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  • drupal themes: .info file: how do I add more than 1 css file / js file to my theme?

    - by egarcia
    I'm creating a new Drupal theme. Until now, I only needed to include a single css file and a single js file. So my theme.info file had something like this: stylesheets[all][] = css/style.css scripts[] = js/script.js Now I must include jquery and jquery-ui in order to use a calendar date. These come with 2 new javascript files, and 1 additonal css file that I must add to the site. The calendar input form is going to be used in all pages (on a side block) so it is ok for me to load the extra css/javascript on all pages. I think the easiest thing would be to reference them on the .info file itself. At first I tried to just put them there with separate spaces: stylesheets[all][] = css/style.css css/ui-lightness/jquery-ui-1.8.1.custom.css scripts[] = js/jquery-1.4.2.min.js js/jquery-ui-1.8.1.custom.min.js js/reservations.js I emptied drupal's cache and... none of them loaded. I then tried separating each file with a comma, and flushing the cache again. Same result. I've browsed some drupal pages, but could not find how to add several javascript/css files on one theme (they always seem to add just 1 of each). So, how do I include several css/javascript files on the .info file?

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  • PHP: parse $_FILES[] data in multidimesional array

    - by superUntitled
    Example form here: http://jsfiddle.net/superuntitled/uaTtx/1/ I have a form that allows for dynamic duplication of the form fields. The form allows for file uploads and text input, so the data is sent in both $_POST and $_FILES arrays. The the initial set of inputs look like this: <input type="text" name="question[1][text]" /> <input type="file" name="question[1][file]" /> <input type="text" class="a" name="answer[1][text][]" /> <input type="file" name="answer[1][file][]" /> When duplicated the fields are incremented, they look like this: <input type="text" name="question[2][text]" /> <input type="file" name="question[2][file]" /> <input type="text" class="a" name="answer[2][text][]" /> <input type="file" name="answer[2][file][]" /> To complicate matters, the "answer" form fields can also be duplicated (thus the [] at the end of the 'answer' name array. How can I parse the posted $_FILES array? I have tried something like this: foreach ($_FILES['question'] as $p_num) { echo $p_num['file']['name']; foreach ($_FILES['answer'] as $a_num) { echo $a_num['file']['name']; } } but I get an "Undefined index: file... " error. How can I parse out the posted values.

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  • Reading and writing to files simultaneously?

    - by vipersnake005
    Moved the question here. Suppose, I want to store 1,000,000,000 integers and cannot use my memory. I would use a file(which can easily handle so much data ). How can I let it read and write and the same time. Using fstream file("file.txt', ios::out | ios::in ); doesn't create a file, in the first place. But supposing the file exists, I am unable to use to do reading and writing simultaneously. WHat I mean is this : Let the contents of the file be 111111 Then if I run : - #include <fstream> #include <iostream> using namespace std; int main() { fstream file("file.txt",ios:in|ios::out); char x; while( file>>x) { file<<'0'; } return 0; } Shouldn't the file's contents now be 101010 ? Read one character and then overwrite the next one with 0 ? Or incase the entire contents were read at once into some buffer, should there not be atleast one 0 in the file ? 1111110 ? But the contents remain unaltered. Please explain. Thank you.

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  • Windows 7: Can't see ISO file in C:\

    - by cbp
    I used DVD shrink to create an ISO file and saved it into C:\ The ISO file is visible with some programs but not with others. The file is not hidden as far as I am aware. But it cannot be seen by Windows Explorer, DVD Decrypter or a bunch of other programs. If I search for the file using Windows 7's Start Menu search tool, I can see the file and I can right click and select Properties. The Properties window appears OK, but if I try to change tabs on the property window, I receive an error message as though the file is not there. DVD Shrink can still open the file OK. I can also find the file using Agent Ransack (a file searching tool), but then I cannot open it. What gives?

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  • optimizing file share performance on Win2k8?

    - by Kirk Marple
    We have a case where we're accessing a RAID array (drive E:) on a Windows Server 2008 SP2 x86 box. (Recently installed, nothing other than SQL Server 2005 on the server.) In one scenario, when directly accessing it (E:\folder\file.xxx) we get 45MBps throughput to a video file. If we access the same file on the same array, but through UNC path (\server\folder\file.xxx) we get about 23MBps throughput with the exact same test. Obviously the second test is going through more layers of the stack, but that's a major performance hit. What tuning should we be looking at for making the UNC path be closer in performance to the direct access case? Thanks, Kirk (corrected: it is CIFS not SMB, but generalized title to 'file share'.) (additional info: this happens during the read from a single file, not an issue across multiple connections. the file is on the local machine, but exposed via file share. so client and file server are both same Windows 2008 server.)

