Search Results

Search found 22501 results on 901 pages for 'form upload'.

Page 15/901 | < Previous Page | 11 12 13 14 15 16 17 18 19 20 21 22  | Next Page >

  • Silverlight file upload - file is in use by another process (Excel, Word)

    - by walkor
    Hi, all. I have a problem with uploading of the file in Silverlight application. Here is a code sample. In case when this file is opened in other application (excel or word for example) it fails to open it, otherwise it's working fine. I'm using OpenFileDialog to choose the file and pass it to this function. private byte[] GetFileContent(FileInfo file) { var result = new byte[] {}; try { using (var fs = file.OpenRead()) { result = new byte[file.Length]; fs.Read(result, 0, (int)file.Length); } } catch (Exception e) { // File is in use } return result; } Is there any way i can access this file or should i just notify the user that the file is locked?

    Read the article

  • Export GridView to TXT, then upload file to server

    Basically what I want to do is export an array (or GridView) to a file called "getpathin.dat". That is easy, but the method I am using downloads the file to my computer, which is what I don't want. I want to write an array to either a PRE-EXISTING file that is on the server OR create a new file on the server in a folder, and this new file will contain either the array or the gridview, which I have stored in the array. i'll be doing this in... Visual Studio 2008/SQL Server using C#

    Read the article

  • Help with PHP simplehtmldom - Modifiying a form.

    - by onemyndseye
    Ive gotten some great help here and I am so close to solving my problem that I can taste it. But I seem to be stuck. I need to scrape a simple form from a local webserver and only return the lines that match a users local email (i.e. onemyndseye@localhost). simplehtmldom makes easy work of extracting the correct form element: foreach($html->find('form[action*="delete"]') as $form) echo $form; Returns: <form action="/delete" method="post"> <input type="checkbox" id="D1" name="D1" /><a href="http://www.linux.com/rss/feeds.php"> http://www.linux.com/rss/feeds.php </a> [email: onemyndseye@localhost (Default) ]<br /> <input type="checkbox" id="D2" name="D2" /><a href="http://www.ubuntu.com/rss.xml"> http://www.ubuntu.com/rss.xml </a> [email: onemyndseye@localhost (Default) ]<br /> However I am having trouble making the next step. Which is returning lines that contain 'onemyndseye@localhost' and removing it so that only the following is returned: <input type="checkbox" id="D1" name="D1" /><a href="http://www.linux.com/rss/feeds.php">http://www.linux.com/rss/feeds.php</a> <br /> <input type="checkbox" id="D2" name="D2" /><a href="http://www.ubuntu.com/rss.xml">http://www.ubuntu.com/rss.xml</a> <br /> Thanks to the wonderful users of this site Ive gotten this far and can even return just the links but I am having trouble getting the rest... Its important that the complete <input> tags are returned EXACTLY as shown above as the id and name values will need to be passed back to the original form in post data later on. Thanks in advance!

    Read the article

  • asp code for upload data

    - by vicky
    hello everyone i have this code for uploading an excel file and save the data into database.I m not able to write the code for database entry. someone please help <% if (Request("FileName") <> "") Then Dim objUpload, lngLoop Response.Write(server.MapPath(".")) If Request.TotalBytes > 0 Then Set objUpload = New vbsUpload For lngLoop = 0 to objUpload.Files.Count - 1 'If accessing this page annonymously, 'the internet guest account must have 'write permission to the path below. objUpload.Files.Item(lngLoop).Save "D:\PrismUpdated\prism_latest\Prism\uploadxl\" Response.Write "File Uploaded" Next Dim FSYSObj, folderObj, process_folder process_folder = server.MapPath(".") & "\uploadxl" set FSYSObj = server.CreateObject("Scripting.FileSystemObject") set folderObj = FSYSObj.GetFolder(process_folder) set filCollection = folderObj.Files Dim SQLStr SQLStr = "INSERT ALL INTO TABLENAME " for each file in filCollection file_name = file.name path = folderObj & "\" & file_name Set objExcel_chk = CreateObject("Excel.Application") Set ws1 = objExcel_chk.Workbooks.Open(path).Sheets(1) row_cnt = 1 'for row_cnt = 6 to 7 ' if ws1.Cells(row_cnt,col_cnt).Value <> "" then ' col = col_cnt ' end if 'next While (ws1.Cells(row_cnt, 1).Value <> "") for col_cnt = 1 to 10 SQLStr = SQLStr & "VALUES('" & ws1.Cells(row_cnt, 1).Value & "')" next row_cnt = row_cnt + 1 WEnd 'objExcel_chk.Quit objExcel_chk.Workbooks.Close() set ws1 = nothing objExcel_chk.Quit Response.Write(SQLStr) 'set filobj = FSYSObj.GetFile (sub_fol_path & "\" & file_name) 'filobj.Delete next End if End If plz tell me how to save the following excel data to the oracle databse.any help would be appreciated

    Read the article

  • PHP FTP Upload thousands of files

    - by user275074
    Hi, I've written a small FTP class which I used to move files from a local server to a remote server. It does this by checking an array of local files with an array of files on the remote server. If the file exists on the remote server, it won't bother uploading it. The script works fine for small amounts of files, but I've noticed that the local server can have as many as 3000+ image files to transfer, this seems to cause the script to flop and only transfer a 100 or so. How can I modify the script to handle potentially thousands of image transfer files?

