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  • SQL SERVER – FIX: ERROR Msg 5169, Level 16: FILEGROWTH cannot be greater than MAXSIZE for file

    - by pinaldave
    I am writing this blog post right after I resolve this error for one of the system. Recently one of the my friend who is expert in infrastructure as well private cloud was working on SQL Server installation. Please note he is seriously expert in what he does but he has never worked SQL Server before and have absolutely no experience with its installation. He was modifying database file and keep on getting following error. As soon as he saw me he asked me where is the maxfile size setting so he can change. Let us quickly re-create the scenario he was facing. Error Message: Msg 5169, Level 16, State 1, Line 1 FILEGROWTH cannot be greater than MAXSIZE for file ‘NewDB’. Creating Scenario: CREATE DATABASE [NewDB] ON PRIMARY (NAME = N'NewDB', FILENAME = N'D:\NewDB.mdf' , SIZE = 4096KB, FILEGROWTH = 1024KB, MAXSIZE = 4096KB) LOG ON (NAME = N'NewDB_log', FILENAME = N'D:\NewDB_log.ldf', SIZE = 1024KB, FILEGROWTH = 10%) GO Now let us see what exact command was creating error for him. USE [master] GO ALTER DATABASE [NewDB] MODIFY FILE ( NAME = N'NewDB', FILEGROWTH = 1024MB ) GO Workaround / Fix / Solution: The reason for the error is very simple. He was trying to modify the filegrowth to much higher value than the maximum file size specified for the database. There are two way we can fix it. Method 1: Reduces the filegrowth to lower value than maxsize of file USE [master] GO ALTER DATABASE [NewDB] MODIFY FILE ( NAME = N'NewDB', FILEGROWTH = 1024KB ) GO Method 2: Increase maxsize of file so it is greater than new filegrowth USE [master] GO ALTER DATABASE [NewDB] MODIFY FILE ( NAME = N'NewDB', FILEGROWTH = 1024MB, MAXSIZE = 4096MB) GO I think this blog post will help everybody who is facing similar issues. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Error Messages, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Changing the BizTalk message output file name

    - by Bill Osuch
    By default, BizTalk creates the filename of the message dropped to a send port as %MessageID%, which is the unique identifier (GUID) of the message. What if you want to create your own filename? To start, create a simple schema, and a basic orchestration that will receive the message and send it right back out, like this: If you deploy this and wire up the ports, you can drop an xml file into your receive port and have it come out at your send port named something like {7A63CAF8-317B-49D5-871F-9FD57910C3A0}.xml. Now, we'll create a new message with a custom filename. First, create a new orchestration variable called NewFileName, of the type System.String. Next, create a second message using the same schema as the message you're receiving in the Receive shape. Now, drag a Construct Message shape to the orchestration. In the shape's properties, set Messages Constructed to be the new message you just created. Double click the Message Assignment shape (inside the Construct shape...) and paste in the following code: Message_2 = Message_1;   NewFileName = Message_1(FILE.ReceivedFileName); NewFileName = NewFileName.Replace(".xml","_"); NewFileName = NewFileName + "output_" + System.DateTime.Now.Year.ToString() + "-" + System.DateTime.Now.Month.ToString();   Message_2(FILE.ReceivedFileName) = NewFileName; Here we make a copy of the received message, get it's original file name (ReceivedFileName), replace its extension with an underscore, and date-stamp it. Finally, add a Send shape and a Port to the surface, and configure them to send the message you just created. You should wind up with an orchestration like this: Deploy it, and create a new send port. It should be just about identical to the first send port, except this time the file name will be "%SourceFileName%.xml" (without the quotes of course). Fire up the application, drop in a test file, and you should now get both the xml file named with a GUID, and a second file named something along the lines of "MySchemaTestFile_output_2011-6.xml".

