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  • ASP.NET MVC - using model property as form, how can I post to action?

    - by Ryan Peters
    Consider the following model: public class BandProfileModel { public BandModel Band { get; set; } public IEnumerable<Relationship> Requests { get; set; } } and the following form: <% using (Html.BeginForm()) { %> <%: Html.EditorFor(m => m.Band) %> <input type="submit" value="Save Band" /> <% } %> which posts to the following action: public ActionResult EditPost(BandProfileModel m, string band) { // stuff is done here, but m is null? return View(m); } Basically, I only have one property on my model that is used in the form. The other property in BandProfleModel is just used in the UI for other data. I'm trying to update just the Band property, but for each post, the argument "m" is always null (specifically, the .Band property is null). It's posting just fine to the action, so it isn't a problem with my route. Just the data is null. The ID and name attributes of the fields are BAND_whatever and Band.whatever (whatever being a property of Band), so it seems like it would work... What am I doing wrong? How can I use just one property as part of a form, post back, and have values populated via the model binder for my BandProfileModel property in the action? Thanks.

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  • How do I add a confirmation popup on a button (GET POST action in MVC)?

    - by user54197
    I have a get/post/JSON function on an aspx page. This page adds data entered in a textbox to a table populated by javascript. When the user select the submit button. If the textbox is not empty, have a popup button telling the user the data in the textbox is not saved in the table. How do I have a confirm "ok/cancel" popup display on the post action in the Controller? I made a quick summary of what my code looks like. ... <% using (Html.BeginForm("AddName", "Name", FormMethod.Post, new { id = "AddNameForm" })) { %> ... <table id="displayNameTable" width= "100%"> <tr> <th colspan="3">Names Already Added</th> </tr> <tr> <td style="font-size:smaller;" class="name"></td> </tr> </table> ... <input name="Name" id="txtInjuryName" type="text" value="<%=test.Name %>" /> ... <input type="submit" name="add" value="Add"/> <% } %> <form id="form1" runat="server"> string confirmNext = ""; if (test.Name == "") { confirmNext = "return confirm('It seems you have a name not added.\n\nAre Continue?')"; }%> <input type="submit" name="getNext" value="Next" onclick="<%=confirmNext%>" /> </form>

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  • How to bind form collection back to custom model object that uses 2 custom objects in asp.net mvc?

    - by baijajusav
    What I'm trying to do is rather basic, but I might have my facts mixed up. I have a details page that has a custom class as it's Model. The custom class uses 2 custom objects with yet another custom object a property of one of the 2. The details page outputs a fair amount of information, but allows the user to post a comment. When the user clicks the post button, the page gets posted to a Details action that looks something like this: [AcceptVerbs(HttpVerbs.Post)] public ActionResult Details(VideoDetailModel vidAndComment) { ....} The only fields on the form that is posted are CommentText and VideoId. Here is what the VideoDetailModel looks like. public class VideoDetailModel { public VideoDetailModel() { Video = new VideoDTO(); Comment = new CommentDTO(); } public VideoDetailModel(VideoDTO vid) { Video = vid; Comment = new CommentDTO(); } public VideoDTO Video { get; set; } public CommentDTO Comment { get; set; } } VideoDTO has a few properties, but the ones I need are VideoId. CommentDTO's pertinent properties include CommentText (which is posting correctly) and a UserDTO object that contains a userId property. Everything other than the CommentText value is not being posted. I also have the following line on the ascx page, but the model value never gets posted to the controller. Html.Hidden("Model.Video.VideoId", Model.Video.VideoId); I'm really not sure what I'm missing here. I suppose if I added more form fields for the properties I need, they would get posted, but I only need 1 form entry field for the CommentText. If I could get the same Model objects value that were sent to the page to post with the page, that would help. I'll be happy to make any clarifications needed here. I'm just at loss as to what's going on.

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  • How do I make a jQuery POST function open the new page?

