Search Results

Search found 68301 results on 2733 pages for 'file shredding'.

Page 247/2733 | < Previous Page | 243 244 245 246 247 248 249 250 251 252 253 254  | Next Page >

  • Accessing property file through batch script

    - by Farid
    Hi everyone, I'm trying to write a batch script, this script is responsible to launch a jar with one parameters. This parameter indicate to my jar wich property file to use in order to setup some configuration. Then the script will zip the results produced by the jar and send them to a location. But in order to set the name of the zip file I would need to be able to read the property file directly from the batch, is there a way to do so ? Thanks and regards, F

    Read the article

  • File.mkdir is not working and I can't understand why

    - by gotch4
    Hello, I've this brief snippet: String target = baseFolder.toString() + entryName; target = target.substring(0, target.length() - 1); File targetdir = new File(target); if (!targetdir.mkdirs()) { throw new Exception("Errore nell'estrazione del file zip"); } doesn't mattere if I leave the last char (that is usually a slash). It's done this way to work on both unix and windows. The path is actually obtained from the URI of the base folder. As you can see from baseFolder.toString() (baseFolder is of type URI and is correct). The base folder actually exists. I can't debug this because all I get is true or false from mkdir, no other explanations.The weird thing is that baseFolder is created as well with mkdir and in that case it works. Now I'm under windows. the value of target just before the creation of targetdir is "file:/C:/Users/dario/jCommesse/jCommesseDB" if I cut and paste it (without the last entry) in windows explore it works...

    Read the article

  • IIS Directory listing doesn't reconize .mkv files

    - by Buckley
    Hello I use the directory listing function in IIS to upload a bunch of files for friends and family to easy access and download. My problem is that .mkv files it lists but when you click it i get a 'The page cannot be found'. Ive tried relocating the file and renaming it but i get the same error each time. Why does it do this? Its only my .mkv files everything else works perfectly. Thanks in advance.

    Read the article

  • xml file reading

    - by Dilse Naaz
    How to read an xml file in silverlight using webclient. I have one task, in this the xml file that i stored in my machine will read and the content od the xml file will display on a data grid. how it will be done?

    Read the article

  • problems with unpickling a 80 megabyte file in python

    - by tipu
    I am using the pickle module to read and write large amounts of data to a file. After writing to the file a 80 megabyte pickled file, I load it in a SocketServer using class MyTCPHandler(SocketServer.BaseRequestHandler): def handle(self): print("in handle") words_file_handler = open('/home/tipu/Dropbox/dev/workspace/search/words.db', 'rb') words = pickle.load(words_file_handler) tweets = shelve.open('/home/tipu/Dropbox/dev/workspace/search/tweets.db', 'r'); results_per_page = 25 query_details = self.request.recv(1024).strip() query_details = eval(query_details) query = query_details["query"] page = int(query_details["page"]) - 1 return_ = [] booleanquery = BooleanQuery(MyTCPHandler.words) if query.find("(") > -1: result = booleanquery.processAdvancedQuery(query) else: result = booleanquery.processQuery(query) result = list(result) i = 0 for tweet_id in result and i < 25: #return_.append(MyTCPHandler.tweets[str(tweet_id)]) return_.append(tweet_id) i += 1 self.request.send(str(return_)) However the file never seems to load after the pickle.load line and it eventually halts the connection attempt. Is there anything I can do to speed this up?

    Read the article

  • Where will log4net create this log file?