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  • undelete big files - mission impossible?

    - by johnrembo
    Hi, I've accidentaly deleted outlook.pst (6.7GB) file, while there was only 400MB free space left on primary NTFS partition (winxp). I've tried several recovery tools to get this file back. "Ontrack Easy Recovery Pro" found 0 pst files (complete scan mode), while "Recover My Files" in sector scan mode found 5 pst's, but 4 of them of sizes from 3 to 28 KB, while the 5th one - 1Gb. I've managed to succesfuly recover 1Gb pst file, which was 1 year old copy (the one used after the latest windows reinstall). Now, I'm frustrated and confused Why 1 year old file was succesfuly recovered if there were only 400MB left on primary partition? Where's 6.7GB file gone? I did some reading (i.e. here), and it seems that there's almost no probability to retrieve the file I'm looking for, but wait - none of recovery tools i've used found zero-sized pst file, moreover - if due to fragmentation a file might be corrupted - we could use scanpst.exe to fix some errors and survive with 10 or 100 emails missing - whatever. Could you please recommend some more sophisticated recovery tools for this particular task? Appretiate your help - thanks in advance

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  • Is it possible to download extremely large files intelligently or in parts via SSH from Linux to Windows?

    - by Andrew
    I have a ~35 GB file on a remote Linux Ubuntu server. Locally, I am running Windows XP, so I am connecting to the remote Linux server using SSH (specifically, I am using a Windows program called SSH Secure Shell Client version 3.3.2). Although my broadband internet connection is quite good, my download of the large file often fails with a Connection Lost error message. I am not sure, but I think that it fails because perhaps my internet connection goes out for a second or two every several hours. Since the file is so large, downloading it may take 4.5 to 5 hours, and perhaps the internet connection goes out for a second or two during that long time. I think this because I have successfully downloaded files of this size using the same internet connection and the same SSH software on the same computer. In other words, sometimes I get lucky and the download finishes before the internet connection drops for a second. Is there any way that I can download the file in an intelligent way -- whereby the operating system or software "knows" where it left off and can resume from the last point if a break in the internet connection occurs? Perhaps it is possible to download the file in sections? Although I do not know if I can conveniently split my file into multiple files -- I think this would be very difficult, since the file is binary and is not human-readable. As it is now, if the entire ~35 GB file download doesn't finish before the break in the connection, then I have to start the download over and overwrite the ~5-20 GB chunk that was downloaded locally so far. Do you have any advice? Thanks.

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  • Command line switching

    - by Larry
    I have read through some suggestions but I am just not technical enough to get this I think. I am a CAD designer and each file has 5 files associated with it. I have 3 sets of 5 files, and each set needs to go into its own zip file, placed on a separate server. For example: "C:\Program Files\7-zip\7z.exe" a file1.zip "O:\server2\map files\BC\BC.d*"-0 "C:\Program Files\7-zip\7z.exe" a file2.zip "O:\server2\map files\BC\ON.d*"-0 "C:\Program Files\7-zip\7z.exe" a file3.zip "O:\server2\map files\BC\AB.d*"-0 and I am in directory "S:\server\map files\provinces" (for example). These lines run within an existing batch file and by the time it reaches the 3 lines above, it's in the S: directory sample above. So it's looking on my pc for the 7-zip program, creating 3 zip file names which it does, but places those zip files on a separate server which it doesn't and the first zip file also includes all the other 10 files, the second zip file the same plus the first zip file, and the third the same with the other two zip files making me think the code isn't recognizing the part after file1.zip where I am trying to tell it what files to include and where to place the zip files. Ultimately, I want to either have the system create a new zip file if the old one was deleted, or copy the new files into the existing zip and overwrite any older files, and for these zip files to be placed in a separate location which is where we share our files with other personnel from within our company. The S: drive is for all originals, and O: is for sharing. Is there a list of all switching options with many different samples?

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  • how to write or what is the concept behind the file unlocker program

    - by Jach Many
    Recently i was trying to delete a file thinking that i had closed the program which is manipulating the file but it did not delete because the program was still running. I misunderstood that file as an unwanted file. So i used the file unlocker program to find which process is manipulating that file. That program really worked well by showing the process which was handling that file. and that file was http://download.cnet.com/Unlocker/3000-2248_4-10493998.html. What i want to know is i would like to write a program in win32 C or .net to mimic the same process. Just to find which process is handling which file. and if possible to close it. Or i want to know the concept behind that. I know this cannot be explained in a few paragraphs yet if i could get some references or external links to references then that could be nice.