    - by ciclistadan
    I know that a submit button in HTML can submit a form which opens the target page, but how do I cause a jQuery ajax call POST information to a new page and display the new page. I am submitting information that is gathered by clicking elements (which toggle a new class) and then all items with this new class are added to an array and POSTed to a new page. I can get it to POST the data but it seems to be working functioning in an ajax non-refreshing manner, not submitting the page and redirecting to the new page. how might I go about doing this? here's the script section: //onload function $(function() { //toggles items to mark them for purchase //add event handler for selecting items $(".line").click(function() { //get the lines item number var item = $(this).toggleClass("select").attr("name"); }); $('#process').click(function() { var items = []; //place selected numbers in a string $('.line.select').each(function(index){ items.push($(this).attr('name')); }); $.ajax({ type: 'POST', url: 'additem.php', data: 'items='+items, success: function(){ $('#menu').hide(function(){ $('#success').fadeIn(); }); } }); }); return false; }); any pointers would be great!! thanks

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  • Am I under risk of CSRF attacks in a POST form that doesn't require the user to be logged in?

    - by Monika Sulik
    I'm probably being a total noob here, but I'm still uncertain about what a CSRF (Cross-Site Request Forgery) attack is exactly. So lets look at three situations... 1) I have a POST form that I use to edit data on my site. I want this data to be edited only by users that are logged in. 2) I have a site, which can be used by both users who are logged in as well as guests. Parts of the site are for logged in users only, but there are also POST forms that can be used by all users - anonymous and not (for example a standard contact form). Should the contact form be safeguarded against CSRF attacks? 3) I have a site which doesn't have an authentication system at all (well, perhaps that's unrealistic, so lets say it has an admin site which is separate from the rest of it and the admin part is properly safeguarded). The main part of the site is only used by anonymous users. Do the POST forms on it need to be safeguarded? In the case of 1) the answer is clearly yes. But in the case of 2 and 3 I don't know (and is the difference between 2 and 3 even significant?).

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  • How do I insert an input's value into a separate form?

    - by ryan
    I'm making a tool for my university that allows students to create a list of classes by doing the following: Searching by course title for their course via an autocomplete input field. Adding that course to a separate form that upon being submitted creates a course list. I am trying to link the autocomplete and the course list form with an 'add to course list' button that inserts a hidden input field into the course list form which can subsequently be submitted by a 'create course list' button. My question is this: How do I take the value of the autocomplete input and insert it into the course list form without using AJAX? So far I have something like the following: <%= text_field_with_auto_complete :course, :title, :size => 40 %> <%= link_to_function "Add to Course List" do |page| page.insert_html :top, :course_list, hidden_field(:courses, :course, {:value => "$('course_title').value"}) %> <% form_for(@course_list) do |f|%> <div id="course_list">Insert selected courses here.</div> <% end %>

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  • can anyone post their windows 7 UILanguages\en-US key?

    - by Sholom
    Hi I am having issues getting Windows 7 to change my system language to English. I followed the normal process but it's not completely changing it. I want to verify if my [HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\MUI\UILanguages\en-US] key is set correctly. Current values are: "LCID"=dword:00000409 "Type"=dword:00000091 can anyone with Windows 7 with english as the system language post their said key values? thanks

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  • Control 'ctl00_TextBox1' of type 'TextBox' must be placed inside a form tag with runat=server.