    - by Blankman
    When I set the file value to 'logs\log-file.txt' where exactly will it create this folder? in the /bin directory? My web.config looks like: <log4net> <appender name="FileAppender" type="log4net.Appender.FileAppender"> <file value="logs\log-file.txt" /> <appendToFile value="true" /> <lockingModel type="log4net.Appender.FileAppender+MinimalLock" /> <layout type="log4net.Layout.PatternLayout"> <conversionPattern value="%date [%thread] %-5level %logger [%property{NDC}] - %message%newline" /> </layout> </appender> </log4net> is this the correct way to log: ILog logger = LogManager.GetLogger(typeof(CCController)); logger.Error("Some Page", ex); // where ex is the exception instance

    Read the article

  • Crystal Reports "File Break"

    - by Chris B. Behrens
    I'm generating a Crystal Reports report which will ultimately need to be split into thousands of pdf files. What would be ideal would be if Crystal Reports had something like a "file break", like a page break, that you could insert into the file at the appropriate places. I will need reasonably fine control over the file names, as well....something like "fileName_{CustomerId}_{CustomerIsLocal}.pdf". I'm presuming a third-party piece of software will probably be needed. Thoughts? TIA.

    Read the article

  • Loading an external .htm file into a div with javascript

    - by Mads Friis
    So I got this code Javascript: <script type="text/javascript"> $(function(){ $('.ajax') .click(function(e){ e.preventDefault() $('#content').load( 'file.htm' ) }) }) </script> html: <a href="file.htm" class="ajax">Link</a> it works perfectly in firefox, but nothing happens when I click the link in chrome and IE simply opens a new window with the file. any advice?

    Read the article

  • Setting classpath and installing database via batch file

    - by Supereme
    Hi, I want my classpath to be set via a batch file. I'm working on Windows XP. I have two questions: My first question: I made a batch file in which I typed "set classpath = C:\WINDOWS\system32\;.;C:\jdk1.5.0\lib\tools.jar;C:\poi-3.6\poi-3.6-20091214.jar;C:\poi-3.6\poi-contrib-3.6-20091214.jar;C:\poi-3.6\poi-ooxml-3.6-20091214.jar;C:\poi-3.6\poi-ooxml-schemas-3.6-20091214.jar;C:\poi-3.6\poi-scratchpad-3.6-20091214.jar;E:\jdbc\postgresql-8.2-505.jdbc3.jar;C:\xmlbeans-2.5.0\lib\jsr173_1.0_api.jar;C:\xmlbeans-2.5.0\lib\resolver.jar;C:\xmlbeans-2.5.0\lib\xbean.jar;C:\xmlbeans-2.5.0\lib\xbean_xpath.jar;C:\xmlbeans-2.5.0\lib\xmlbeans-qname.jar;C:\xmlbeans-2.5.0\lib\xmlpublic.jar;C:\dom4j-1.6.1\dom4j-1.6.1.jar; exit" When I tried to run this file it ran but when I went into control panel systemadvancedenvironment variables and then selected classpath, it didn't show me the classpath I did set. What is the correct way to set the classpath via batch file? My second question: Is there any way by which we can install database via batch file say for eg: postgresql8.2? Thank you.

    Read the article

  • display records which exist in file2 but not in file1

    - by Phoenix
    log file1 contains records of customers(name,id,date) who visited yesterday log file2 contains records of customers(name,id,date) who visited today How would you display customers who visited yesterday but not today? Constraint is: Don't use auxiliary data structure because file contains millions of records. [So, no hashes] Is there a way to do this using Unix commands ??

    Read the article

  • Define Servlet Context in WAR-File

    - by er4z0r
    Hi, How can I tell e.g. Tomcat to use a specific context path when given my WAR-File? Example: I have a war file created by maven build and the resulting name of the file is rather long. So I do not want the tomcat manager application to use the filename of the war as the context. Supplying a context.xml in META-INF did not produce the desired results I also found this in the documentation for the path attribute of Context: The value of this field must not be set except when statically defining a Context in server.xml, as it will be inferred from the filenames used for either the .xml context file or the docBase. So it does not seem to be the right way to tell the application-server what the path for my WAR should be. Any more hints?

    Read the article

  • Strip the last line from a text file

    - by fraXis
    Hello, I need to strip the last line from a text file. I know how to open and save text files in C#, but how would I strip the last line of the text file? The text file will always be different sizes (some have 80 lines, some have 20). Can someone please show me how to do this? Thanks.

