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  • Windows 7 delayed file delete

    - by GregoryM
    I'm stuck with a pretty rare problem that happens on Windows 7 OS only. Every time I'm deleting the file with *.exe extension through explorer, the file doesn't get deleted immediately. I'm forced to wait for around one-two minutes before the system will delete the file. The main problem is that I cannot develop in such situation, because every time I build my solution, the old executable gets 'deleted', but is still there. So the new one cannot be created by Visual Studio. This problem breaks the Steam update progress and a few other installers functionality too. Fresh installed Win7 doesn't have this kind of trouble, so I guess this must be some bad registry entries or some services. Browsing the internet for solutions I found only this: http://www.sevenforums.com/software/72091-several-minute-delay-when-deleting-any-exe-file.html. But the solution the author found is not working (change the userName :)). Is there any ideas how to find what causes this to happen? BTW: when I place the file into Trash bin, no delay occurs. When I delete file with Total Commander - no delay too. Tech details: Windows 7 x64 Ultimate. UPD: maybe some shadow copying or system restore services (though I have the system restore turned off) block the files? Can't even guess...

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  • How to simulate inner join on very large files in java (without running out of memory)

    - by Constantin
    I am trying to simulate SQL joins using java and very large text files (INNER, RIGHT OUTER and LEFT OUTER). The files have already been sorted using an external sort routine. The issue I have is I am trying to find the most efficient way to deal with the INNER join part of the algorithm. Right now I am using two Lists to store the lines that have the same key and iterate through the set of lines in the right file once for every line in the left file (provided the keys still match). In other words, the join key is not unique in each file so would need to account for the Cartesian product situations ... left_01, 1 left_02, 1 right_01, 1 right_02, 1 right_03, 1 left_01 joins to right_01 using key 1 left_01 joins to right_02 using key 1 left_01 joins to right_03 using key 1 left_02 joins to right_01 using key 1 left_02 joins to right_02 using key 1 left_02 joins to right_03 using key 1 My concern is one of memory. I will run out of memory if i use the approach below but still want the inner join part to work fairly quickly. What is the best approach to deal with the INNER join part keeping in mind that these files may potentially be huge public class Joiner { private void join(BufferedReader left, BufferedReader right, BufferedWriter output) throws Throwable { BufferedReader _left = left; BufferedReader _right = right; BufferedWriter _output = output; Record _leftRecord; Record _rightRecord; _leftRecord = read(_left); _rightRecord = read(_right); while( _leftRecord != null && _rightRecord != null ) { if( _leftRecord.getKey() < _rightRecord.getKey() ) { write(_output, _leftRecord, null); _leftRecord = read(_left); } else if( _leftRecord.getKey() > _rightRecord.getKey() ) { write(_output, null, _rightRecord); _rightRecord = read(_right); } else { List<Record> leftList = new ArrayList<Record>(); List<Record> rightList = new ArrayList<Record>(); _leftRecord = readRecords(leftList, _leftRecord, _left); _rightRecord = readRecords(rightList, _rightRecord, _right); for( Record equalKeyLeftRecord : leftList ){ for( Record equalKeyRightRecord : rightList ){ write(_output, equalKeyLeftRecord, equalKeyRightRecord); } } } } if( _leftRecord != null ) { write(_output, _leftRecord, null); _leftRecord = read(_left); while(_leftRecord != null) { write(_output, _leftRecord, null); _leftRecord = read(_left); } } else { if( _rightRecord != null ) { write(_output, null, _rightRecord); _rightRecord = read(_right); while(_rightRecord != null) { write(_output, null, _rightRecord); _rightRecord = read(_right); } } } _left.close(); _right.close(); _output.flush(); _output.close(); } private Record read(BufferedReader reader) throws Throwable { Record record = null; String data = reader.readLine(); if( data != null ) { record = new Record(data.split("\t")); } return record; } private Record readRecords(List<Record> list, Record record, BufferedReader reader) throws Throwable { int key = record.getKey(); list.add(record); record = read(reader); while( record != null && record.getKey() == key) { list.add(record); record = read(reader); } return record; } private void write(BufferedWriter writer, Record left, Record right) throws Throwable { String leftKey = (left == null ? "null" : Integer.toString(left.getKey())); String leftData = (left == null ? "null" : left.getData()); String rightKey = (right == null ? "null" : Integer.toString(right.getKey())); String rightData = (right == null ? "null" : right.getData()); writer.write("[" + leftKey + "][" + leftData + "][" + rightKey + "][" + rightData + "]\n"); } public static void main(String[] args) { try { BufferedReader leftReader = new BufferedReader(new FileReader("LEFT.DAT")); BufferedReader rightReader = new BufferedReader(new FileReader("RIGHT.DAT")); BufferedWriter output = new BufferedWriter(new FileWriter("OUTPUT.DAT")); Joiner joiner = new Joiner(); joiner.join(leftReader, rightReader, output); } catch (Throwable e) { e.printStackTrace(); } } } After applying the ideas from the proposed answer, I changed the loop to this private void join(RandomAccessFile left, RandomAccessFile right, BufferedWriter output) throws Throwable { long _pointer = 0; RandomAccessFile _left = left; RandomAccessFile _right = right; BufferedWriter _output = output; Record _leftRecord; Record _rightRecord; _leftRecord = read(_left); _rightRecord = read(_right); while( _leftRecord != null && _rightRecord != null ) { if( _leftRecord.getKey() < _rightRecord.getKey() ) { write(_output, _leftRecord, null); _leftRecord = read(_left); } else if( _leftRecord.getKey() > _rightRecord.getKey() ) { write(_output, null, _rightRecord); _pointer = _right.getFilePointer(); _rightRecord = read(_right); } else { long _tempPointer = 0; int key = _leftRecord.getKey(); while( _leftRecord != null && _leftRecord.getKey() == key ) { _right.seek(_pointer); _rightRecord = read(_right); while( _rightRecord != null && _rightRecord.getKey() == key ) { write(_output, _leftRecord, _rightRecord ); _tempPointer = _right.getFilePointer(); _rightRecord = read(_right); } _leftRecord = read(_left); } _pointer = _tempPointer; } } if( _leftRecord != null ) { write(_output, _leftRecord, null); _leftRecord = read(_left); while(_leftRecord != null) { write(_output, _leftRecord, null); _leftRecord = read(_left); } } else { if( _rightRecord != null ) { write(_output, null, _rightRecord); _rightRecord = read(_right); while(_rightRecord != null) { write(_output, null, _rightRecord); _rightRecord = read(_right); } } } _left.close(); _right.close(); _output.flush(); _output.close(); } UPDATE While this approach worked, it was terribly slow and so I have modified this to create files as buffers and this works very well. Here is the update ... private long getMaxBufferedLines(File file) throws Throwable { long freeBytes = Runtime.getRuntime().freeMemory() / 2; return (freeBytes / (file.length() / getLineCount(file))); } private void join(File left, File right, File output, JoinType joinType) throws Throwable { BufferedReader leftFile = new BufferedReader(new FileReader(left)); BufferedReader rightFile = new BufferedReader(new FileReader(right)); BufferedWriter outputFile = new BufferedWriter(new FileWriter(output)); long maxBufferedLines = getMaxBufferedLines(right); Record leftRecord; Record rightRecord; leftRecord = read(leftFile); rightRecord = read(rightFile); while( leftRecord != null && rightRecord != null ) { if( leftRecord.getKey().compareTo(rightRecord.getKey()) < 0) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.LeftExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, leftRecord, null); } leftRecord = read(leftFile); } else if( leftRecord.getKey().compareTo(rightRecord.getKey()) > 0 ) { if( joinType == JoinType.RightOuterJoin || joinType == JoinType.RightExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, null, rightRecord); } rightRecord = read(rightFile); } else if( leftRecord.getKey().compareTo(rightRecord.getKey()) == 0 ) { String key = leftRecord.getKey(); List<File> rightRecordFileList = new ArrayList<File>(); List<Record> rightRecordList = new ArrayList<Record>(); rightRecordList.add(rightRecord); rightRecord = consume(key, rightFile, rightRecordList, rightRecordFileList, maxBufferedLines); while( leftRecord != null && leftRecord.getKey().compareTo(key) == 0 ) { processRightRecords(outputFile, leftRecord, rightRecordFileList, rightRecordList, joinType); leftRecord = read(leftFile); } // need a dispose for deleting files in list } else { throw new Exception("DATA IS NOT SORTED"); } } if( leftRecord != null ) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.LeftExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, leftRecord, null); } leftRecord = read(leftFile); while(leftRecord != null) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.LeftExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, leftRecord, null); } leftRecord = read(leftFile); } } else { if( rightRecord != null ) { if( joinType == JoinType.RightOuterJoin || joinType == JoinType.RightExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, null, rightRecord); } rightRecord = read(rightFile); while(rightRecord != null) { if( joinType == JoinType.RightOuterJoin || joinType == JoinType.RightExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, null, rightRecord); } rightRecord = read(rightFile); } } } leftFile.close(); rightFile.close(); outputFile.flush(); outputFile.close(); } public void processRightRecords(BufferedWriter outputFile, Record leftRecord, List<File> rightFiles, List<Record> rightRecords, JoinType joinType) throws Throwable { for(File rightFile : rightFiles) { BufferedReader rightReader = new BufferedReader(new FileReader(rightFile)); Record rightRecord = read(rightReader); while(rightRecord != null){ if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.RightOuterJoin || joinType == JoinType.FullOuterJoin || joinType == JoinType.InnerJoin ) { write(outputFile, leftRecord, rightRecord); } rightRecord = read(rightReader); } rightReader.close(); } for(Record rightRecord : rightRecords) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.RightOuterJoin || joinType == JoinType.FullOuterJoin || joinType == JoinType.InnerJoin ) { write(outputFile, leftRecord, rightRecord); } } } /** * consume all records having key (either to a single list or multiple files) each file will * store a buffer full of data. The right record returned represents the outside flow (key is * already positioned to next one or null) so we can't use this record in below while loop or * within this block in general when comparing current key. The trick is to keep consuming * from a List. When it becomes empty, re-fill it from the next file until all files have * been consumed (and the last node in the list is read). The next outside iteration will be * ready to be processed (either it will be null or it points to the next biggest key * @throws Throwable * */ private Record consume(String key, BufferedReader reader, List<Record> records, List<File> files, long bufferMaxRecordLines ) throws Throwable { boolean processComplete = false; Record record = records.get(records.size() - 1); while(!processComplete){ long recordCount = records.size(); if( record.getKey().compareTo(key) == 0 ){ record = read(reader); while( record != null && record.getKey().compareTo(key) == 0 && recordCount < bufferMaxRecordLines ) { records.add(record); recordCount++; record = read(reader); } } processComplete = true; // if record is null, we are done if( record != null ) { // if the key has changed, we are done if( record.getKey().compareTo(key) == 0 ) { // Same key means we have exhausted the buffer. // Dump entire buffer into a file. The list of file // pointers will keep track of the files ... processComplete = false; dumpBufferToFile(records, files); records.clear(); records.add(record); } } } return record; } /** * Dump all records in List of Record objects to a file. Then, add that * file to List of File objects * * NEED TO PLACE A LIMIT ON NUMBER OF FILE POINTERS (check size of file list) * * @param records * @param files * @throws Throwable */ private void dumpBufferToFile(List<Record> records, List<File> files) throws Throwable { String prefix = "joiner_" + files.size() + 1; String suffix = ".dat"; File file = File.createTempFile(prefix, suffix, new File("cache")); BufferedWriter writer = new BufferedWriter(new FileWriter(file)); for( Record record : records ) { writer.write( record.dump() ); } files.add(file); writer.flush(); writer.close(); }