    - by Hiru
    When a form with a run at server is added there will be two forms with runat server and another error occurs. Can some one give me an idea. Thankx in advance. The details of the error are as follows. Control 'ctl00_TextBox1' of type 'TextBox' must be placed inside a form tag with runat=server. Description: An unhandled exception occurred during the execution of the current web request. Please review the stack trace for more information about the error and where it originated in the code. Exception Details: System.Web.HttpException: Control 'ctl00_TextBox1' of type 'TextBox' must be placed inside a form tag with runat=server. Source Error: An unhandled exception was generated during the execution of the current web request. Information regarding the origin and location of the exception can be identified using the exception stack trace below. Stack Trace: [HttpException (0x80004005): Control 'ctl00_TextBox1' of type 'TextBox' must be placed inside a form tag with runat=server.] System.Web.UI.Page.VerifyRenderingInServerForm(Control control) +2052287 System.Web.UI.WebControls.TextBox.AddAttributesToRender(HtmlTextWriter writer) +49 System.Web.UI.WebControls.WebControl.RenderBeginTag(HtmlTextWriter writer) +17 System.Web.UI.WebControls.TextBox.Render(HtmlTextWriter writer) +17 System.Web.UI.Control.RenderControlInternal(HtmlTextWriter writer, ControlAdapter adapter) +25 System.Web.UI.Control.RenderControl(HtmlTextWriter writer, ControlAdapter adapter) +121 System.Web.UI.Control.RenderControl(HtmlTextWriter writer) +22 System.Web.UI.Control.RenderChildrenInternal(HtmlTextWriter writer, ICollection children) +199 System.Web.UI.Control.RenderChildren(HtmlTextWriter writer) +20 System.Web.UI.Control.Render(HtmlTextWriter writer) +7 System.Web.UI.Control.RenderControlInternal(HtmlTextWriter writer, ControlAdapter adapter) +25 System.Web.UI.Control.RenderControl(HtmlTextWriter writer, ControlAdapter adapter) +121 System.Web.UI.Control.RenderControl(HtmlTextWriter writer) +22 System.Web.UI.Control.RenderChildrenInternal(HtmlTextWriter writer, ICollection children) +199 System.Web.UI.Control.RenderChildren(HtmlTextWriter writer) +20 System.Web.UI.Page.Render(HtmlTextWriter writer) +26 System.Web.UI.Control.RenderControlInternal(HtmlTextWriter writer, ControlAdapter adapter) +25 System.Web.UI.Control.RenderControl(HtmlTextWriter writer, ControlAdapter adapter) +121 System.Web.UI.Control.RenderControl(HtmlTextWriter writer) +22 System.Web.UI.Page.ProcessRequestMain(Boolean includeStagesBeforeAsyncPoint, Boolean includeStagesAfterAsyncPoint) +2558 Version Information: Microsoft .NET Framework Version:2.0.50727.1873; ASP.NET Version:2.0.50727.1433

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  • UICollectionView with one static cell and N dynamic ones from a fetchresultscontroller exception

    - by nflacco
    I'm trying to make a UITableView that shows a blog post and comments for that post. My setup is a tableview in storyboard with two dynamic prototype cells. The first cell is for the post and should never change. The second cell represents the 0 to N comments. My cellForRowAtIndexPath method shows the post cell properly, but fails to get the comment at the given index path (though if I comment out the fetch I get the appropriate number of comment cells with a green background that I set as a visual debug thing). let comment = fetchedResultController.objectAtIndexPath(indexPath) as Comment I get the following exception on this line: 2014-08-24 15:06:40.712 MessagePosting[21767:3266409] *** Terminating app due to uncaught exception 'NSRangeException', reason: '*** -[__NSArrayM objectAtIndex:]: index 1 beyond bounds [0 .. 0]' *** First throw call stack: ( 0 CoreFoundation 0x0000000101aa43e5 __exceptionPreprocess + 165 1 libobjc.A.dylib 0x00000001037f9967 objc_exception_throw + 45 2 CoreFoundation 0x000000010198f4c3 -[__NSArrayM objectAtIndex:] + 227 3 CoreData 0x00000001016e4792 -[NSFetchedResultsController objectAtIndexPath:] + 162 Section and cell setup: override func tableView(tableView: UITableView!, numberOfRowsInSection section: Int) -> Int { // #warning Incomplete method implementation. // Return the number of rows in the section. switch section { case 0: return 1 default: if let realPost:Post = post { return fetchedResultController.sections[0].numberOfObjects } else { return 0 } } } override func tableView(tableView: UITableView!, cellForRowAtIndexPath indexPath: NSIndexPath!) -> UITableViewCell! { switch indexPath.section { case 0: let cell = tableView.dequeueReusableCellWithIdentifier(postViewCellIdentifier, forIndexPath: indexPath) as UITableViewCell cell.backgroundColor = lightGrey if let realPost:Post = self.post { cell.textLabel.text = realPost.text } return cell default: let cell = tableView.dequeueReusableCellWithIdentifier(commentCellIdentifier, forIndexPath: indexPath) as UITableViewCell cell.backgroundColor = UIColor.greenColor() let comment = fetchedResultController.objectAtIndexPath(indexPath) as Comment // <---------------------------- :( cell.textLabel.text = comment.text return cell } } FRC: func controllerDidChangeContent(controller: NSFetchedResultsController!) { tableView.reloadData() } func getFetchedResultController() -> NSFetchedResultsController { fetchedResultController = NSFetchedResultsController(fetchRequest: taskFetchRequest(), managedObjectContext: managedObjectContext, sectionNameKeyPath: nil, cacheName: nil) return fetchedResultController } func taskFetchRequest() -> NSFetchRequest { if let realPost:Post = self.post { let fetchRequest = NSFetchRequest(entityName: "Comment") let sortDescriptor = NSSortDescriptor(key: "date", ascending: false) fetchRequest.predicate = NSPredicate(format: "post = %@", realPost) fetchRequest.sortDescriptors = [sortDescriptor] return fetchRequest } else { return NSFetchRequest(entityName: "") } }

