Search Results

Search found 76075 results on 3043 pages for 'file path'.

Page 143/3043 | < Previous Page | 139 140 141 142 143 144 145 146 147 148 149 150  | Next Page >

  • php file path of server

    - by Jacksta
    I am trying to get this script to work. it opens up a directry and lists the files in the directory. I have copied this code from somewhere else and the problem is that this php file is hosted on an apache server not my localhost. what is the correct $dir_name = "c:/"; to use? The file is in this directory /domains/domainxxxx.com.au/public_html/lsitfiles.php so would I use domainxxxx.com.au/public_html/lsitfiles.php ? <?php $dir_name = "c:/"; $dir = opendir($dir_name); $file_list = "<ul>"; while ($file_name = readdir($dir)) { if(($file_name != ".") && (file_name != "..")) { $file_list .= "<li>$file_name"; } } $file_list.= "<ul>"; closedir($dir); ?> <HTML> <BODY> <p>Files in: <? echo "$dir_name"; ?></p> <? echo "$file_list"; ?> </BODY> </HTML>

    Read the article

  • Specifiy package path for Androd:entries

    - by Priyank
    Hi. I am using following code in preferences page in android to show a list of items. The list and values are located in a file at location "app/res/xml/time.xml" <ListPreference android:title="Time unit list" android:summary="Select the time unit" android:dependency="Main_Option" android:key="listPref" android:defaultValue="1" android:entries="?xml:time/timet" android:entryValues="@xml:time/timet_values" /> The code for the time.xml is as follow: <?xml version="1.0" encoding="utf-8"?> <resources> <string-array name="timet"> <item>seconds</item> <item>minutes</item> <item>hours</item> </string-array> <string-array name="timet_values"> <item>3600</item> <item>60</item> <item>1</item> </string-array> </resources> I am not able to reference these values in my preference xml file. (The code snippet above). It gives an error. How can I specify packaged path for the List preferences entry and entry_values Any help is appreciated. Cheers

    Read the article

  • How to define custom path to Interop *.dll

    - by NoviceAndNovice
    Well, I have an ActiveX (*.ocx) component, and i use it in a managed C++/CLI project: write a managed wrapper around ActiveX component[ NET has a great Interop services : provides me genarated dll so i can easily use it in my managed code] The problem is that Visual Studio (2008) automatically copy the generated Interop *.dll to the directory where my *.exe file stay.But i want put all my genarated Interop *.dll to a folder ... Suppose My directory structure is so: D:\MyProject\Output\MyProject.exe //My mamanged exe D:\MyProject\Output\Interop.XXXLib.1.0.dll // *Interop .dll I want to put Interop.XXXLib.1.0.dll into new folder D:\MyProject\Output\Interops and use it from that directory...How Can i do it? Best Wishes PS: What I found so far was using using codeBase/ probing tags in my app.config file such as <?xml version="1.0"?> <configuration> <runtime> <assemblyBinding xmlns="urn:schemas-microsoft-com.asm.v1"> <probing privatePath="Interops" /> </assemblyBinding> </runtime> </configuration> But i did not work in C++/CLI

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.)

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • creating objects from trivial graph format text file. java. dijkstra algorithm.