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  • Pass string between two threads in java

    - by geeta
    I have to search a string in a file and write the matched lines to another file. I have a thread to read a file and a thread to write a file. I want to send the stringBuffer from read thread to write thread. Please help me to pass this. I amm getting null value passed. write thread: class OutputThread extends Thread{ /****************** Writes the line with search string to the output file *************/ Thread runner1,runner; File Out_File; public OutputThread() { } public OutputThread(Thread runner,File Out_File) { runner1 = new Thread(this,"writeThread"); // (1) Create a new thread. this.Out_File=Out_File; this.runner=runner; runner1.start(); // (2) Start the thread. } public void run() { try{ BufferedWriter bufferedWriter=new BufferedWriter(new FileWriter(Out_File,true)); System.out.println("inside write"); synchronized(runner){ System.out.println("inside wait"); runner.wait(); } System.out.println("outside wait"); // bufferedWriter.write(line.toString()); Buffer Buf = new Buffer(); bufferedWriter.write(Buf.buffers); System.out.println(Buf.buffers); bufferedWriter.flush(); } catch(Exception e){ System.out.println(e); e.printStackTrace(); } } } Read Thraed: class FileThread extends Thread{ Thread runner; File dir; String search_string,stats; File Out_File,final_output; StringBuffer sb = new StringBuffer(); public FileThread() { } public FileThread(CountDownLatch latch,String threadName,File dir,String search_string,File Out_File,File final_output,String stats) { runner = new Thread(this, threadName); // (1) Create a new thread. this.dir=dir; this.search_string=search_string; this.Out_File=Out_File; this.stats=stats; this.final_output=final_output; this.latch=latch; runner.start(); // (2) Start the thread. } public void run() { try{ Enumeration entries; ZipFile zipFile; String source_file_name = dir.toString(); File Source_file = dir; String extension; OutputThread out = new OutputThread(runner,Out_File); int dotPos = source_file_name.lastIndexOf("."); extension = source_file_name.substring(dotPos+1); if(extension.equals("zip")) { zipFile = new ZipFile(source_file_name); entries = zipFile.entries(); while(entries.hasMoreElements()) { ZipEntry entry = (ZipEntry)entries.nextElement(); if(entry.isDirectory()) { (new File(entry.getName())).mkdir(); continue; } searchString(runner,entry.getName(),new BufferedInputStream(zipFile.getInputStream(entry)),Out_File,final_output,search_string,stats); } zipFile.close(); } else { searchString(runner,Source_file.toString(),new BufferedInputStream(new FileInputStream(Source_file)),Out_File,final_output,search_string,stats); } } catch(Exception e){ System.out.println(e); e.printStackTrace(); } } /********* Reads the Input Files and Searches for the String ******************************/ public void searchString(Thread runner,String Source_File,BufferedInputStream in,File output_file,File final_output,String search,String stats) { int count = 0; int countw = 0; int countl=0; String s; String[] str; String newLine = System.getProperty("line.separator"); try { BufferedReader br2 = new BufferedReader(new InputStreamReader(in)); //OutputFile outfile = new OutputFile(); BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(output_file,true)); Buffer Buf = new Buffer(); //StringBuffer sb = new StringBuffer(); StringBuffer sb1 = new StringBuffer(); while((s = br2.readLine()) != null ) { str = s.split(search); count = str.length-1; countw += count; if(s.contains(search)){ countl++; sb.append(s); sb.append(newLine); } if(countl%100==0) { System.out.println("inside count"); Buf.setBuffers(sb.toString()); sb.delete(0,sb.length()); System.out.println("outside notify"); synchronized(runner) { runner.notify(); } //outfile.WriteFile(sb,bufferedWriter); //sb.delete(0,sb.length()); } } } synchronized(runner) { runner.notify(); } br2.close(); in.close(); if(countw == 0) { System.out.println("Input File : "+Source_File ); System.out.println("Word not found"); System.exit(0); } else { System.out.println("Input File : "+Source_File ); System.out.println("Matched word count : "+countw ); System.out.println("Lines with Search String : "+countl); System.out.println("Output File : "+output_file.toString()); System.out.println(); } } catch(Exception e){ System.out.println(e); e.printStackTrace(); } } }

