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  • Image upload - Latin chars problem

    - by Holian
    Dear Gods! I use this script to upload images to serveR: <?php if (($_FILES["image_upload_box"]["type"] == "image/jpeg" || $_FILES["image_upload_box"]["type"] == "image/pjpeg" && ($_FILES["image_upload_box"]["size"] < 2000000)) { $max_upload_width = 450; $max_upload_height = 450; if(isset($_REQUEST['max_width_box']) and $_REQUEST['max_width_box']!='' and $_REQUEST['max_width_box']<=$max_upload_width){ $max_upload_width = $_REQUEST['max_width_box']; } if(isset($_REQUEST['max_height_box']) and $_REQUEST['max_height_box']!='' and $_REQUEST['max_height_box']<=$max_upload_height){ $max_upload_height = $_REQUEST['max_height_box']; } if($_FILES["image_upload_box"]["type"] == "image/jpeg" || $_FILES["image_upload_box"]["type"] == "image/pjpeg"){ $image_source = imagecreatefromjpeg($_FILES["image_upload_box"]["tmp_name"]); } $remote_file =$directory."/".$_FILES["image_upload_box"]["name"]; imagejpeg($image_source,$remote_file,100); chmod($remote_file,0644); list($image_width, $image_height) = getimagesize($remote_file); if($image_width>$max_upload_width || $image_height >$max_upload_height){ $proportions = $image_width/$image_height; if($image_width>$image_height){ $new_width = $max_upload_width; $new_height = round($max_upload_width/$proportions); } else{ $new_height = $max_upload_height; $new_width = round($max_upload_height*$proportions); } $new_image = imagecreatetruecolor($new_width , $new_height); $image_source = imagecreatefromjpeg($remote_file); imagecopyresampled($new_image, $image_source, 0, 0, 0, 0, $new_width, $new_height, $image_width, $image_height); imagejpeg($new_image,$remote_file,100); imagedestroy($new_image); } imagedestroy($image_source); }else{ something.... } ?> This is works well, till i upload a photo with latin chars in filename. For example the filename: kék hegyek.jpg. After upload file name will be: KA©k hegyek.jpg How can i solve this? Thank you

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  • Still don't understand file upload-folder permissions

    - by Camran
    I have checked out articles and tutorials. I don't know what to do about the security of my picture upload-folder. It is pictures for classifieds which should be uploaded to the folder. This is what I want: Anybody may upload images to the folder. The images will be moved to another folder, by another php-code later on (automatic). Only I may manually remove them, as well as another php file on the server which automatically empties the folder after x-days. What should I do here? The images are uploaded via a php-upload script. This script checks to see if the extension of the file is actually a valid image-file. When I try this: chmod 755 images the images wont be uploaded. But like this it works: chmod 777 images But 777 is a security risk right? Please give me detailed information... The Q is, what to do to solve this problem, not info about what permissions there are etc etc... Thanks If you need more info let me know...

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  • My C# UploadFile method successfully uploads a file, but then my UI hangs...

    - by kyrathaba
    I have a simple WinForms test application in C#. Using the following method, I'm able to upload a file when I invoke the method from my button's Click event handler. The only problem is: my Windows Form "freezes". I can't close it using the Close button. I have to end execution from within the IDE (Visual C# 2010 Express edition). Here are the two methods: public void UploadFile(string FullPathFilename) { string filename = Path.GetFileName(FullPathFilename); try { FtpWebRequest request = (FtpWebRequest)WebRequest.Create(_remoteHost + filename); request.Method = WebRequestMethods.Ftp.UploadFile; request.Credentials = new NetworkCredential(_remoteUser, _remotePass); StreamReader sourceStream = new StreamReader(FullPathFilename); byte[] fileContents = Encoding.UTF8.GetBytes(sourceStream.ReadToEnd()); request.ContentLength = fileContents.Length; Stream requestStream = request.GetRequestStream(); requestStream.Write(fileContents, 0, fileContents.Length); FtpWebResponse response = (FtpWebResponse)request.GetResponse(); response.Close(); requestStream.Close(); sourceStream.Close(); } catch (Exception ex) { MessageBox.Show(ex.Message, "Upload error"); } finally { } } which gets called here: private void btnUploadTxtFile_Click(object sender, EventArgs e) { string username = "my_username"; string password = "my_password"; string host = "ftp://mywebsite.com"; try { clsFTPclient client = new clsFTPclient(host + "/httpdocs/non_church/", username, password); client.UploadFile(Path.GetDirectoryName(Application.ExecutablePath) + "\\myTextFile.txt"); } catch (Exception ex) { MessageBox.Show(ex.Message, "Upload problem"); } }