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  • Flash AS3 load file xml

    - by Elias
    Hello, I'm just trying to load an xml file witch can be anywere in the hdd, this is what I have done to browse it, but later when I'm trying to load the file it would only look in the same path of the swf file here is the code package { import flash.display.Sprite; import flash.events.; import flash.net.; public class cargadorXML extends Sprite { public var cuadro:Sprite = new Sprite(); public var file:FileReference; public var req:URLRequest; public var xml:XML; public var xmlLoader:URLLoader = new URLLoader(); public function cargadorXML() { cuadro.graphics.beginFill(0xFF0000); cuadro.graphics.drawRoundRect(0,0,100,100,10); cuadro.graphics.endFill(); cuadro.addEventListener(MouseEvent.CLICK,browser); addChild(cuadro); } public function browser(e:Event) { file = new FileReference(); file.addEventListener(Event.SELECT,bien); file.browse(); } public function bien(e:Event) { xmlLoader.addEventListener(Event.COMPLETE, loadXML); req=new URLRequest(file.name); xmlLoader.load(req); } public function loadXML(e:Event) { xml=new XML(e.target.data); //xml.name=file.name; trace(xml); } } } when I open a xml file that isnt it the same directory as the swf, it gives me an unfound file error. is there anything I can do? cause for example for mp3 there is an especial class for loading the file, see http://www.flexiblefactory.co.uk/flexible/?p=46 thanks

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  • Unable to access Java-created file -- sometimes

    - by BlairHippo
    In Java, I'm working with code running under WinXP that creates a file like this: public synchronized void store(Properties props, byte[] data) { try { File file = filenameBasedOnProperties(props); if ( file.exists() ) { return; } File temp = File.createTempFile("tempfile", null); FileOutputStream out = new FileOutputStream(temp); out.write(data); out.flush(); out.close(); file.getParentFile().mkdirs(); temp.renameTo(file); } catch (IOException ex) { // Complain and whine and stuff } } Sometimes, when a file is created this way, it's just about totally inaccessible from outside the code (though the code responsible for opening and reading the file has no problem), even when the application isn't running. When accessed via Windows Explorer, I can't move, rename, delete, or even open the file. Under Cygwin, I get the following when I ls -l the directory: ls: cannot access [big-honkin-filename] total 0 ?????????? ? ? ? ? ? [big-honkin-filename] As implied, the filenames are big, but under the 260-character max for XP (though they are slightly over 200 characters). To further add to the sense the my computer just wants me to feel stupid, sometimes the files created by this code are perfectly normal. The only pattern I've spotted is that once one file in the directory "locks", the rest are screwed. Anybody ever run into something like this before, or have any insights into what's going on here?