    - by user560084
    i want to create objects, vertex and edge, from trivial graph format txt file. one of programmers here suggested that i use trivial graph format to store data for dijkstra algorithm. the problem is that at the moment all the information, e.g., weight, links, is in the sourcecode. i want to have a separate text file for that and read it into the program. i thought about using a code for scanning through the text file by using scanner. but i am not quite sure how to create different objects from the same file. could i have some help please? the file is v0 Harrisburg v1 Baltimore v2 Washington v3 Philadelphia v4 Binghamton v5 Allentown v6 New York # v0 v1 79.83 v0 v5 81.15 v1 v0 79.75 v1 v2 39.42 v1 v3 103.00 v2 v1 38.65 v3 v1 102.53 v3 v5 61.44 v3 v6 96.79 v4 v5 133.04 v5 v0 81.77 v5 v3 62.05 v5 v4 134.47 v5 v6 91.63 v6 v3 97.24 v6 v5 87.94 and the dijkstra algorithm code is Downloaded from: http://en.literateprograms.org/Special:Downloadcode/Dijkstra%27s_algorithm_%28Java%29 */ import java.util.PriorityQueue; import java.util.List; import java.util.ArrayList; import java.util.Collections; class Vertex implements Comparable<Vertex> { public final String name; public Edge[] adjacencies; public double minDistance = Double.POSITIVE_INFINITY; public Vertex previous; public Vertex(String argName) { name = argName; } public String toString() { return name; } public int compareTo(Vertex other) { return Double.compare(minDistance, other.minDistance); } } class Edge { public final Vertex target; public final double weight; public Edge(Vertex argTarget, double argWeight) { target = argTarget; weight = argWeight; } } public class Dijkstra { public static void computePaths(Vertex source) { source.minDistance = 0.; PriorityQueue<Vertex> vertexQueue = new PriorityQueue<Vertex>(); vertexQueue.add(source); while (!vertexQueue.isEmpty()) { Vertex u = vertexQueue.poll(); // Visit each edge exiting u for (Edge e : u.adjacencies) { Vertex v = e.target; double weight = e.weight; double distanceThroughU = u.minDistance + weight; if (distanceThroughU < v.minDistance) { vertexQueue.remove(v); v.minDistance = distanceThroughU ; v.previous = u; vertexQueue.add(v); } } } } public static List<Vertex> getShortestPathTo(Vertex target) { List<Vertex> path = new ArrayList<Vertex>(); for (Vertex vertex = target; vertex != null; vertex = vertex.previous) path.add(vertex); Collections.reverse(path); return path; } public static void main(String[] args) { Vertex v0 = new Vertex("Nottinghill_Gate"); Vertex v1 = new Vertex("High_Street_kensignton"); Vertex v2 = new Vertex("Glouchester_Road"); Vertex v3 = new Vertex("South_Kensignton"); Vertex v4 = new Vertex("Sloane_Square"); Vertex v5 = new Vertex("Victoria"); Vertex v6 = new Vertex("Westminster"); v0.adjacencies = new Edge[]{new Edge(v1, 79.83), new Edge(v6, 97.24)}; v1.adjacencies = new Edge[]{new Edge(v2, 39.42), new Edge(v0, 79.83)}; v2.adjacencies = new Edge[]{new Edge(v3, 38.65), new Edge(v1, 39.42)}; v3.adjacencies = new Edge[]{new Edge(v4, 102.53), new Edge(v2, 38.65)}; v4.adjacencies = new Edge[]{new Edge(v5, 133.04), new Edge(v3, 102.53)}; v5.adjacencies = new Edge[]{new Edge(v6, 81.77), new Edge(v4, 133.04)}; v6.adjacencies = new Edge[]{new Edge(v0, 97.24), new Edge(v5, 81.77)}; Vertex[] vertices = { v0, v1, v2, v3, v4, v5, v6 }; computePaths(v0); for (Vertex v : vertices) { System.out.println("Distance to " + v + ": " + v.minDistance); List<Vertex> path = getShortestPathTo(v); System.out.println("Path: " + path); } } } and the code for scanning file is import java.util.Scanner; import java.io.File; import java.io.FileNotFoundException; public class DataScanner1 { //private int total = 0; //private int distance = 0; private String vector; private String stations; private double [] Edge = new double []; /*public int getTotal(){ return total; } */ /* public void getMenuInput(){ KeyboardInput in = new KeyboardInput; System.out.println("Enter the destination? "); String val = in.readString(); return val; } */ public void readFile(String fileName) { try { Scanner scanner = new Scanner(new File(fileName)); scanner.useDelimiter (System.getProperty("line.separator")); while (scanner.hasNext()) { parseLine(scanner.next()); } scanner.close(); } catch (FileNotFoundException e) { e.printStackTrace(); } } public void parseLine(String line) { Scanner lineScanner = new Scanner(line); lineScanner.useDelimiter("\\s*,\\s*"); vector = lineScanner.next(); stations = lineScanner.next(); System.out.println("The current station is " + vector + " and the destination to the next station is " + stations + "."); //total += distance; //System.out.println("The total distance is " + total); } public static void main(String[] args) { /* if (args.length != 1) { System.err.println("usage: java TextScanner2" + "file location"); System.exit(0); } */ DataScanner1 scanner = new DataScanner1(); scanner.readFile(args[0]); //int total =+ distance; //System.out.println(""); //System.out.println("The total distance is " + scanner.getTotal()); } }

    Read the article

  • Why does this IF statement fail?