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  • Java Client-Server problem when sending multiple files

    - by Jim
    Client public void transferImage() { File file = new File(ServerStats.clientFolder); String[] files = file.list(); int numFiles = files.length; boolean done = false; BufferedInputStream bis; BufferedOutputStream bos; int num; byte[] byteArray; long count; long len; Socket socket = null ; while (!done){ try{ socket = new Socket(ServerStats.imgServerName,ServerStats.imgServerPort) ; InputStream inStream = socket.getInputStream() ; OutputStream outStream = socket.getOutputStream() ; System.out.println("Connected to : " + ServerStats.imgServerName); BufferedReader inm = new BufferedReader(new InputStreamReader(inStream)); PrintWriter out = new PrintWriter(outStream, true /* autoFlush */); for (int itor = 0; itor < numFiles; itor++) { String fileName = files[itor]; System.out.println("transfer: " + fileName); File sentFile = new File(fileName); len = sentFile.length(); len++; System.out.println(len); out.println(len); out.println(sentFile); //SENDFILE bis = new BufferedInputStream(new FileInputStream(fileName)); bos = new BufferedOutputStream(socket.getOutputStream( )); byteArray = new byte[1000000]; count = 0; while ( count < len ){ num = bis.read(byteArray); bos.write(byteArray,0,num); count++; } bos.close(); bis.close(); System.out.println("file done: " + itor); } done = true; }catch (Exception e) { System.err.println(e) ; } } } Server public static void main(String[] args) { BufferedInputStream bis; BufferedOutputStream bos; int num; File file = new File(ServerStats.serverFolder); if (!(file.exists())){ file.mkdir(); } try { int i = 1; ServerSocket socket = new ServerSocket(ServerStats.imgServerPort); Socket incoming = socket.accept(); System.out.println("Spawning " + i); try { try{ if (!(file.exists())){ file.mkdir(); } InputStream inStream = incoming.getInputStream(); OutputStream outStream = incoming.getOutputStream(); BufferedReader inm = new BufferedReader(new InputStreamReader(inStream)); PrintWriter out = new PrintWriter(outStream, true /* autoFlush */); String length2 = inm.readLine(); System.out.println(length2); String filename = inm.readLine(); System.out.println("Filename = " + filename); out.println("ACK: Filename received = " + filename); //RECIEVE and WRITE FILE byte[] receivedData = new byte[1000000]; bis = new BufferedInputStream(incoming.getInputStream()); bos = new BufferedOutputStream(new FileOutputStream(ServerStats.serverFolder + "/" + filename)); long length = (long)Integer.parseInt(length2); length++; long counter = 0; while (counter < length){ num = bis.read(receivedData); bos.write(receivedData,0,num); counter ++; } System.out.println(counter); bos.close(); bis.close(); File receivedFile = new File(filename); long receivedLen = receivedFile.length(); out.println("ACK: Length of received file = " + receivedLen); } finally { incoming.close(); } } catch (IOException e){ e.printStackTrace(); } } catch (IOException e1){ e1.printStackTrace(); } } The code is some I found, and I have slightly modified it, but I am having problems transferring multiple images over the server. Output on Client: run ServerQueue.Client Connected to : localhost transfer: Picture 012.jpg 1312743 java.lang.ArrayIndexOutOfBoundsException Connected to : localhost transfer: Picture 012.jpg 1312743 Cant seem to get it to transfer multiple images. But bothsides I think crash or something because the file never finishes transfering

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  • 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.

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  • How to Upload a file from client to server using OFBIZ?