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  • How to use cURL to FTPS upload to SecureTransport (hint: SITE AUTH and client certificates)

    - by Seamus Abshere
    I'm trying to connect to SecureTransport 4.5.1 via FTPS using curl compiled with gnutls. You need to use --ftp-alternative-to-user "SITE AUTH" per http://curl.haxx.se/mail/lib-2006-07/0068.html Do you see anything wrong with my client certificates? I try with # mycert.crt -----BEGIN CERTIFICATE----- ... -----END CERTIFICATE----- # mykey.pem -----BEGIN RSA PRIVATE KEY----- ... -----END RSA PRIVATE KEY----- And it says "530 No client certificate presented": myuser@myserver ~ $ curl -v --ftp-ssl --cert mycert.crt --key mykey.pem --ftp-alternative-to-user "SITE AUTH" -T helloworld.txt ftp://ftp.example.com:9876/upload/ * About to connect() to ftp.example.com port 9876 (#0) * Trying 1.2.3.4... connected * Connected to ftp.example.com (1.2.3.4) port 9876 (#0) < 220 msn1 FTP server (SecureTransport 4.5.1) ready. > AUTH SSL < 334 SSLv23/TLSv1 * found 142 certificates in /etc/ssl/certs/ca-certificates.crt > USER anonymous < 331 Password required for anonymous. > PASS [email protected] < 530 Login incorrect. > SITE AUTH < 530 No client certificate presented. * Access denied: 530 * Closing connection #0 curl: (67) Access denied: 530 I also tried with a pk8 version... # openssl pkcs8 -in mykey.pem -topk8 -nocrypt > mykey.pk8 -----BEGIN CERTIFICATE----- ... -----END CERTIFICATE----- ...but got exactly the same result. What's the trick to sending a client certificate to SecureTransport?

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  • VSFTPD does not allow upload with virtual users

    - by Mr. Squig
    I am attempting to setup VSFTPD with virtual users on a server running Ubuntu 12.04. I have configured the server to allow for virtual users to login, but I am having trouble getting it to allow uploads. My vsftpd.conf is as follows: listen=YES anonymous_enable=NO local_enable=YES write_enable=YES local_umask=022 anon_upload_enable=YES dirmessage_enable=YES use_localtime=YES xferlog_enable=YES connect_from_port_20=YES chroot_local_user=YES virtual_use_local_privs=YES guest_enable=YES guest_username=virtual user_sub_token=$USER local_root=/var/www/$USER hide_ids=YES secure_chroot_dir=/var/run/vsftpd/empty pam_service_name=vsftpd rsa_cert_file=/etc/ssl/private/vsftpd.pem /etc/pam.d/vsftpd contains: auth required pam_pwdfile.so pwdfile /etc/vsftpd.passwd crypt=hash account required pam_permit.so crypt=hash I have two virtual users set up, one of which has the same name as a local user. They each have a directory in /var/www/ owned by 'virtual'. As I understand it, when a virtual user logs in this way they will appear to the system as the user virtual. Using this configuration user can log on, but cannot upload files. The error given in /var/log/vsftpd.log is: Tue Nov 20 19:49:00 2012 [pid 2] CONNECT: Client "96.233.116.53" Tue Nov 20 19:49:07 2012 [pid 1] [zac] OK LOGIN: Client "96.233.116.53" Tue Nov 20 19:49:11 2012 [pid 2] CONNECT: Client "96.233.116.53" Tue Nov 20 19:49:11 2012 [pid 1] [zac] OK LOGIN: Client "96.233.116.53" Tue Nov 20 19:49:11 2012 [pid 3] [zac] FAIL CHMOD: Client "96.233.116.53", "/test.ppm 644" I have tried changing the permissions of these directories in all sorts of ways, but nothing seem to work. I have a feeling that it is something simple related to permissions. Any ideas?