    - by ChosenOne
    If variable path is empty, and editor.Text is not empty, the SaveFileDialog should be displayed. Now, why on earth is this damn thing failing??? I have tried this with many different variations of code with the same result: FAIL: if(path.Length >= 1) // path contains a path. Save changes instead of creating NEW file. { File.WriteAllText(path, content); } else { // no path defined. Create new file and write to it. using(SaveFileDialog saver = new SaveFileDialog()) { if(saver.ShowDialog() == DialogButtons.OK) { File.WriteAllText(saver.Filename, content); } } } At the top of code file I have: path = String.Empty; So why the heck it this failing every single time, even after trying all of the below variations? if(path.Length > 1) // path contains a path. Save changes instead of creating NEW file. { File.WriteAllText(path, content); } else { // no path defined. Create new file and write to it. using(SaveFileDialog saver = new SaveFileDialog()) { if(saver.ShowDialog() == DialogButtons.OK) { File.WriteAllText(saver.Filename, content); } } } and if(String.IsNullOrEmpty(path)) // path contains a path. Save changes instead of creating NEW file. { File.WriteAllText(path, content); } else { // no path defined. Create new file and write to it. using(SaveFileDialog saver = new SaveFileDialog()) { if(saver.ShowDialog() == DialogButtons.OK) { File.WriteAllText(saver.Filename, content); } } } and if(String.IsNullOrWhiteSpace(path)) // path contains a path. Save changes instead of creating NEW file. { File.WriteAllText(path, content); } else { // no path defined. Create new file and write to it. using(SaveFileDialog saver = new SaveFileDialog()) { if(saver.ShowDialog() == DialogButtons.OK) { File.WriteAllText(saver.Filename, content); } } } This is making me very angry. How could this fail? Setting a break point reveals that path is definitely null/"".

    Read the article

  • 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

    Read the article

  • 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(); } } }

    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

  • SSIS - Access Denied with UNC paths - The file name is a device or contains invalid characters

    - by simonsabin
    I spent another day tearing my hair out yesterday trying to resolve an issue with SSIS packages runnning in SQLAgent (not got much left at the moment, maybe I should contact the SSIS team for a wig). My situation was that I am deploying packages to a development server, and to provide isolation I was running jobs with a proxy account that only had access to the development servers. Proxies are an awesome feature and mean that you should never have to "just run the job as sysadmin". The issue I was facing was that the job step was failing. The job step was a simple execution of the package.The following errors appeared in my log file. I always check the "Log step output in history" for a job step, this ensures you get all the output from the command that you run. I'll blog about this later. If looking at the output in sysdtslog90 then you will have an entry with datacode -1073573533 and error message File or directory "<filename>" represented by connection "<connection>" does not exist.  Not exactly helpful. If you get the output from the console then you will also get these errors. 0xC0202070 "The file name property is not valid. The file name is a device or contains invalid characters." 0xC001401E "specified in the connection was not valid." It appears this error is due to the use of a UNC path and the account runnnig the package not having access to all the folders in the path. Solution To solve this you need to ensure that the proxy account has access to ALL folders in the path you are accessing. To check this works, logon as the relevant proxy user, or run a command window as the specified user. Then try and do net use \\server\share and then do a dir for each folder in the path and check you have access. If these work and you still have the problem then you have some other problem, sorry. The following are posts on experts exchange that also discuss this,http://www.experts-exchange.com/Microsoft/Development/MS-SQL-Server/SSIS/Q_24056047.htmlhttp://www.experts-exchange.com/Microsoft/Development/MS-SQL-Server/SSIS/Q_23968903.html This blog had a post about it being a 64 bit issue. That definitely wasn't the issue for me as I was on a 32 bit server http://blogs.perkinsconsulting.com/post/64-bit-SQL-Server-2005-SSIS-and-UNC-paths-Part-2.aspx  

    Read the article

  • 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.

    Read the article

  • Oracle EBS R12.1.1 system09.dbf file corruption Bug

    - by longchun.zhu
    ??????,??????????????????,???? ?????????????.. ???????????,??????,???????????? After Installing or Upgrading Perform the following steps after installing or upgrading to Release 12.1.1 and before allowing users to access the system. Manually fix database dbf file If you installed 12.1.1 with a startCD of 12.1.1.9 or earlier (see Oracle Applications Release Notes, Release 12.1.1 My Oracle Support Document 798258.1), you must run the following sql commands to fix a particular corrupted dbf file: $ sqlplus/nolog sql connect / as sysdba sql alter database datafile '[full path of system09.dbf]' resize 1000M; sql alter database datafile '[full path of system09.dbf]' resize 1500M;

    Read the article

  • Is using dirname(__FILE__) a good practice?