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  • FileOpenPicker/FileSavePicker doesn't allow *.* wildcard file associations

    - by mbrit
    On Twitter, Matthias Jauernig commented that the FileOpenPicker and FileSavePicker doesn't allow *.* wildcard file associations. I was relaxed about this and wrote back that it was related to sandboxing implying it was a "good thing", however as Matthias commented back, perhaps it's not.In Metro-style the sandboxing works that if something gives you a file (e.g. the picker, or a share operation), you can access it regardless of where on the system. If you find the file yourself, you have to declare the type.The reason why I think it's related to sandboxing is because if you work with files programmatically you have to be explicit about the file types. This is to stop malware that you think is only interested in - say .PDF files, scanning and uploading any .EML files that it can find on the machine. It follows then on the pickers that restriction would continue. It allow's the retail store team to validate that an app is likely to behave itself. If it's an app that works with images, locking down the picker so that it can only access image file types makes sense.However Matthias mentioned that he has an app that should allow files of any arbitrary file. That fits more into the "if the user selects it, it must be OK" camp than the "programmatic scanning" camp. So now I'm left wondering why the picker doesn't allow any type to be selected.I think then maybe the decision comes down to simplicity. A lot of the decisions in Metro-style design relate to ideas about "zero intimidation". Allow the user to select any file is too much like Old Windows, and not enough like Reimagined Windows. What happens in Matthias's app if the user selects Explorer.exe as the file he or she wants to work with? I guess it's fine if you expect your user to know what they're doing (Old Windows), but not so fine if you're expecting a three year old to work with it (Reimagined Windows).

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  • BizTalk 2009 - Error when Testing Map with Flat File Source Schema

    - by StuartBrierley
    I have recently been creating some flat file schemas using the BizTalk Server 2009 Flat File Schema Wizard.  I have then been mapping these flat file schemas to a "normal" xml schema format. I have not previsouly had any cause to map flat files and ran into some trouble when testing the first of these flat file maps; with an instance of the flat file as the source it threw an XSL transform error: Test Map.btm: error btm1050: XSL transform error: Unable to write output instance to the following <file:///C:\Documents and Settings\sbrierley\Local Settings\Temp\_MapData\Test Mapping\Test Map_output.xml>. Data at the root level is invalid. Line 1, position 1. Due to the complexity of the map in question I decided to created a small test map using the same source and destination schemas to see if I could pinpoint the problem.  Although the source message instance vaildated correctly against the flat file schema, when I then tested this simplified map I got the same error. After a time of fruitless head scratching and some serious google time I figured out what the problem was. Looking at the map properties I noticed that I had the test map input set to "XML" - for a flat file instance this should be set to "native".

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  • Nested namespaces, correct static library design issues

    - by PeterK
    Hello all, I'm currently in the process of developing a fairly large static library which will be used by some tools when it's finished. Now since this project is somewhat larger than anything i've been involved in so far, I realized its time to think of a good structure for the project. Using namespaces is one of those logical steps. My current approach is to divide the library into parts (which are not standalone, but their purpose calls for such a separation). I have a 'core' part which now just holds some very common typedefs and constants (used by many different parts of the library). Other parts are for example some 'utils' (hash etc.), file i/o and so on. Each of these parts has its own namespace. I have nearly finished the 'utils' part and realized that my approach probably is not the best. The problem (if we want to call it so) is that in the 'utils' namespace i need something from the 'core' namespace which results in including the core header files and many using directives. So i began to think that this probably is not a good thing and should be changed somehow. My first idea is to use nested namespaces as to have something like core::utils. Since this will require some heavy refactoring i want to ask here first. What do you think? How would you handle this? Or more generally: How to correctly design a static library in terms of namespaces and code organization? If there are some guidelines or articles about it, please mentoin them too. Thanks. Note: i'm quite sure that there are more good approaches than just one. Feel free to post your ideas, suggestions etc. Since i'm designing this library i want it to be really good. The goal is to make it as clean and FAST as possible. The only problem is that i will have to integrate a LOT of existing code and refactor it, which will really be a painful process (sigh) - thats why good structure is so important)

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  • small code redundancy within while-loops (doesn't feel clean)

    - by wallacoloo
    So, in Python (though I think it can be applied to many languages), I find myself with something like this quite often: the_input = raw_input("what to print?\n") while the_input != "quit": print the_input the_input = raw_input("what to print?\n") Maybe I'm being too picky, but I don't like how the line the_input = raw_input("what to print?\n") has to get repeated. It decreases maintainability and organization. But I don't see any workarounds for avoiding the duplicate code without further decreasing the problem. In some languages, I could write something like this: while ((the_input=raw_input("what to print?\n")) != "quit") { print the_input } This is definitely not Pythonic, and Python doesn't even allow for assignment within loop conditions AFAIK. This valid code fixes the redundancy, while 1: the_input = raw_input("what to print?\n") if the_input == "quit": break print the_input But doesn't feel quite right either. The while 1 implies that this loop will run forever; I'm using a loop, but giving it a fake condition and putting the real one inside it. Am I being too picky? Is there a better way to do this? Perhaps there's some language construct designed for this that I don't know of?