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  • How to compare two file structures in PHP?

    - by OM The Eternity
    I have a function which gives me the complete file structure upto n-level, function getDirectory($path = '.', $ignore = '') { $dirTree = array (); $dirTreeTemp = array (); $ignore[] = '.'; $ignore[] = '..'; $dh = @opendir($path); while (false !== ($file = readdir($dh))) { if (!in_array($file, $ignore)) { if (!is_dir("$path/$file")) { //display of file and directory name with their modification time $stat = stat("$path/$file"); $statdir = stat("$path"); $dirTree["$path"][] = $file. " === ". date('Y-m-d H:i:s', $stat['mtime']) . " Directory == ".$path."===". date('Y-m-d H:i:s', $statdir['mtime']) ; } else { $dirTreeTemp = getDirectory("$path/$file", $ignore); if (is_array($dirTreeTemp))$dirTree = array_merge($dirTree, $dirTreeTemp); } } } closedir($dh); return $dirTree; } $ignore = array('.htaccess', 'error_log', 'cgi-bin', 'php.ini', '.ftpquota'); //function call $dirTree = getDirectory('.', $ignore); //file structure array print print_r($dirTree); Now here my requirement is , I have two sites The Development/Test Site- where i do testing of all the changes The Production Site- where I finally post all the changes as per test in development site Now, for example, I have tested an image upload in the Development/test site, and i found it appropriate to publish on Production site then i will completely transfer the Development/Test DB detail to Production DB, but now I want to compare the files structure as well to transfer the corresponding image file to Production folder. There could be the situation when I update the image by editing the image and upload it with same name, now in this case the image file would be already present there, which will restrict the use of "file_exist" logic, so for these type of situations....HOW CAN I COMPARE THE TWO FILE STRUCTURE TO GET THE SYNCHRONIZATION DONE AS PER REQUIREMENT??

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  • Modifying File while in use using Java

    - by Marquinio
    Hi all, I have this recurrent Java JAR program tasks that tries to modify a file every 60seconds. Problem is that if user is viewing the file than Java program will not be able to modify the file. I get the typical IOException. Anyone knows if there is a way in Java to modify a file currently in use? Or anyone knows what would be the best way to solve this problem? I was thinking of using the File canRead(), canWrite() methods to check if file is in use. If file is in use then I'm thinking of making a backup copy of data that could not be written. Then after 60 seconds add some logic to check if backup file is empty or not. If backup file is not empty then add its contents to main file. If empty then just add new data to main file. Of course, the first thing I will always do is check if file is in use. Thanks for all your ideas.

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  • File mkdirs() method not working in android/java

    - by Leif Andersen
    I've been pulling out my hair on this for a while now. The following method is supposed to download a file, and save it to the location specified on the hard drive. private static void saveImage(Context context, boolean backgroundUpdate, URL url, File file) { if (!Tools.checkNetworkState(context, backgroundUpdate)) return; // Get the image try { // Make the file file.getParentFile().mkdirs(); // Set up the connection URLConnection uCon = url.openConnection(); InputStream is = uCon.getInputStream(); BufferedInputStream bis = new BufferedInputStream(is); // Download the data ByteArrayBuffer baf = new ByteArrayBuffer(50); int current = 0; while ((current = bis.read()) != -1) { baf.append((byte) current); } // Write the bits to the file OutputStream os = new FileOutputStream(file); os.write(baf.toByteArray()); os.close(); } catch (Exception e) { // Any exception is probably a newtork faiilure, bail return; } } Also, if the file doesn't exist, it is supposed to make the directory for the file. (And if there is another file already in that spot, it should just not do anything). However, for some reason, the mkdirs() method never makes the directory. I've tried everything from explicit parentheses, to explicitly making the parent file class, and nothing seems to work. I'm fairly certain that the drive is writable, as it's only called after that has already been determined, also that is true after running through it while debugging. So the method fails because the parent directories aren't made. Can anyone tell me if there is anything wrong with the way I'm calling it? Also, if it helps, here is the source for the file I'm calling it in: https://github.com/LeifAndersen/NetCatch/blob/master/src/net/leifandersen/mobile/android/netcatch/services/RSSService.java Thank you

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  • How to easily input and display isolated japanese radicals on MacOS X?