    - by webose
    Looking at the code of Joomla I see that in the first line of the index, it defines the base path of installation with dirname(__FILE__). Is this a possible risk for the site? If a non controlled error message show the internal path of the Joomla directory, because of, for example a failed include, can it be used to perform some kind of attack to the site? If yes, is it convenient to use this function?

    Read the article

  • an error in .profile file prevents the system from booting?

    - by nafaa
    I tryied to set the JAVA_HOME environmet variable in the ~/.profile file and I made a mistake JAVA_HOME="/usr/lib/jvm/jdk1.6.0_37" PATH=$JAVA_HOME:PATH the error is in the PATH referencing. I put PATH rather than $PATH. this prevents me from login. I tried to edit the file using the recovery mode but it says that the file system is read only. so any suggestions how to deal with this issue. thanks

    Read the article

  • an error in .profile ubuntu file prevents the system from booting?

    - by nafaa
    I tryied to set the JAVA_HOME environmet variable in the ~/.profile file and I made a mistake JAVA_HOME="/usr/lib/jvm/jdk1.6.0_37" PATH=$JAVA_HOME:PATH the error is in the PATH referencing. I put PATH rather than $PATH. this prevents me from login. I tried to edit the file using the recovery mode but it says that the file system is read only. so any suggestions how to deal with this issue. thanks

    Read the article

  • Python progression path - From apprentice to guru

    - by Morlock
    Hi all, I've been learning, working, and playing with Python for a year and a half now. As a biologist slowly making the turn to bio-informatics, this language has been a the very core of all the major contributions I have made in the lab. (bash and R scripts have helped some too. My C++ capabilities are very not functional yet). I more or less fell in love with the way Python permits me to express beautiful solutions and also with the semantics of the language that allows such a natural flow from thoughts to workable code. What I would like to know from you is your answer to a kind of question I have seldom seen in this or other forums. Let me sum up what I do NOT want to ask first ;) I don't want to know how to QUICKLY learn Python Nor do I want to find out the best way to get acquainted with the language Finally, I don't want to know a 'one trick that does it all' approach. What I do want to know your opinion about, is: What are the steps YOU would recommend to a Python journeyman, from apprenticeship to guru status (feel free to stop wherever your expertise dictates it), in order that one IMPROVES CONSTANTLY, becoming a better and better Python coder, one step at a time. The kind of answers I would enjoy (but feel free to surprise the readership :P ), is formatted more or less like this: Read this (eg: python tutorial), pay attention to that kind of details Code for so manytime/problems/lines of code Then, read this (eg: this or that book), but this time, pay attention to this Tackle a few real-life problems Then, proceed to reading Y. Be sure to grasp these concepts Code for X time Come back to such and such basics or move further to... (you get the point :) This process depicts an iterative Learn/Code cycle, and I really care about knowing your opinion on what exactly one should pay attention to, at various stages, in order to progress CONSTANTLY (with due efforts, of course). If you come from a specific field of expertise, discuss the path you see as appropriate in this field. Thanks a lot for sharing your opinions and good Python coding!

    Read the article

  • Maintaining ISAPI Rewrite Path with the ASP.NET tilde (~)

    - by Adam
    My team is upgrading from ASP.NET 3.5 to ASP.NET 4.0. We are currently using Helicon ISAPI Rewrite to map http://localhost/<account-name>/default.aspx to http://localhost/<virtual-directory>/default.aspx?AccountName=<account-name> where <account-name> is a query string variable and <virtual-directory> is a virtual directory (naturally). Before the upgrade the tilde (~) resolved to http://localhost/<account-name>/... (which I want it to do) and after the upgrade the tilde resolves to http://localhost/<virtual-directory>/... which results in an error because the <account-name> query string is required. I'd like to avoid going down the road of replacing everything with relative paths because there are several features in our system that use the entire URL instead of just the relative path. For what it's worth I'm using IIS7 in Windows 7, Visual Studio 2010 with ASP.NET 4.0 and the 64 bit Helicon ISAPI Rewrite. If I switch back to the ASP.NET 3.5 version then it still works fine (leading me to believe nothing changed in IIS unless it's within the 4.0 app pool - when I switch back and forth between 3.5 and 4.0 I have to change the app pool in IIS). Any ideas? Thanks in advance!

    Read the article

< Previous Page | 139 140 141 142 143 144 145 146 147 148 149 150  | Next Page >