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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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  • Server not accepting uploads

    - by Tatu Ulmanen
    I'm having a strange problem with my VPS: I can download files from it, I can use PuTTy to connect to it and all behaves normally. But sometimes, when I try to upload a file to the server or save a file via SFTP, the connection inexplicably fails. I am using jEdit to edit files remotely via SFTP. When it works, it works fine. When it doesn't, I get an error message: Cannot save: java.io.IOException: inputstream is closed Cannot save: java.io.IOException: 4: I can see that a temporary save file (#file.php#save#) is created on the server with a filesize of 0. So the connection works, but when it comes to sending the actual data, something fails. The same thing with WinSCP, but the error is different: Copying file fatally failed. Copying files to remote side failed. And I can always browse the server with PuTTy without a problem. I see nothing abnormal in any log files. Auth.log shows this when I try to save: sshd[32638]: Accepted password for - from - port 62272 ssh2 sshd[32638]: pam_unix(sshd:session): session opened for user - by (uid=0) sshd[32640]: subsystem request for sftp sshd[32638]: pam_unix(sshd:session): session closed for user - When I wait for a while (say, an hour), everything works fine again. It can't be a temporary ban, as I am still allowed to connect to the server, right? I know this may not be enough info to solve the problem, but I am grateful for any clues or bits of information that might help me. What are the possible causes for this kind of behaviour, what log files can I check for clues etc.. I'm running out of ideas!

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  • different user group can not upload file in the server

    - by Dallal
    I have a CentOS server running in Thailand, and I'm in Canada. The guy at the computer center who set up the server for me doesn't really understand much about linux and left me off an issue to solve myself. I just moved from Mac Server to Linux server, and the first thing I'm facing a problem now is `file name` has failed to upload due to an error The uploaded file could not be moved to `location name` So what happen is that I knew from my experiences of these problem is all about permissions. So I go ahead and checked on my whole folder and found that everything in the folder permission is like myusername mygroupname then I checked the httpd file in the server and it is default to apache apache. My question is that how can I make my user to be in the same group with apache group so that I don't have to have any problem about uploading, changing data in my file....? But without having to affect other user in the same server. I'm holding Administrator account, but not root account, but I can change stuff on the server root no problem. When I was with godaddy.com there never been any problem about the permission and I wish I know how they configure that :(

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  • Yii file upload issue

    - by user1289853
    Couple of months ago,I developed a simple app using YII,one of the feature was to upload file. The feature was working well in my dev machine,couple of days ago client found the file upload feature is not working in his server since deployment. And after that I test my dev machine that was not working too. My controller looks: public function actionEntry() { if (!Yii::app()->user->isGuest) { $model = new TrackForm; if (isset($_POST['TrackForm'])) { $entry = new Track; try { $entry->product_image = $_POST['TrackForm']['product_image']; $entry->product_image = CUploadedFile::getInstance($model, 'product_image'); if ($entry->save()) { if ($entry->product_image) { $entry->product_image->saveAs($entry->product_image->name, '/trackshirt/uploads'); } } $this->render('success', array('model' => $model)); // redirect to success page } } catch (Exception $e) { echo 'Caught exception: ', $e->getMessage(), "\n"; } } else { $this->render('entry', array('model' => $model)); } } } Model is like below: <?php class Track extends CActiveRecord { public static function model($className=__CLASS__) { return parent::model($className); } public function tableName() { return 'product_details'; } } My view looks: <?php $form = $this->beginWidget('CActiveForm', array( 'id' => 'hide-form', 'enableClientValidation' => true, 'clientOptions' => array( 'validateOnSubmit' => true, ), 'htmlOptions' => array('enctype' => 'multipart/form-data'), )); ?> <p class="auto-style2"><strong>Administration - Add New Product</strong></p> <table align="center" style="width: 650px"><td class="auto-style3" style="width: 250px">Product Image</td> <td> <?php echo $form->activeFileField($model, 'product_image'); ?> </td> </tr> </table> <p class="auto-style1"> <div style="margin-leftL:-100px;"> <?php echo CHtml::submitButton('Submit New Product Form'); ?> </div> <?php $this->endWidget(); ?> Any idea where is the problem?I tried to debug it but every time it returns Null. Thanks.