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  • Inside what the TexBox value is posted back? ViewState or post back data?

    - by burak ozdogan
    In one article I was reading on ViewState, I saw a sentence saying that I should not fall into a mistake to believe that the value of a TextBox is stored in ViewState; it is stored in PostBack data. From here what I understand is when I post back a web form, the input controls values are stored in HTTP Request body. Not in the Viewstate. But as far as I know ViewState values are stored in an hidden field called __VIEWSTATE anyway. Then does it mean that __VIEVSTATE value is not posted in HTTP POST Request body as a postback data? Sounds nonesense to me. In another words, basically if I say the ViewState mechanism for such scenerio works like this, am I seeing it right or skipping something: You enter a value on an empty TextBox and submit the page The value of text box is posted back inside POST HTTP Request body. Nothing inside __VIEWSTATE at this point from the TextBox On the server side, the TextBox is created with the default value on OnInit method of the page The TrackChange property of ViewState is set to true. The posted back data of TextBox is loaded. Because it is different than the TextBox defalut value(because the user entered something), the ViewState of this text box is marked as DIRTY. The new value of the textbox is written into __VIEWSTATE hidden field From now on __VIEWSTATE hiddenfeild contains the last given value of the TextBox The page is sent to the user's browser having the __VIEWSTATE hidden field. But this time containing the last value entered by user which will be ready to be rendered Thanks guys! burak ozdogan

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  • KeyUp processed for wrong control

    - by Mikael
    I have made a simple test application for the issue, two winforms each containing a button. The button on the first form opens the other form when clicked. It also subscribes to keyup events. The second form has its button set as "AcceptButton" and in the Clicked event we sleep for 1s and then set DialogResult to true (the sleep is to simulate some processing done) When enter is used to close this second form the KeyUp event of the button on the first form is triggered, even though the key was released well before the second had passed so the second form was still shown and focused. If any key other then enter is pressed in the second form the event is not triggered for the button on the first form. First form: public Form1() { InitializeComponent(); buttonForm2.KeyUp += new KeyEventHandler(cntKeyUp); } void cntKeyUp(object sender, KeyEventArgs e) { MessageBox.Show(e.KeyCode.ToString()); } private void buttonForm2_Click(object sender, EventArgs e) { using (Form2 f = new Form2()) { f.ShowDialog(); } } Second form: private void button1_Click(object sender, EventArgs e) { Thread.Sleep(1000); this.DialogResult = DialogResult.OK; } Does anyone know why the event is triggered for the button on the non active form and what can be done to stop this from happening?