    Read the article

  • Nginx: check content-length before file upload takes place

    - by robw
    I'm trying to prevent users from uploading (accidentally or maliciously) very large files to my website. I have nginx max_client_body_size set to 4M, but if a file larger than this is uploaded, then it uploads the entire file before returning 413 (entity too large). I want to make nginx check the Content-Length header, so that it rejects the request before it's uploaded. Alternatively, a Rails solution would also be acceptable. Any help appreciated.

    Read the article

  • jQuery/JavaScript Date form validation

    - by Victor Jackson
    I am using the jQuery date picker calendar in a form. Once submitted the form passes params along via the url to a third party site. Everything works fine, except for one thing. If the value inserted into the date field by the datepicker calendar is subsequently deleted, or if the default date, that is in the form on page load, is deleted and the form is submitted I get the following error: "Conversion from string "" to type 'Date' is not valid." The solution to my problem is really simple, I want to validate the text field where the date is submitted and send out a default date (current date) if the field is empty for any reason. The problem is I am terrible at Javascript and cannot figure out how to do this. Here is the form code for my date field. [var('default_date' = date)] <input type="text" id="datepicker" name="txtdate" value="[$default_date]" onfocus="if (this.value == '[$default_date]') this.value = '';" onchange="form.BeginDate.value = this.value; form.EndDate.value = this.value;" /> <input type="hidden" name="BeginDate" value="[$default_date]"/> <input type="hidden" name="EndDate" value="[$default_date]"/>

    Read the article

  • Ajax -jquery image upload to iFrame resize problem

    - by DMort59
    I'm modifiying an ajax like jquery image uploader from /www.atwebresults.com/php_ajax_image_upload/ to see image on screen in iframe. I need the image to come in at 100% so I can use jquery resize. filename=name&maxSize=9999999999&maxW=200&fullPath=http://localhost/nyb/uploads/&relPath=../uploads/&colorR=255&colorG=255&colorB=255&maxH=300"

    Read the article

  • ASP.NET Image Upload Parameter Not Valid. Exception

    - by pennylane
    Hi Guys, Im just trying to save a file to disk using a posted stream from jquery uploadify I'm also getting Parameter not valid. On adding to the error message so i can tell where it blew up in production im seeing it blow up on: var postedBitmap = new Bitmap(postedFileStream) any help would be most appreciated public string SaveImageFile(Stream postedFileStream, string fileDirectory, string fileName, int imageWidth, int imageHeight) { string result = ""; string fullFilePath = Path.Combine(fileDirectory, fileName); string exhelp = ""; if (!File.Exists(fullFilePath)) { try { using (var postedBitmap = new Bitmap(postedFileStream)) { exhelp += "got past bmp creation" + fullFilePath; using (var imageToSave = ImageHandler.ResizeImage(postedBitmap, imageWidth, imageHeight)) { exhelp += "got past resize"; if (!Directory.Exists(fileDirectory)) { Directory.CreateDirectory(fileDirectory); } result = "Success"; postedBitmap.Dispose(); imageToSave.Save(fullFilePath, GetImageFormatForFile(fileName)); } exhelp += "got past save"; } } catch (Exception ex) { result = "Save Image File Failed " + ex.Message + ex.StackTrace; Global.SendExceptionEmail("Save Image File Failed " + exhelp, ex); } } return result; }

    Read the article

  • Multiple file upload with preview

    - by sean717
    Hi, Is there any good control or plug-in for uploading multiple photos with preview? As far as I understand it is impossible to preview photo on local computer using just JavaScript. So it has to use Flash or Java. Thanks, also, I use ASP.NET.