    - by SIVAKUMAR.J
    I'm new to ofbiz so try to keep your answer as simple as possibly. If you can give examples that would be kind. My problem is I created a project inside the ofbiz/hot-deploy folder namely productionmgntSystem. Inside the folder ofbiz\hot-deploy\productionmgntSystem\webapp\productionmgntSystem I created a file app_details_1.ftl. The following are the code of this file <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title>Insert title here</title> <script TYPE="TEXT/JAVASCRIPT" language=""JAVASCRIPT"> function uploadFile() { //alert("Before calling upload.jsp"); window.location='<@ofbizUrl>testing_service1</@ofbizUrl>' } </script> </head> <!-- <form action="<@ofbizUrl>testing_service1</@ofbizUrl>" enctype="multipart/form-data" name="app_details_frm"> --> <form action="<@ofbizUrl>logout1</@ofbizUrl>" enctype="multipart/form-data" name="app_details_frm"> <center style="height: 299px; "> <table border="0" style="height: 177px; width: 788px"> <tr style="height: 115px; "> <td style="width: 103px; "> <td style="width: 413px; "><h1>APPLICATION DETAILS</h1> <td style="width: 55px; "> </tr> <tr> <td style="width: 125px; ">Application name : </td> <td> <input name="app_name_txt" id="txt_1" value=" " /> </td> </tr> <tr> <td style="width: 125px; ">Excell sheet &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;: </td> <td> <input type="file" name="filename"/> </td> </tr> <tr> <td> <!-- <input type="button" name="logout1_cmd" value="Logout" onclick="logout1()"/> --> <input type="submit" name="logout_cmd" value="logout"/> </td> <td> <!-- <input type="submit" name="upload_cmd" value="Submit" /> --> <input type="button" name="upload1_cmd" value="Upload" onclick="uploadFile()"/> </td> </tr> </table> </center> </form> </html> the following coding is present in the file ofbiz\hot-deploy\productionmgntSystem\webapp\productionmgntSystem\WEB-INF\controller.xml ...... ....... ........ <request-map uri="testing_service1"> <security https="true" auth="true"/> <event type="java" path="org.ofbiz.productionmgntSystem.web_app_req.WebServices1" invoke="testingService"/> <response name="ok" type="view" value="ok_view"/> <response name="exception" type="view" value="exception_view"/> </request-map> .......... ............ .......... <view-map name="ok_view" type="ftl" page="ok_view.ftl"/> <view-map name="exception_view" type="ftl" page="exception_view.ftl"/> ................ ............. ............. The following are the coding present in the file ofbiz\hot-deploy\productionmgntSystem\src\org\ofbiz\productionmgntSystem\web_app_req\WebServices1.java package org.ofbiz.productionmgntSystem.web_app_req; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import java.io.DataInputStream; import java.io.FileOutputStream; import java.io.IOException; public class WebServices1 { public static String testingService(HttpServletRequest request, HttpServletResponse response) { //int i=0; String result="ok"; System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- Start"); String contentType=request.getContentType(); System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- contentType : "+contentType); String str=new String(); // response.setContentType("text/html"); //PrintWriter writer; if ((contentType != null) && (contentType.indexOf("multipart/form-data") >= 0)) { System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) after if (contentType != null)"); try { // writer=response.getWriter(); System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - try Start"); DataInputStream in = new DataInputStream(request.getInputStream()); int formDataLength = request.getContentLength(); byte dataBytes[] = new byte[formDataLength]; int byteRead = 0; int totalBytesRead = 0; //this loop converting the uploaded file into byte code while (totalBytesRead < formDataLength) { byteRead = in.read(dataBytes, totalBytesRead,formDataLength); totalBytesRead += byteRead; } String file = new String(dataBytes); //for saving the file name String saveFile = file.substring(file.indexOf("filename=\"") + 10); saveFile = saveFile.substring(0, saveFile.indexOf("\n")); saveFile = saveFile.substring(saveFile.lastIndexOf("\\")+ 1,saveFile.indexOf("\"")); int lastIndex = contentType.lastIndexOf("="); String boundary = contentType.substring(lastIndex + 1,contentType.length()); int pos; //extracting the index of file pos = file.indexOf("filename=\""); pos = file.indexOf("\n", pos) + 1; pos = file.indexOf("\n", pos) + 1; pos = file.indexOf("\n", pos) + 1; int boundaryLocation = file.indexOf(boundary, pos) - 4; int startPos = ((file.substring(0, pos)).getBytes()).length; int endPos = ((file.substring(0, boundaryLocation)).getBytes()).length; //creating a new file with the same name and writing the content in new file FileOutputStream fileOut = new FileOutputStream("/"+saveFile); fileOut.write(dataBytes, startPos, (endPos - startPos)); fileOut.flush(); fileOut.close(); System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - try End"); } catch(IOException ioe) { System.out.println("\n\n\t*********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - Catch IOException"); //ioe.printStackTrace(); return("exception"); } catch(Exception ex) { System.out.println("\n\n\t*********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - Catch Exception"); return("exception"); } } else { System.out.println("\n\n\t********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) else part"); result="exception"; } System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- End"); return(result); } } I want to upload a file to the server. The file is get from user " tag in the "app_details_1.ftl" file & it is updated into the server by using the method "testingService(HttpServletRequest request, HttpServletResponse response)" in the class "WebServices1". But the file is not uploaded. Give me a good solution for uploading a file to the server.