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  • [C#][XNA 3.1] How can I host two different XNA windows inside one Windows Form?

    - by secutos
    I am making a Map Editor for a 2D tile-based game. I would like to host two XNA controls inside the Windows Form - the first to render the map; the second to render the tileset. I used the code here to make the XNA control host inside the Windows Form. This all works very well - as long as there is only one XNA control inside the Windows Form. But I need two - one for the map; the second for the tileset. How can I run two XNA controls inside the Windows Form? While googling, I came across the terms "swap chain" and "multiple viewports", but I can't understand them and would appreciate support. Just as a side note, I know the XNA control example was designed so that even if you ran 100 XNA controls, they would all share the same GraphicsDevice - essentially, all 100 XNA controls would share the same screen. I tried modifying the code to instantiate a new GraphicsDevice for each XNA control, but the rest of the code doesn't work. The code is a bit long to post, so I won't post it unless someone needs it to be able to help me. Thanks in advance.

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  • [EF 4 POCO] Problem with INSERT...

    - by Darmak
    Hi all, I'm so frustrated because of this problem, you have no idea... I have 2 classes: Post and Comment. I use EF 4 POCO support, I don't have foreign key columns in my .edmx model (Comment class doesn't have PostID property, but has Post property) class Comment { public Post post { get; set; } // ... } class Post { public virtual ICollection<Comment> Comments { get; set; } // ... } Can someone tell me why the code below doesn't work? I want to create a new comment for a post: Comment comm = context.CreateObject<Comment>(); Post post = context.Posts.Where(p => p.Slug == "something").SingleOrDefault(); // post != null, so don't worry, be happy // here I set all other comm properties and... comm.Post = post; context.AddObject("Comments", comm); // Exception here context.SaveChanges(); The Exception is: Cannot insert the value NULL into column 'PostID', table 'Blog.Comments'; column does not allow nulls. INSERT fails. ... this 'PostID' column is of course a foreign key to the Posts table. Any help will be appreciated!

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  • Problem with multiple forms on one page

    - by Jon
    I have two forms on a web page. The first is not an actual form since this is a .NET based site. So instead I have the standard input fields, along with asp:Button PostBackURL="http:/www...". One of the fields is "email". That works fine. Then I have an XSLT file with another form, and am using Javascript (via this.form.action, .method, etc) to post that form. Besides that it has the same fields as the form above. The problem is that when someone submits the second form, the script at the PostBackURL returns an error because the "email" field appears empty. So it seems as though the form is submitting the "email" field form the top form instead of the current one. I thought maybe the wrong form is being submitted, but if I remove the "email" field from the top form, and then submit the bottom one, it works fine. Any ideas on how to fix this? Thanks.

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  • Apply the REL attibute automatically to posts using jQuery

    - by Couto
    Edited: I mean grouping as giving the same REL attibute to all IMGs in the same post, but each post has different REL as the example at the end of this question. So, I need to do the following: <div id="Blog1" class="widget Blog"> <div class="blog-posts hfeed"> <div class="post hentry uncustomized-post-template"> <a name="8829400899632947948"/> <div class="post-body entry-content"> <div id="8829400899632947948"> <div class="separator"> <a imageanchor="1" href="/images/outta.png"> <img src="/images/outta.png"/></a></div></div> <div style="clear: both;"/> </div> <div class="post-footer"> </div></div></div></div> I'm using jQuery and Colorbox. The first two DIVs are posts containers. I need to group the IMGs in each <div class="post hentry uncustomized-post-template"> using the REL attribute, like: 1 - Post 1.1 - IMG - REL="group0" 1.2 - IMG - REL="group0" 1.3 - IMG - REL="group0" 1.4 - IMG - REL="group0" 2 - Post 2.1 - IMG - REL="group1" 3 - Post 3.1 - IMG - REL="group2" 3.2 - IMG - REL="group2" I've tryied HAS, PARENT > CHILDREN and CHILDREN() from jQuery and REL: from Colorbox, but it seems I'm lacking somewhere in logic. Could someone help me?