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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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  • ASP.NET MVC File Upload Error - "The input is not a valid Base-64 string"

    - by Justin
    Hey all, I'm trying to add a file upload control to my ASP.NET MVC 2 form but after I select a jpg and click Save, it gives the following error: The input is not a valid Base-64 string as it contains a non-base 64 character, more than two padding characters, or a non-white space character among the padding characters. Here's the view: <% using (Html.BeginForm("Save", "Developers", FormMethod.Post, new {enctype = "multipart/form-data"})) { %> <%: Html.ValidationSummary(true) %> <fieldset> <legend>Fields</legend> <div class="editor-label"> Login Name </div> <div class="editor-field"> <%: Html.TextBoxFor(model => model.LoginName) %> <%: Html.ValidationMessageFor(model => model.LoginName) %> </div> <div class="editor-label"> Password </div> <div class="editor-field"> <%: Html.Password("Password") %> <%: Html.ValidationMessageFor(model => model.Password) %> </div> <div class="editor-label"> First Name </div> <div class="editor-field"> <%: Html.TextBoxFor(model => model.FirstName) %> <%: Html.ValidationMessageFor(model => model.FirstName) %> </div> <div class="editor-label"> Last Name </div> <div class="editor-field"> <%: Html.TextBoxFor(model => model.LastName) %> <%: Html.ValidationMessageFor(model => model.LastName) %> </div> <div class="editor-label"> Photo </div> <div class="editor-field"> <input id="Photo" name="Photo" type="file" /> </div> <p> <%: Html.Hidden("DeveloperID") %> <%: Html.Hidden("CreateDate") %> <input type="submit" value="Save" /> </p> </fieldset> <% } %> And the controller: //POST: /Secure/Developers/Save/ [AcceptVerbs(HttpVerbs.Post)] public ActionResult Save(Developer developer) { //get profile photo. var upload = Request.Files["Photo"]; if (upload.ContentLength > 0) { string savedFileName = Path.Combine( ConfigurationManager.AppSettings["FileUploadDirectory"], "Developer_" + developer.FirstName + "_" + developer.LastName + ".jpg"); upload.SaveAs(savedFileName); } developer.UpdateDate = DateTime.Now; if (developer.DeveloperID == 0) {//inserting new developer. DataContext.DeveloperData.Insert(developer); } else {//attaching existing developer. DataContext.DeveloperData.Attach(developer); } //save changes. DataContext.SaveChanges(); //redirect to developer list. return RedirectToAction("Index"); } Thanks, Justin

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • html5 uploader + jquery drag & drop: how to store file data with FormData?

    - by lauthiamkok
    I am making a html5 drag and drop uploader with jquery, below is my code so far, the problem is that I get an empty array without any data. Is this line incorrect to store the file data - fd.append('file', $thisfile);? $('#div').on( 'dragover', function(e) { e.preventDefault(); e.stopPropagation(); } ); $('#div').on( 'dragenter', function(e) { e.preventDefault(); e.stopPropagation(); } ); $('#div').on( 'drop', function(e){ if(e.originalEvent.dataTransfer){ if(e.originalEvent.dataTransfer.files.length) { e.preventDefault(); e.stopPropagation(); // The file list. var fileList = e.originalEvent.dataTransfer.files; //console.log(fileList); // Loop the ajax post. for (var i = 0; i < fileList.length; i++) { var $thisfile = fileList[i]; console.log($thisfile); // HTML5 form data object. var fd = new FormData(); //console.log(fd); fd.append('file', $thisfile); /* var file = {name: fileList[i].name, type: fileList[i].type, size:fileList[i].size}; $.each(file, function(key, value) { fd.append('file['+key+']', value); }) */ $.ajax({ url: "upload.php", type: "POST", data: fd, processData: false, contentType: false, success: function(response) { // .. do something }, error: function(jqXHR, textStatus, errorMessage) { console.log(errorMessage); // Optional } }); } /*UPLOAD FILES HERE*/ upload(e.originalEvent.dataTransfer.files); } } } ); function upload(files){ console.log('Upload '+files.length+' File(s).'); }; then if I use another method is that to make the file data into an array inside the jquery code, var file = {name: fileList[i].name, type: fileList[i].type, size:fileList[i].size}; $.each(file, function(key, value) { fd.append('file['+key+']', value); }); but where is the tmp_name data inside e.originalEvent.dataTransfer.files[i]? php, print_r($_POST); $uploaddir = './uploads/'; $file = $uploaddir . basename($_POST['file']['name']); if (move_uploaded_file($_POST['file']['tmp_name'], $file)) { echo "success"; } else { echo "error"; } as you can see that tmp_name is needed to upload the file via php... html, <div id="div">Drop here</div>

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