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  • Error when trying to open SQL Maintenance Plan - SSMS 2008

    - by alex
    If I open SSMS on my client machine, connect to our SQL server, and try and open a maintenance plan on there, I get this error: TITLE: Microsoft SQL Server Management Studio Could not load file or assembly 'msddsp, Version=9.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a' or one of its dependencies. The system cannot find the file specified. (Microsoft.DataTransformationServices.Design) ------------------------------ BUTTONS: OK If i try the same thing directly on the server, nothing happens (no errors or anything)

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  • Outlook 2007 - Cannot start Outlook - Cannot open the Outlook window

    - by Dean Perry
    I went to open Outlook 2007 on my Windows 7 32bit machine and it came up with this error: Cannot start Microsoft Office Outlook. Cannot open the Outlook window. The set of folders cannot be opened. The information store could not be opened. I have deleted and created a new profile in Control Panel Mail but it still doesn't want to work properly but it still comes up with this message. Thanks in advance.

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  • Publisher 2007 files created on Vista 32 bit won't open on Vista 64 bit

    - by BBlack
    I created several documents in MS Office Publisher 2007 when I was using Windows Vista 32 bit version. I've recently upgraded to the 64 bit version of Vista and am now having trouble opening the files created on the previous Windows setup. When I try to open my documents, I get this warning: Publisher can not open the document. How do I go about resolving this?

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  • Create or Open an .xlsx file having >256 columns in MS Excel 2003

    - by Daredev
    I'm using Microsoft Office 2003. I have installed 'Microsoft Office Compatibility Pack for Word, Excel, Powerpoint 2007' to support new xml based formats (.docx, .xlsx, .pptx). Now given that I have installed Compatibility pack, can I create or open a Microsoft Excel 2007 file (.xlsx) having more than 256 columns in Excel 2003? If no, then how can I achieve the same. My observation: When I open a .xlsx file in Excel 2003 with compatibility, I can't see more than 256 columns (till Column IV).

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  • Outlook 2003: open email in edit mode

    - by Eleasar
    Is there some macro (or C# code) to open an email automatically in edit mode? I know i can double click an email item, go to edit and edit message - but can i do this to automatically open an email in this way? Or even better if i could show emails in edit mode in the reading pane?

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  • Java compiler error: Can't open input server /Library/InputManagers/Inquisitor

    - by unknown (yahoo)
    I am trying to compile HelloWorld in Java under Mac OS X 10.6 (Snow Leopard) and I get this compiler error: java[51692:903] Can't open input server /Library/InputManagers/Inquisitor It happens when I am using terminal command javac and when I am trying to do this in NetBeans. I was trying to open folder "Inquisitor", but I have no access to folder, even if I login as root user. What is going on?

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  • Sorted list of file names in a folder in VBA?

    - by Karsten W.
    Is there a way to get a sorted list of file names of a folder in VBA? Up to now, I arrived at Dim fso As Object Dim objFolder As Object Dim objFileList As Object Dim vFile As Variant Dim sFolder As String sFolder = "C:\Docs" Set fso = CreateObject("Scripting.FileSystemObject") Set objFolder = fso.GetFolder(sFolder) Set objFileList = objFolder.Files For Each vFile In objFileList ' do something ' Next vFile but it is crucial to be sure the processing order of the for loop is determined by the file names... Any help appreciated!

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  • Unable to open an ra_local session to URL

    - by AntonAL
    I have repository on my Windows machine, using VisualSVN + TortoiseSVN I want to use this repositories on my Mac with Versions.app I create a new Repository bookmark, choose a repository directory and after clicking "Create", get following error: Unable to open an ra_local session to URL Unable to open repository 'file://localhost/Path/to/my/repositories/MyRepo' Expected FS format '2'; found format '4' I can surf through the repository, using svn ls file://localhost/Path/to/my/repositories/MyRepo Also, when i create local repository, using Versions.app, the bookmark is created well and i can work with it Help!

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  • Emacs open files from a filename list

    - by crasic
    I have a largish tex project that is separated into several tex files. Everytime I want to work on it I open emacs and manually C-x C-f all the files that I want to work on. I was wondering if there is a way to open files (from command line) from a file containing a list of filenames, something like filelist.txt: file1.tex file2.tex file3.tex then do cat files | emacs -nw except that emacs doesn't support the command used as it doesn't like that stdin is reassigned. any ideas?

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