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  • CakePHP 1.26: Bug in 'Security' component?

    - by Steve
    Okay, for those of you who may have read this earlier, I've done a little research and completely revamped my question. I've been having a problem where my form requests get blackholed by the Security component, although everything works fine when the Security component is disabled. I've traced it down to a single line in a form: <?php echo $form->create('Audition');?> <fieldset> <legend><?php __('Edit Audition');?></legend> <?php echo $form->input('ensemble'); echo $form->input('position'); echo $form->input('aud_date'); // The following line works fine... echo $form->input('owner'); // ...but the following line blackholes when Security included // and the form is submitted: // echo $form->input('owner', array('disabled'=>'disabled'); ?> </fieldset> <?php echo $form->end('Submit');?> (I've commented out the offending line for clarity) I think I'm following the rules by using the form helper; as far as I can tell, this is a bug in the Security component, but I'm too much of a CakePHP n00b to know for sure. I'd love to get some feedback, and if it's a real bug, I'll submit it to the CakePHP team. I'd also love to know if I'm just being dumb and missing something obvious here.

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  • What's the best way to format this simple HTML form using CSS?

    - by GregH
    I have have a simple HTML form with say four input widgets (see below)...two lines with two widgets on each line. However, when this renders it is pretty ugly. I want the whole form to be indented from the edge of the left page say 40px and I want the left edge of the widgets to line up with each other and the right edge of the labels to line up. I also want to be able to specify a minimum distance between the right edge of the first widget and the label of the widget next to it. How would I do this using CSS? Basically so it looks something like: Name: _____________ Common Names: _____________ Version: ____________ Status: ____________ See current un-formatted HTML below: <form name="detailData"> <div id="dataEntryForm"> <label> Name: <input type="text" class="input_text" name="ddName"/> Common Names: <input type="text" class="input_text" name="ddCommonNames"><P> Version: <input type="text" class="input_text" name="ddVer"/> Status: <select name="ddStatus"><option value="A" selected="selected">Active</option><option value="P">Planned</option><option value="D">Deprecated</option> </label> </div> </form>

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  • jQuery ajax post of jpg image to .net webservice. Image results corrupted

    - by sosergio
    I have a phonegap jquery app that opens the camera and take a picture. I then POST this picture to a .net webservice, wich I've coded. I can't use phonegap FileTransfer because such isn't supported by Bada os, wich is a requirement. I believe I've successfully loaded the image from phonegap FileSystem API, I've attached it into an .ajax type:post, I've even received it from .net side, but when .net save the image into the server, the image results corrupted. It seems to me that two sides of the communication have different data type. Has anyone experience in this? Any help will be appreciated. This is my code: //PHONEGAP CAMERA ACCESS (summed up) navigator.camera.getPicture(onGetPictureSuccess, onGetPictureFail, { quality: 50, destinationType:Camera.DestinationType.FILE_URI }); window.resolveLocalFileSystemURI(imageURI, onResolveFileSystemURISuccess, onResolveFileSystemURIError); fileEntry.file(gotFileSuccess, gotFileError); new FileReader().readAsDataURL(file); //UPLOAD FILE function onDataReadSuccess(evt) { var image_data = evt.target.result; var filename = unique_id(); var filext = "jpg"; $.ajax({ type : 'POST', url : SERVICE_BASE_URL+"/fotos/"+filename+"?ext="+filext, cache: false, timeout: 100000, processData: false, data: image_data, contentType: 'image/jpeg', success : function(data) { console.log("Data Uploaded with success. Message: "+ data); $.mobile.hidePageLoadingMsg(); $.mobile.changePage("ok.html"); } }); } On my .net Web Service this is the method that gets invoked: public string FotoSave(string filename, string extension, Stream fileContent) { string filePath = HttpContext.Current.Server.MapPath("~/foto_data/") + "\\" + filename; FileStream writeStream = new FileStream(filePath, FileMode.OpenOrCreate, FileAccess.Write); int Length = 256; Byte[] buffer = new Byte[Length]; int bytesRead = readStream.Read(buffer, 0, Length); // write the required bytes while (bytesRead > 0) { writeStream.Write(buffer, 0, bytesRead); bytesRead = readStream.Read(buffer, 0, Length); } readStream.Close(); writeStream.Close(); }