    Read the article

  • Drupal: How to Render Results of Form on Same Page as Form

    - by Aaron
    How would I print the results of a form submission on the same page as the form itself? Relevant hook_menu: $items['admin/content/ncbi_subsites/paths'] = array( 'title' => 'Paths', 'description' => 'Paths for a particular subsite', 'page callback' => 'ncbi_subsites_show_path_page', 'access arguments' => array( 'administer site configuration' ), 'type' => MENU_LOCAL_TASK, ); page callback: function ncbi_subsites_show_path_page() { $f = drupal_get_form('_ncbi_subsites_show_paths_form'); return $f; } Form building function: function _ncbi_subsites_show_paths_form() { // bunch of code here $form['subsite'] = array( '#title' => t('Subsites'), '#type' => 'select', '#description' => 'Choose a subsite to get its paths', '#default_value' => 'Choose a subsite', '#options'=> $tmp, ); $form['showthem'] = array( '#type' => 'submit', '#value' => 'Show paths', '#submit' => array( 'ncbi_subsites_show_paths_submit'), ); return $form; } Submit function (skipped validate function for brevity) function ncbi_subsites_show_paths_submit( &$form, &$form_state ) { //dpm ( $form_state ); $subsite_name = $form_state['values']['subsite']; $subsite = new Subsite( $subsite_name ); //y own class that I use internally in this module $paths = $subsite->normalized_paths; // build list $list = theme_item_list( $paths ); } If I print that $list variable, it is exactly what I want, but I am not sure how to get it into the page with the original form page built from 'ncbi_subsites_show_path_page'. Any help is much appreciated!

    Read the article

  • Detecting file upload size on the client side?

    - by DisgruntledGoat
    I'm using PHP for file uploads. In the PHP manual it shows an example using a MAX_FILE_SIZE hidden field, saying that it will detect on the client side (i.e. the browser) whether the file is too large or not. I've just tried the example in Firefox, Chrome and IE and it doesn't work. The file is always uploaded, even if it is way larger than the specified hidden field. Incidentally, if the file is larger than MAX_FILE_SIZE then calling move_uploaded_file doesn't work, so it seems the variable is having an effect server-side, but not client-side.

    Read the article

  • How to call windows form from native cpp

    - by Jenuel
    I have an existing project using c/c++ .NET. Currently I have been given a task to create a windows form from my existing code. So i have add new project windows form application in the existing c/c++ projects.form.h, form.cpp has been automatically created. Now I am having problem to call the window from my c files. Even i could not call the form.h file from my c program. Is there any solution for this problem. Listed here is the coding.... login.c int LoginMain(int id,int task) { LoginClear(); LoginEntry(id,task); dp_in = 1; Rep(); //I WOULD LIKE TO CALL THE FORM AT THIS STAGE Cashier(); dp_in = 0; Login(); return(0); } form.cpp [STAThreadAttribute] int main(array ^args) { // Enabling Windows XP visual effects before any controls are created Application::EnableVisualStyles(); Application::SetCompatibleTextRenderingDefault(false); // Create the main window and run it Application::Run(gcnew Form1()); return 0; }

    Read the article

  • Upload files from website

    - by Andrew
    All I want to do is allow the user to browse through their folders to look for a file, select it and then press submit which saves the file to my server as well as the path to the saved file. How would someone do this? (Some sort of tutorial website would help a lot)

    Read the article

  • Image upload - Return URL

    - by Qmal
    Hello I build a script that does image uploading and resizing and it all works well, but how can I get the URL from image afterwards? I don't want my Image Source in HTML be like "../img/cat/1.png/" I want it to be like "http://MyIP/img/cat/1.png" I understand that I can just make a variable like $myHost = "http://blabla.com"; and add strip the ".." at the beginning but then it's not so good if I want to use it on other site because I need to replace this all the time. Maybe there is any other way?

    Read the article

  • Java - Display % of upload done

    - by tr-raziel
    I have a java applet for uploading files to server. I want to display the % of data sent but when I use ObjectOutputStream.write() it just writes to the buffer, does not wait until the data has actually been sent. How can I achieve this. Perhaps I need to use thread synchronization or something. Any clues would be most helpful. This is the code I'm using right now: try{ for(File file : ficheiros){ FileInputStream stream = new FileInputStream (file); int bytesRead1 = 0;; int off1 = 0; int len1 = 100000; if(file.length() < 100000) len1 = new Long(file.length()).intValue(); byte[] bytes1 = new byte[len1]; while (off1 < file.length()) { bytes1 = new byte[len1]; if((file.length() - off1) < len1){ len1 = (new Long(file.length()).intValue() - off1); bytes1 = new byte[len1]; } if((bytesRead1 = stream.read(bytes1)) != -1){ //I want this to block until all data has been sent outputToServlet.write(bytes1, 0, bytesRead1 ); System.out.println("off1: " + off1); off1 = off1 + len1; outputToServlet.flush(); } sent += len1; if(sent>totalLength) sent = (int)totalLength; updateFeedback(sent,totalLength,false);//calls method to display % } updateFeedback(-1,-1,true); } }catch(Exception e){ e.printStackTrace(); } Thanks

    Read the article

  • i get error when i try to upload a file?