    - by ogerard
    (This question was considered as off-topic on Japanese.SE and more suitable for SuperUser). I like to write computer notes about what I learn in Japanese. From time to time, I would like to be able to include in my text a given radical, say kokoro ?, which takes several graphic forms when used as an element in a more complex kanji, for instance . I did not succeed on my system (Mac OS X 10.7) to find the glyphs for these variants conveniently and exactly as I would like them (I would also be interested about how to do this on Windows 7 or Linux). I first tried the name of the kanji from which the radical is derived. Then I tried to use the japanese name for them, such as risshinben (as in ?) and shitagokoro (as in ?), hoping that the hiragana or katakana input would recognize them and propose me their representation, but it did not work. So I looked into the Full Japanese Character Table, under the "by radical" tab, and found at least a version of each of them : ? (CJK 5FC4) and ? (CJK 38FA) with the correct kun readings. I have them now as favorites but do I need to do that for all radicals? Do I need to register all of them in a user dictionary? I would imagine that I am not the only one who wants to do use them. Besides, the versions I have found are not suited for all occasions: they are centered on a standard kanji square. If I want them to appear near to a placeholder, or demonstrate their proportion to the rest of a typical kanji, I have to make complicated adjustments, depending on my use and the kind of radical. More generally are there computer tools for japanese dictionary editors and japanese teachers I could use on Mac OS X? (I could not add relevant tags such as : japanese or ideogram, please feel free to edit)

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  • Unset the system immutable bit in Mac OS X

    - by skylarking
    In theory I believe you can unlock and remove the system immutable bit with: chflags noschg /Path/To/File But how can you do this when you've set the bit as root? I have a file that is locked, and even running this command as root will not work as the operation is not permitted. I tried logging in as Single-User mode to no avail. I seem to remember that even though you are in as root you are in at level '1'. And to be able to remove the system-immutable flag you need to be logged in at level '0'. Does this have something to do with this issue?

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  • Reliable file copy (move) process - mostly Unix/Linux

    - by mfinni
    Short story : We have a need for a rock-solid reliable file mover process. We have source directories that are often being written to that we need to move files from. The files come in pairs - a big binary, and a small XML index. We get a CTL file that defines these file bundles. There is a process that operates on the files once they are in the destination directory; that gets rid of them when it's done. Would rsync do the best job, or do we need to get more complex? Long story as follows : We have multiple sources to pull from : one set of directories are on a Windows machine (that does have Cygwin and an SSH daemon), and a whole pile of directories are on a set of SFTP servers (Most of these are also Windows.) Our destinations are a list of directories on AIX servers. We used to use a very reliable Perl script on the Windows/Cygwin machine when it was our only source. However, we're working on getting rid of that machine, and there are other sources now, the SFTP servers, that we cannot presently run our own scripts on. For security reasons, we can't run the copy jobs on our AIX servers - they have no access to the source servers. We currently have a homegrown Java program on a Linux machine that uses SFTP to pull from the various new SFTP source directories, copies to a local tmp directory, verifies that everything is present, then copies that to the AIX machines, and then deletes the files from the source. However, we're finding any number of bugs or poorly-handled error checking. None of us are Java experts, so fixing/improving this may be difficult. Concerns for us are: With a remote source (SFTP), will rsync leave alone any file still being written? Some of these files are large. From reading the docs, it seems like rysnc will be very good about not removing the source until the destination is reliably written. Does anyone have experience confirming or disproving this? Additional info We will be concerned about the ingestion process that operates on the files once they are in the destination directory. We don't want it operating on files while we are in the process of copying them; it waits until the small XML index file is present. Our current copy job are supposed to copy the XML file last. Sometimes the network has problems, sometimes the SFTP source servers crap out on us. Sometimes we typo the config files and a destination directory doesn't exist. We never want to lose a file due to this sort of error. We need good logs If you were presented with this, would you just script up some rsync? Or would you build or buy a tool, and if so, what would it be (or what technologies would it use?) I (and others on my team) are decent with Perl.

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  • Limit the file inputs cloned in a form with Jquery

    - by Philip
    Hi, i use this Jquery function to clone the file input fields in my form, $(function() { var scntDiv = $('#clone'); var i = $('#clone p').size() + 1; $('#addImg').live('click', function() { $('<p><label for="attach"><input type="file" name="attachment_'+ i +'" /> <a href="#" id="remImg">Remove</a></label></p>').appendTo(scntDiv); i++; return false; }); $('#remImg').live('click', function() { if( i > 2 ) { $(this).parents('p').remove(); i--; } return false; }); }); is it possible to limit the fields that can be cloned? lets say a number of 4 fields? thanks a lot, Philip

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  • midi input in python

    - by Nicola Montecchio
    Hello I'm coding a demo in python and I need to read a MIDI file in python (no real-time stuff is needed). In particular, I'm looking for a library which preserves channel information. The most promising libraries I found are: http://code.google.com/p/midiutil/ http://www.mxm.dk/products/public/pythonmidi Any experience with those? Thanks a lot Nicola Montecchio

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  • Copy mdf file and use it in run time

    - by Anibas
    After I copy mdf file (and his log file) I tries to Insert data. I receive the following message: "An attempt to attach an auto-named database for file [fileName].mdf failed. A database with the same name exists, or specified file cannot be opened, or it is located on UNC share. When I copied the file manual everything worked normally. Is it correct the order File.Copy leaves the file engaged?