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  • I am confused about how to use @SessionAttributes

    - by yusaku
    I am trying to understand architecture of Spring MVC. However, I am completely confused by behavior of @SessionAttributes. Please look at SampleController below , it is handling post method by SuperForm class. In fact, just field of SuperForm class is only binding as I expected. However, After I put @SessionAttributes in Controller, handling method is binding as SubAForm. Can anybody explain me what happened in this binding. ------------------------------------------------------- @Controller @SessionAttributes("form") @RequestMapping(value = "/sample") public class SampleController { @RequestMapping(method = RequestMethod.GET) public String getCreateForm(Model model) { model.addAttribute("form", new SubAForm()); return "sample/input"; } @RequestMapping(method = RequestMethod.POST) public String register(@ModelAttribute("form") SuperForm form, Model model) { return "sample/input"; } } ------------------------------------------------------- public class SuperForm { private Long superId; public Long getSuperId() { return superId; } public void setSuperId(Long superId) { this.superId = superId; } } ------------------------------------------------------- public class SubAForm extends SuperForm { private Long subAId; public Long getSubAId() { return subAId; } public void setSubAId(Long subAId) { this.subAId = subAId; } } ------------------------------------------------------- <form:form modelAttribute="form" method="post"> <fieldset> <legend>SUPER FIELD</legend> <p> SUPER ID:<form:input path="superId" /> </p> </fieldset> <fieldset> <legend>SUB A FIELD</legend> <p> SUB A ID:<form:input path="subAId" /> </p> </fieldset> <p> <input type="submit" value="register" /> </p> </form:form>

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  • jQuery: How to use modifier keys on form submit?

    - by Svish
    Say I have a form that looks like this: [ Animal name input field ] Add button If I type a name and hit enter, an animal with the given name is added to a table. Works fine. What I would like now is to call the current way of working "quick add" and add a new feature called "slow add", which I am not quite sure how to do. Basically what I want is that if for example the shift key is held down when enter or the button is clicked, I want the form submit method to do something slightly different. In my case I want it to open up a form where more details on the animal can be added before it is added to the table. Problem is I'm not quite sure how to do this. I have tried add a FireBug console.info(eventData) in my current submit function and I have found that the eventData contains an altKey, shiftKey and controlKey property, but they are always undefined even when I hold those keys down. So, does anyone know how I can do something special in my submit handler when certain modifier keys were pressed when the form was submitted?

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  • WindowsFormsApplicationBase SplashScreen makes login form ignore keypresses until I click on it - how to debug?

    - by Tom Bushell
    My WinForms app has a simple modal login form, invoked at startup via ShowDialog(). When I run from inside Visual Studio, everything works fine. I can just type in my User ID, hit the Enter key, and get logged in. But when I run a release build directly, everything looks normal (the login form is active, there's a blinking cursor in the User ID MaskedEditBox), but all keypresses are ignored until I click somewhere on the login form. Very annoying if you are used to doing everything from the keyboard. I've tried to trace through the event handlers, and to set the focus directly with code, to no avail. Any suggestions how to debug this (outside of Visual Studio), or failing that - a possible workaround? Edit Here's the calling code, in my Main Form: private void OfeMainForm_Shown(object sender, EventArgs e) { OperatorLogon(); } private void OperatorLogon() { // Modal dialogs should be in a "using" block for proper disposal using (var logonForm = new C21CfrLogOnForm()) { var dr = logonForm.ShowDialog(this); if (dr == DialogResult.OK) SaveOperatorId(logonForm.OperatorId); else Application.Exit(); } } Edit 2 Didn't think this was relevant, but I'm using Microsoft.VisualBasic.ApplicationServices.WindowsFormsApplicationBase for it's splash screen and SingleInstanceController support. I just commented out the splash screen code, and the problem has disappeared. So that's opened up a whole new line of inquiry... Edit 3 Changed title to reflect better understanding of the problem