    - by getaway
    I keep getting an error:Notice: Undefined index: on line 35 line 35: $handle = new Upload($_FILES['my_field']); this is my input field <input type="file" size="32" name="my_field" value="" /> I do not understand this error, thanks!!! EDIT: <form name="upload" id="upload" enctype="multipart/form-data" method="post" action="actions/upload.php" /> <p><input type="file" size="32" name="my_field" value="" /></p> <p class="button"><input type="hidden" name="action" value="image" /> <br> <input style="margin-left:224px;" type="submit" name="submit" value="upload" />

    Read the article

  • Creando un File Upload

    - by jaullo
    Para iniciar hablaremos un poco sobre el control File Upload, de esta forma daremos una idea general de que es y como trabaja. El File Upload es un control de asp.net que permite que los usuarios seleccionen un archivo de cualquier ubicación en el equipo y lo suban a un directorio predeterminado a traves de una página asp.net. En principio este control esta limitado para no permitir subir archivos de mas de 4 MB. Sin embargo, desde el webconfig de nuestra aplicacón podremos cambiar ese valor, ya sea para aumentarlo o bien para disminuirlo. Nuestro ejemplo, se enfocará en crear un webcontrol que permita seleccionar un archivo y guardarlo, asi que empecemos. Lo primero será agregar a nuestra página un webcontrol que llamaremos Upload.ascx Posteriormente en nuestro webcontrol, agregamos el siguiente código: <table style="width: 100%">         <tr>             <td colspan="3">             <div align="center">                  <asp:Label ID="Label1" runat="server" Text="File Upload"></asp:Label>              </div>             </td>                    </tr>         <tr>             <td style="width: 456px" rowspan="2">                                                             &nbsp;</td>             <td style="width: 386px">                                <div align="center">                         <asp:FileUpload ID="FileUpload1" runat="server" Height="24px" Width="243px" />                         <span id="Span1" runat="server" />                            </div>                      </td>             <td rowspan="2">                                                             </td>         </tr>         <tr>             <td style="width: 386px">                 <div align="center">                      <asp:ImageButton Id="btnupload" runat="server" OnClick="btnupload_Click"                     ImageUrl="~/Styles/img/upload.png" style="text-align: center" />           </div>                  </td>         </tr>         <tr>             <td colspan="3">                 &nbsp;</td>         </tr>     </table>  De esta forma nuestro control deberá verse algo así   Por último en el code behin de nuestro control agregamos el código a nuestro boton, el cual será el encargado de leer el archivo que se encuentra en el File Upload y guardarlo en la ruta especificada.  Protected Sub btnupload_Click(ByVal sender As Object, ByVal e As System.Web.UI.ImageClickEventArgs) Handles btnupload.Click         If FileUpload1.HasFile Then             Dim fileExt As String             fileExt = System.IO.Path.GetExtension(FileUpload1.FileName)             If (fileExt = ".exe") Then                 Label1.Text = "You can´t upload .exe file!"             Else                 Try                     FileUpload1.SaveAs(decrpath & _                        FileUpload1.FileName)                     Label1.Text = "File name: " & _                       FileUpload1.PostedFile.FileName & "<br>" & _                       "File Size: " & _                       FileUpload1.PostedFile.ContentLength & " kb<br>" & _                       "Content type: " & _                       FileUpload1.PostedFile.ContentType                 Catch ex As Exception                     Label1.Text = "ERROR: " & ex.Message.ToString()                 End Try             End If         Else             Label1.Text = "You have not specified a file!"         End If            End Sub   Como vemos en el código anterior tambien hemos agregado otros elementos los cuales nos dirán el nombre del archivo, el tipo de contenido y el tamaño en kb una vez que el archivo ha sido súbido al servidor. Por último deben tomar en cuenta que decrpath es la ruta en donde será subido el archivo, la cual deben variar a su gusto.

    Read the article

  • php image upload library

    - by Tchalvak
    I'm looking to NOT reinvent the wheel with an image upload system in php. Is there a good standalone library that allows for: Creating the system for uploading, renaming, and perhaps resizing images. Creating the front-end, html and/or javascript for the form upload parts of image upload. I've found a lot of image gallery systems, but no promising standalone system & ui libraries, ready for insertion into custom php.

    Read the article

  • how to upload & preview multiple images at single input and store in to php mysql [closed]

    - by Nilesh Sonawane
    This is nilesh , i am newcomer in this field , i need the script for when i click the upload button then uploaded images it should preview and store into db like wise i want to upload 10 images at same page using php mysql . #div { border:3px dashed #CCC; width:500px; min-height:100px; height:auto; text-align:center: } Multi-Images Uploader '.$f.''; } } } ? </div> <br> <font color='#3d3d3d' size='small'>By: Ahmed Hussein</font> this script select multiple images and then uplod , but i need to upload at a time only one image which preview and store into database like wise min 10 image user can upload .......

    Read the article

< Previous Page | 11 12 13 14 15 16 17 18 19 20 21 22  | Next Page >