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  • Designing a database file format

    - by RoliSoft
    I would like to design my own database engine for educational purposes, for the time being. Designing a binary file format is not hard nor the question, I've done it in the past, but while designing a database file format, I have come across a very important question: How to handle the deletion of an item? So far, I've thought of the following two options: Each item will have a "deleted" bit which is set to 1 upon deletion. Pro: relatively fast. Con: potentially sensitive data will remain in the file. 0x00 out the whole item upon deletion. Pro: potentially sensitive data will be removed from the file. Con: relatively slow. Recreating the whole database. Pro: no empty blocks which makes the follow-up question void. Con: it's a really good idea to overwrite the whole 4 GB database file because a user corrected a typo. I will sell this method to Twitter ASAP! Now let's say you already have a few empty blocks in your database (deleted items). The follow-up question is how to handle the insertion of a new item? Append the item to the end of the file. Pro: fastest possible. Con: file will get huge because of all the empty blocks that remain because deleted items aren't actually deleted. Search for an empty block exactly the size of the one you're inserting. Pro: may get rid of some blocks. Con: you may end up scanning the whole file at each insert only to find out it's very unlikely to come across a perfectly fitting empty block. Find the first empty block which is equal or larger than the item you're inserting. Pro: you probably won't end up scanning the whole file, as you will find an empty block somewhere mid-way; this will keep the file size relatively low. Con: there will still be lots of leftover 0x00 bytes at the end of items which were inserted into bigger empty blocks than they are. Rigth now, I think the first deletion method and the last insertion method are probably the "best" mix, but they would still have their own small issues. Alternatively, the first insertion method and scheduled full database recreation. (Probably not a good idea when working with really large databases. Also, each small update in that method will clone the whole item to the end of the file, thus accelerating file growth at a potentially insane rate.) Unless there is a way of deleting/inserting blocks from/to the middle of the file in a file-system approved way, what's the best way to do this? More importantly, how do databases currently used in production usually handle this?

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  • Parse a text file into multiple text file

    - by Vijay Kumar Singh
    I want to get multiple file by parsing a input file Through Java. The Input file contains many fasta format of thousands of protein sequence and I want to generate raw format(i.e., without any comma semicolon and without any extra symbol like "", "[", "]" etc) of each protein sequence. A fasta sequence starts form "" symbol followed by description of protein and then sequence of protein. For example ? lcl|NC_000001.10_cdsid_XP_003403591.1 [gene=LOC100652771] [protein=hypothetical protein LOC100652771] [protein_id=XP_003403591.1] [location=join(12190..12227,12595..12721,13403..13639)] MSESINFSHNLGQLLSPPRCVVMPGMPFPSIRSPELQKTTADLDHTLVSVPSVAESLHHPEITFLTAFCL PSFTRSRPLPDRQLHHCLALCPSFALPAGDGVCHGPGLQGSCYKGETQESVESRVLPGPRHRH Like above formate the input file contains 1000s of protein sequence. I have to generate thousands of raw file containing only individual protein sequence without any special symbol or gaps. I have developed the code for it in Java but out put is : Cannot open a file followed by cannot find file. Please help me to solve my problem. Regards Vijay Kumar Garg Varanasi Bharat (India) The code is /*Java code to convert FASTA format to a raw format*/ import java.io.*; import java.util.*; import java.util.regex.*; import java.io.FileInputStream; // java package for using regular expression public class Arrayren { public static void main(String args[]) throws IOException { String a[]=new String[1000]; String b[][] =new String[1000][1000]; /*open the id file*/ try { File f = new File ("input.txt"); //opening the text document containing genbank ids FileInputStream fis = new FileInputStream("input.txt"); //Reading the file contents through inputstream BufferedInputStream bis = new BufferedInputStream(fis); // Writing the contents to a buffered stream DataInputStream dis = new DataInputStream(bis); //Method for reading Java Standard data types String inputline; String line; String separator = System.getProperty("line.separator"); // reads a line till next line operator is found int i=0; while ((inputline=dis.readLine()) != null) { i++; a[i]=inputline; a[i]=a[i].replaceAll(separator,""); //replaces unwanted patterns like /n with space a[i]=a[i].trim(); // trims out if any space is available a[i]=a[i]+".txt"; //takes the file name into an array try // to handle run time error /*take the sequence in to an array*/ { BufferedReader in = new BufferedReader (new FileReader(a[i])); String inline = null; int j=0; while((inline=in.readLine()) != null) { j++; b[i][j]=inline; Pattern q=Pattern.compile(">"); //Compiling the regular expression Matcher n=q.matcher(inline); //creates the matcher for the above pattern if(n.find()) { /*appending the comment line*/ b[i][j]=b[i][j].replaceAll(">gi",""); //identify the pattern and replace it with a space b[i][j]=b[i][j].replaceAll("[a-zA-Z]",""); b[i][j]=b[i][j].replaceAll("|",""); b[i][j]=b[i][j].replaceAll("\\d{1,15}",""); b[i][j]=b[i][j].replaceAll(".",""); b[i][j]=b[i][j].replaceAll("_",""); b[i][j]=b[i][j].replaceAll("\\(",""); b[i][j]=b[i][j].replaceAll("\\)",""); } /*printing the sequence in to a text file*/ b[i][j]=b[i][j].replaceAll(separator,""); b[i][j]=b[i][j].trim(); // trims out if any space is available File create = new File(inputline+"R.txt"); try { if(!create.exists()) { create.createNewFile(); // creates a new file } else { System.out.println("file already exists"); } } catch(IOException e) // to catch the exception and print the error if cannot open a file { System.err.println("cannot create a file"); } BufferedWriter outt = new BufferedWriter(new FileWriter(inputline+"R.txt", true)); outt.write(b[i][j]); // printing the contents to a text file outt.close(); // closing the text file System.out.println(b[i][j]); } } catch(Exception e) { System.out.println("cannot open a file"); } } } catch(Exception ex) // catch the exception and prints the error if cannot find file { System.out.println("cannot find file "); } } } If you provide me correct it will be much easier to understand.