    Read the article

  • Python IOError: Not a gzipped file (Gzip and Blowfish Encrypt/Compress)

    - by notbad.jpeg
    I'm having some problems with python's built-in library gzip. Looked through almost every other stack question about it, and none of them seem to work. MY PROBLEM IS THAT WHEN I TRY TO DECOMPRESS I GET THE IOError I'm Getting: Traceback (most recent call last): File "mymodule.py", line 61, in return gz.read() File "/usr/lib/python2.7/gzip.py", line 245, readself._read(readsize) File "/usr/lib/python2.7/gzip.py", line 287, in _readself._read_gzip_header() File "/usr/lib/python2.7/gzip.py", line 181, in _read_gzip_header raise IOError, 'Not a gzipped file'IOError: Not a gzipped file This is my code to send it over SMB, it might not make sense why i do things, but it's normally in a while loop and memory efficient, I just simplified it. buffer = cStringIO.StringIO(output) #output is from a subprocess call small_buffer = cStringIO.StringIO() small_string = buffer.read() #need a string to write to buffer gzip_obj = gzip.GzipFile(fileobj=small_buffer,compresslevel=6, mode='wb') gzip_obj.write(small_string) compressed_str = small_buffer.getvalue() blowfish = Blowfish.new('abcd', Blowfish.MODE_ECB) remainder = '|'*(8 - (len(compressed_str) % 8)) compressed_str += remainder encrypted = blowfish.encrypt(compressed_str) #i send it over smb, then retrieve it later Then this is the code that retrieves it: #buffer is a cStringIO object filled with data from smb retrieval decrypter = Blowfish.new('abcd', Blowfish.MODE_ECB) value = buffer.getvalue() decrypted = decrypter.decrypt(value) buff = cStringIO.StringIO(decrypted) buff.seek(0) gz = gzip.GzipFile(fileobj=buff) return gz.read() Here's the problem return gz.read()

    Read the article

  • How to create a class in dbml file dynamically

    - by Naseem
    Hi, I'm using linq to sql and I need to have a class in the dbml file which some of its properties are creating dynamically . Is there any way to have a class in dbml file with some pre defined properties and some dynamic properties . Or is there any way to create a class in dbml file dynamically? Thank you,

