad" ideas, we're frozen into that. So, if you have a "graphic office" that can't handle directly php code, how can we interface from form(s) to controller(s) ? UPDATE It seems that Symfony2 use a different convention for form's field name and lookup once you've submitted that. In particular, if my form's name is addUser and a field is named userName, the field's name will be AddUser[username] so maybe it have a "dynamic" lookup method that will extract form's name, field's name, concat them and lookup for values. Is it possible?

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  • Can JQuery.Validate plugin prevent submission of an Ajax form

    - by berko
    I am using the JQuery form plugin (http://malsup.com/jquery/form/) to handle the ajax submission of a form. I also have JQuery.Validate (http://docs.jquery.com/Plugins/Validation) plugged in for my client side validation. What I am seeing is that the validation fails when I expect it to however it does not stop the form from submitting. When I was using a traditional form (i.e. non-ajax) the validation failing prevented the form for submitting at all.... which is my desired behaviour. I know that the validation is hooked up correctly as the validation messages still appear after the ajax submit has happened. So what I am I missing that is preventing my desired behaviour? Sample code below.... <form id="searchForm" method="post" action="/User/GetDetails"> <input id="username" name="username" type="text" value="user.name" /> <input id="submit" name="submit" type="submit" value="Search" /> </form> <div id="detailsView"> </div> <script type="text/javascript"> var options = { target: '#detailsView' }; $('#searchForm').ajaxForm(options); $('#searchForm').validate({ rules: { username: {required:true}}, messages: { username: {required:"Username is a required field."}} }); </script>

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  • Data not transfred from form to mysql table (updating of data is not happening)

    - by Jimson
    Hi all and thanks in advance to all for this I tired and was unable to find the answer i am looking for an answer. my problem is that i am unable to update the values enterd in the form. I have attached all the files i'm using MYSQL database to fetch data. what happens is that i'm able to add and delete records from form using ajax and PHP scripts to MYSQL database, but i am not able to update data which was retrived from database. the file structure is as follows index.php is a file with ajax functions where it displays form for adding new data to MYSQL using save.php file and list of all records are view without refrishing page (calling load-list.php to view all records from index.php works fine, and save.php to save data from form) - *Delete*is an ajax function called from index.php to delete record from mysql database (function calling delete.php works fine) - Update is an ajax function called from index.php to update data using update-form.php by retriving specific record from mysql tabel, (works fine) Problem lies in updating data from update-form.php to update.php (in which update query is wrriten for mysql) i had tried in many ways at last i had figured out that data is not being transfred from update-form.php to update.php there is a small problem in jquery ajax function where it is not transfering data to update.php page. can any one correct this ????? i will be greatfull to them..... please find the link below for all files link to get my form files

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