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  • Remove Trailing Slash From Batch File Input

    - by Brook
    I have a batch file that I want to improve. Instead of requiring a user to provide a folder path without a trailing slash, is there an easy way for me to just remove the last character from the path if there is a slash on the end? :START @echo What folder do you want to process? (Provide a path without a closing backslash) set /p datapath= ::Is string empty? IF X%datapath% == X GOTO:START ::Does string have a trailing slash? IF %datapath:~-1%==\ GOTO:START

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  • Reading data from text file in C

    - by themake
    I have a text file which contains words separated by space. I want to take each word from the file and store it. So i have opened the file but am unsure how to assign the word to a char. FILE *fp; fp = fopen("file.txt", "r"); //then i want char one = the first word in the file char two = the second word in the file

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  • opening and viewing a file in php

    - by Christian Burgos
    how do i open/view for editing an uploaded file in php? i have tried this but it doesn't open the file. $my_file = 'file.txt'; $handle = fopen($my_file, 'r'); $data = fread($handle,filesize($my_file)); i've also tried this but it wont work. $my_file = 'file.txt'; $handle = fopen($my_file, 'w') or die('Cannot open file: '.$my_file); $data = 'This is the data'; fwrite($handle, $data); what i have in mind is like when you want to view an uploaded resume,documents or any other ms office files like .docx,.xls,.pptx and be able to edit them, save and close the said file. edit: latest tried code... <?php // Connects to your Database include "configdb.php"; //Retrieves data from MySQL $data = mysql_query("SELECT * FROM employees") or die(mysql_error()); //Puts it into an array while($info = mysql_fetch_array( $data )) { //Outputs the image and other data //Echo "<img src=localhost/uploadfile/images".$info['photo'] ."> <br>"; Echo "<b>Name:</b> ".$info['name'] . "<br> "; Echo "<b>Email:</b> ".$info['email'] . " <br>"; Echo "<b>Phone:</b> ".$info['phone'] . " <hr>"; //$file=fopen("uploadfile/images/".$info['photo'],"r+"); $file=fopen("Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt","r") or exit("unable to open file");; } ?> i am getting the error: Warning: fopen(Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt): failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/uploadfile/view.php on line 17 unable to open file the file is in that folder, i don't know it wont find it.

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  • Input-type-file path, where is it stored on AJAX request ?!?

    - by Sheavi
    Hi, I have been monitoring the parameters a website receives when a file is uploaded (via an input type="file"). Surprisingly, the parameter and its value were looking like this : parameter: upfile value: filename="this is the name of the uploaded file.png" Content-type: image/x-png Now in this POST request to the server page, the file name and its type is passed into a parameter, but what about the path to that filename? Where is that path stored so that the server page can upload the file at the good location? Also, I would like to know if it would be possible by any way to specify a path, NOT to the input type="file" since its impossible, but to the server (though this question probably depends a lot on how the server-side page is scripted). Thank you for your answers.