    Read the article

  • File Upload drops with no reason

    - by sufoid
    Hallo I want to make an file upload. The script should take the image, resize it and upload it. But it seems that there is any unknown to me error in the upload. Here the code define ("MAX_SIZE","2000"); // maximum size for uploaded images define ("WIDTH","107"); // width of thumbnail define ("HEIGHT","107"); // alternative height of thumbnail (portrait 107x80) define ("WIDTH2","600"); // width of (compressed) photo define ("HEIGHT2","600"); // alternative height of (compressed) photo (portrait 600x450) if (isset($_POST['Submit'])) { // iterate thorugh all upload fields foreach ($_FILES as $key => $value) { //read name of user-file $image = $_FILES[$key]['name']; // if it is not empty if ($image) { $filename = stripslashes($_FILES[$key]['name']); // get original name of file from clients machine $extension = getExtension($filename); // get extension of file in lower case format $extension = strtolower($extension); // if extension not known, output error // otherwise continue if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { echo '<div class="failure">Fehler bei Datei '. $_FILES[$key]['name'] .': Unbekannter Dateityp: Es können nur Dateien vom Typ .gif, .jpg oder .png hochgeladen werden.</div>'; } else { // get size of image in bytes // $_FILES[\'image\'][\'tmp_name\'] >> temporary filename of file in which the uploaded file was stored on server $size = getimagesize($_FILES[$key]['tmp_name']); $sizekb = filesize($_FILES[$key]['tmp_name']); // if image size exceeds defined maximum size, output error // otherwise continue if ($sizekb > MAX_SIZE*1024) { echo '<div class="failure">Fehler bei Datei '. $_FILES[$key]['name'] .': Die Datei konnte nicht hochgeladen werden: die Dateigröße überschreitet das Limit von 2MB.</div>'; } else { $rand = md5(rand() * time()); // create random file name $image_name = $rand.'.'.$extension; // unique name (random number) // new name contains full path of storage location (images folder) $consname = "photos/".$image_name; // path to big image $consname2 = "photos/thumbs/".$image_name; // path to thumbnail $copied = copy($_FILES[$key]['tmp_name'], $consname); $copied = copy($_FILES[$key]['tmp_name'], $consname2); $sql="INSERT INTO photos (galery_id, photo, thumb) VALUES (". $id .", '$consname', '$consname2')" or die(mysql_error()); $query = mysql_query($sql) or die(mysql_error()); // if image hasnt been uploaded successfully, output error // otherwise continue if (!$copied) { echo '<div class="failure">Fehler bei Datei '. $_FILES[$key]['name'] .': Die Datei konnte nicht hochgeladen werden.</div>'; } else { $thumb_name = $consname2; // path for thumbnail for creation & storage // call to function: create thumbnail // parameters: image name, thumbnail name, specified width and height $thumb = make_thumb($consname,$thumb_name,WIDTH,HEIGHT); $thumb = make_thumb($consname,$consname,WIDTH2,HEIGHT2); } } } } } // current image could be uploaded successfully echo '<div class="success">'. $success .' Foto(s) erfolgreich hochgeladen!</div>'; showForm(); // call to function: create upload form }

    Read the article

  • Document ranking strategy for P2P file sharing system.

    - by ablmf
    Recently, I got a task of building a P2P file sharing system. There is one requirement : the system should have a document ranking algorithm so that it could be used help users find more valuable files. Several strategy might be useful : force user to give a score to a file before it was downloaded The documents containing certain key words would get higher rank Manager could modify file ranking manually the more a file was downloaded, it would get a higher rank. Do you know any other strategy or methods that also appropriate? Or is there any real world example?

    Read the article

  • IIS 6.0 - ASP Error 0126 Include file not found

    - by André
    Hello, I have a Win Sever 2003 running IIS 6.0 which has only my main website on it and now I am trying to setup a test website which currently is an exact duplicate of the main site. When accessing my main site everything works fine, and has done for a long time. If I access the test site (through 'test.' subdomain) I get this error: Active Server Pages error 'ASP 0126' Include file not found /html/shop/asp/admin/inc/incWeeklySpecialswide3.asp, line 71 The include file '/html/asp/quickfindwithSuburbs31.asp' was not found. The file actually exists, and the paths are correct. I have enabled Parent Paths, replaced the include file path to the full path (http://foo.com/html/asp -etc.), removing the ' / ' at the start of the path and changing the code from ' include ' to ' virtual '. Thanks in advance.

    Read the article

  • Make CMD sensitive to use with MySQL?

    - by acidzombie24
    I have a cleanup scrip in a bat file and i wanted to do the below but the problem is i get a mysql error saying unknown database and showing it as testdb. I guess i can change my code testdb but i would like to know. How do i have windows cmd use case sensitivity so i can execute queries properly instead of always in lower case? mysql.exe -u root -q "drop database TestDB; create database TestDB;"

    Read the article

  • Create an upload form, select a folder and get a specific file in folder

    - by aladine
    Hi, To upload a single file, it is very simple to use this HTML script: <p> <input type="file" name="input.txt" /> In this question, my task is to select a folder and then get the input.txt inside that folder. The server will response whether input.txt is available or not and to upload it to web server. Is there any way to select a folder instead selecting a file in the input form. Thanks.

    Read the article

< Previous Page | 243 244 245 246 247 248 249 250 251 252 253 254  | Next Page >