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  • How should I organize my Java GUI?

    - by Spencer
    I'm creating a game in Java for fun and I'm trying to decide how to organize my classes for the GUI. So far, all the classes with only the swing components and layout (no logic) are in a package called "ui". I now need to add listeners (i.e. ActionListener) to components (i.e. button). The listeners need to communicate with the Game class. Currently I have: Game.java - creates the frame add panels to it import javax.swing.; import ui.; public class Game { private JFrame frame; Main main; Rules rules; Game() { rules = new Rules(); frame = new JFrame(); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); main = new Main(); frame.setContentPane(main.getContentPane()); show(); } void show() { frame.pack(); frame.setLocationRelativeTo(null); frame.setVisible(true); } public static void main(String[] args) { new Game(); } } Rules.java - game logic ui package - all classes create new panels to be swapped out with the main frame's content pane Main.java (Main Menu) - creates a panel with components Where do I now place the functionality for the Main class? In the game class? Separate class? Or is the whole organization wrong? Thanks

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  • Automatically Organize Tags in Tax/Folksonomy

    - by Rob Wilkerson
    I'm working on a process that will perform natural language processing (NLP) on one--and potentially several--of our content rich sites. What I'd like to do once the NLP is complete is to automatically organize the output (generally a set of terms that you might think of as tags given the prevalence of that metaphor) into some kind of standard or generally accepted organizational structure. In a perfect world, I'd really like this to be crowd sourced under the folksonomy concept (as opposed to a taxonomy) since the ultimate goal is to target/appeal to real people rather than "domain experts", but I'm open to ideas and best practices. For the obvious purpose of scalability, I'd like to automate the population of this tax/folksonomy so that "some guy" in the team/organization isn't responsible for looking at a bunch of words (with or without context) and arbitrarily fleshing out the contextual components of the tree. I have a few ideas for doing this that require some research to establish viability, but I have exactly zero practical experience with this sort of thing so the ideas really just boil down to stuff I made up that might perform some role in accomplishing the task. Imagining that others have vastly more experience with this sort of thing, I'm hoping that I can stand on your shoulders. Thanks for your thoughts and insights.

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  • How to obtain position in file (byte-position) from java scanner?

    - by september2010
    How to obtain a position in file (byte-position) from the java scanner? Scanner scanner = new Scanner(new File("file")); scanner.useDelimiter("abc"); scanner.hasNext(); String result = scanner.next(); and now: how to get the position of result in file (in bytes)? Using scanner.match().start() is not the answer, because it gives the position within internal buffer.

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  • Python: shutil.rmtree fails on Windows with 'Access is denied'

    - by Sridhar Ratnakumar
    In Python, when running shutil.rmtree over a folder that contains a read-only file, the following exception is printed: File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\Python26\lib\shutil.py", line 221, in rmtree onerror(os.remove, fullname, sys.exc_info()) File "C:\Python26\lib\shutil.py", line 219, in rmtree os.remove(fullname) WindowsError: [Error 5] Access is denied: 'build\\tcl\\tcl8.5\\msgs\\af.msg' Looking in File Properties dialog I noticed that af.msg file is set to be read-only. So the question is: what is the simplest workaround/fix to get around this problem - given that my intention is to do an equivalent of rm -rf build/ but on Windows? (without having to use third-party tools like unxutils or cygwin - as this code is targeted to be run on a bare Windows install with Python 2.6 w/ PyWin32 installed)

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  • Extract wav file from video file

    - by Nikos Steiakakis
    I am developing an application in which I need to extract the audio from a video. The audio needs to be extracted in .wav format but I do not have a problem with the video format. Any format will do, as long as I can extract the audio in a wav file. Currently I am using Windows Media Player COM control in a windows form to play the videos, but any other embedded player will do as well. Any suggestions on how to do this? Thanks

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  • Locking behaviour is different via network shares