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  • Uploading multiple files asynchronously by blueimp jquery-fileupload

    - by Ryo
    I'm using jQuery File Upload library (http://github.com/blueimp/jQuery-File-Upload), and I've been stuck figuring out how to use the library satisfying the following conditions. The page has multiple file input fields surrounded by a form tag. Users can attach multiple files to each input field All files are sent to a server when the button is clicked, not when files are attached to the input fields. Upload is done asynchronously Say the page has 3 input fields with their name attributes being "file1[]", "file2[]" and "file3[]", the request payload should be like {file1: [ array of files on file1[] ], file2: [ array of files on file2[] ], ...} Here's jsFiddle, it's behaving weird so far in that it sends post request twice and the first one is cancelled. http://jsfiddle.net/BAQtG/24/ The core part of js code looks like this. $(document).ready(function(){ var filesList = [] var elem = $("form") file_upload = elem.fileupload({ formData:{extra:1}, autoUpload: false, fileInput: $("input:file"), }).on("fileuploadadd", function(e, data){ filesList.push(data.files[0]) }); $("button").click(function(){ file_upload.fileupload('send', {files:filesList} ) }) }) Anybody have idea how to get this to work? Updates Now thanks to @CBroe 's comment, the issue that request is sent twice is fixed. However the keys of request parameter is not correctly set. Here's updated jsFiddle. http://jsfiddle.net/BAQtG/27/

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  • Copied a file with winscp; only winscp can see it

    - by nilbus
    I recently copied a 25.5GB file from another machine using WinSCP. I copied it to C:\beth.tar.gz, and WinSCP can still see the file. However no other app (including Explorer) can see the file. What might cause this, and how can I fix it? The details that might or might not matter WinSCP shows the size of the file (C:\beth.tar.gz) correctly as 27,460,124,080 bytes, which matches the filesize on the remote host Neither explorer, cmd (command line prompt w/ dir C:\), the 7Zip archive program, nor any other File Open dialog can see the beth.tar.gz file under C:\ I have configured Explorer to show hidden files I can move the file to other directories using WinSCP If I try to move the file to Users/, UAC prompts me for administrative rights, which I grant, and I get this error: Could not find this item The item is no longer located in C:\ When I try to transfer the file back to the remote host in a new directory, the transfer starts successfully and transfers data The transfer had about 30 minutes remaining when I left it for the night The morning after the file transfer, I was greeted with a message saying that the connection to the server had been lost. I don't think this is relevant, since I did not tell it to disconnect after the file was done transferring, and it likely disconnected after the file transfer finished. I'm using an old version of WinSCP - v4.1.8 from 2008 I can view the file properties in WinSCP: Type of file: 7zip (.gz) Location: C:\ Attributes: none (Ready-only, Hidden, Archive, or Ready for indexing) Security: SYSTEM, my user, and Administrators group have full permissions - everything other than "special permissions" is checked under Allow for all 3 users/groups (my user, Administrators, SYSTEM) What's going on?!

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  • C++, Ifstream opens local file but not file on HTTP Server

    - by fammi
    Hi, I am using ifstream to open a file and then read from it. My program works fine when i give location of the local file on my system. for eg /root/Desktop/abc.xxx works fine But once the location is on the http server the file fails to open. for eg http://192.168.0.10/abc.xxx fails to open. Is there any alternate for ifstream when using a URL address? thanks. part of the code where having problem: bool readTillEof = (endIndex == -1) ? true : false; // Open the file in binary mode and seek to the end to determine file size ifstream file ( fileName.c_str ( ), ios::in|ios::ate|ios::binary ); if ( file.is_open ( ) ) { long size = (long) file.tellg ( ); long numBytesRead; if ( readTillEof ) { numBytesRead = size - startIndex; } else { numBytesRead = endIndex - startIndex + 1; } // Allocate a new buffer ptr to read in the file data BufferSptr buf (new Buffer ( numBytesRead ) ); mpStreamingClientEngine->SetResponseBuffer ( nextRequest, buf ); // Seek to the start index of the byte range // and read the data file.seekg ( startIndex, ios::beg ); file.read ( (char *)buf->GetData(), numBytesRead ); // Pass on the data to the SCE // and signal completion of request mpStreamingClientEngine->HandleDataReceived( nextRequest, numBytesRead); mpStreamingClientEngine->MarkRequestCompleted( nextRequest ); // Close the file file.close ( ); } else { // Report error to the Streaming Client Engine // as unable to open file AHS_ERROR ( ConnectionManager, " Error while opening file \"%s\"\n", fileName.c_str ( ) ); mpStreamingClientEngine->HandleRequestFailed( nextRequest, CONNECTION_FAILED ); } }

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