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  • Multiple File upload doesnot work in CI

    - by sabuj
    Hi I have used Jquery Plugin to upload multiple file with CI. It works fine for one file but if i try to upload more than one file it doesnot work it gives the result like this"menu_back9.pngmenu_back9.png.jpg". function do_upload() { $config['upload_path'] = './uploads/'; // server directory $config['allowed_types'] = 'gif|jpg|png'; // by extension, will check for whether it is an image $config['max_size'] = '1000'; // in kb $config['max_width'] = '1024'; $config['max_height'] = '768'; $this-load-library('upload', $config); $this-load-library('Multi_upload'); $files = $this-multi_upload-go_upload(); if ( ! $files ) { $error = array('error' => $this->upload->display_errors()); $this->load->view('admin/add_digital_product', $error); } else { foreach($files as $file) { $pic_name = $file['name']; echo $pic_name; } exit; $data = array( 'dg_image_1'=$picture, 'dgproduct_name'=$this-input-post('dgproduct_name',TRUE), 'dgproduct_description'=$this-input-post('dgproduct_description',TRUE), 'url_additional'=$this-input-post('url_additional',TRUE), 'url_stored'=$this-input-post('url_stored',TRUE), 'delivery_format'=$this-input-post('delivery_format',TRUE), 'item_size'=$this-input-post('item_size',TRUE), 'price'=$this-input-post('price',TRUE), 'item_code'=$this-input-post('item_code',TRUE), 'payment_pro_code'=$this-input-post('payment_pro_code',TRUE), 'delivery_time'=$this-input-post('delivery_time',TRUE), 'thankyou_message'=$this-input-post('thankyou_message',TRUE), ); $this-db-insert('sm_digital_product',$data); redirect(site_url().'admin/admins'); } } I used avobe code If any body know please tell me. I am stack there.

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  • In CakePHP, how do I create a form to list models withing a date range?

    - by anonymous coward
    I have a very simple model that includes the auto-filled field much like 'created'. (DateTime format). I'd like to use the Form helpers if possible, to validate the date fields and whatnot. I'd like a simple form with a "Begin Date" (YMD, 12 hours), and an "End Date" (same format). There is already a controller action set up as follows: function view_between($start_date = null, $end_date = null) { // ... stuff that works correctly when the URL is manually entered. } Have I defined the controller wrong, or how can I pass these values into that function? The reason I'm stuck is because I tried adding a $form->input('my_datetime_field' ...) twice, but obviously the name/id were the same on the respective elements. I have also tried using $form->dateTime(...) with similar results. I'm not sure how to uniquely identify a Begin and End date selection, when it should interact with a single field. Am I going about this wrong? A kind shove in the right direction should suffice.

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  • C# - Is it possible to start my project with a class instead of a form?

    - by Irro
    I want my project to be started through an class instead of a form, is there any way to do this? Or to be more precise is there any good way to make sure that the first class, except Program, that is started isn't a form-class. I tried to change to my class in Program.main() but it looks like Application.run() needs a ApplicationContext. I guess that I could change the Program-class to start another class and let that class start the form with Application.run() but I think that it will cause a lot of problem since I don't want the same form to be started first each time and Application.run() have to be used at least once and at most once. So I think it will be hard to keep track of if Application.run() has been used or not. Another question that might be even more important; Is this a good way to do things in .net? The reason I want to do so is because I want to create some sort of MVC project where the class I want to start with is the controller and all forms I'll use will be views.

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