    - by MattH
    I have been trying to lock a file so that other cloned services cannot access the file. I then read the file, and then move the file when finished. The Move is allowed by using FileShare.Delete. However in later testing, we found that this approach does not work if we are looking at a network share. I appreciate my approach may not have been the best, but my specific question is: Why does the below demo work against the local file, but not against the network file? The more specific you can be the better, as I've found very little information in my searches that indicates network shares behave differently to local disks. string sourceFile = @"C:\TestFile.txt"; string localPath = @"C:\MyLocalFolder\TestFile.txt"; string networkPath = @"\\MyMachine\MyNetworkFolder\TestFile.txt"; File.WriteAllText(sourceFile, "Test data"); if (!File.Exists(localPath)) File.Copy(sourceFile, localPath); foreach (string path in new string[] { localPath, networkPath }) { using (FileStream fsLock = File.Open(path, FileMode.Open, FileAccess.ReadWrite, (FileShare.Read | FileShare.Delete))) { string target = path + ".out"; File.Move(path, target); //This is the point of failure, when working with networkPath if (File.Exists(target)) File.Delete(target); } if (!File.Exists(path)) File.Copy(sourceFile, path); }

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  • shutil.rmtree fails on Windows with 'Access is denied'

    - by Sridhar Ratnakumar
    In Python, when running shutil.rmtree over a folder that contains a read-only file, the following exception is printed: File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 216, in rmtree rmtree(fullname, ignore_errors, onerror) File "C:\ActivePython32Python26\lib\shutil.py", line 221, in rmtree onerror(os.remove, fullname, sys.exc_info()) File "C:\ActivePython32Python26\lib\shutil.py", line 219, in rmtree os.remove(fullname) WindowsError: [Error 5] Access is denied: 'build\\pyhg_trunk-win32-x86-hgtip27\\image\\feature-core\\INSTALLDIR\\tcl\\tcl8.5\\msgs\\af.msg' Looking in File Properties dialog I noticed that af.msg file is set to be read-only. So the question is: what is the simplest workaround/fix to get around this problem - given that my intention is to do an equivalent of rm -rf build/ but on Windows? (without having to use unxutils or cygwin)

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  • Can't read excel file after creating it using File.WriteAllText() function

    - by Srikanth Mattihalli
    public void ExportDataSetToExcel(DataTable dt) { HttpResponse response = HttpContext.Current.Response; response.Clear(); response.Charset = "utf-8"; response.ContentEncoding = Encoding.GetEncoding("utf-8"); response.ContentType = "application/vnd.ms-excel"; Random Rand = new Random(); int iNum = Rand.Next(10000, 99999); string extension = ".xls"; string filenamepath = AppDomain.CurrentDomain.BaseDirectory + "graphs\\" + iNum + ".xls"; string file_path = "graphs/" + iNum + extension; response.AddHeader("Content-Disposition", "attachment;filename=\"" + iNum + "\""); string query = "insert into graphtable(graphtitle,graphpath,creategraph,year) VALUES('" + iNum.ToString() + "','" + file_path + "','" + true + "','" + DateTime.Now.Year.ToString() + "')"; try { int n = connect.UpdateDb(query); if (n > 0) { resultLabel.Text = "Merge Successfull"; } else { resultLabel.Text = " Merge Failed"; } resultLabel.Visible = true; } catch { } using (StringWriter sw = new StringWriter()) { using (HtmlTextWriter htw = new HtmlTextWriter(sw)) { // instantiate a datagrid DataGrid dg = new DataGrid(); dg.DataSource = dt; //ds.Tables[0]; dg.DataBind(); dg.RenderControl(htw); File.WriteAllText(filenamepath, sw.ToString()); // File.WriteAllText(filenamepath, sw.ToString(), Encoding.UTF8); response.Write(sw.ToString()); response.End(); } } } Hi all, I have created an excel sheet from datatable using above function. I want to read the excel sheet programatically using the below connectionstring. This string works fine for all other excel sheets but not for the one i created using the above function. I guess it is because of excel version problem. OleDbConnection conn= new OleDbConnection("Data Source='" + path +"';provider=Microsoft.Jet.OLEDB.4.0;Extended Properties=Excel 8.0;";); Can anyone suggest a way by which i can create an excel sheet such that it is readable again using above query. I cannot use Microsoft InterOp library as it is not supported by my host.

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  • Code to update HyperV Export file

    - by Andy Schneider
    I am using the HyperV Module from Codeplex to do a "config only" export from a 2008R2 Hyper-V server. In order to import the configuration on another HyperV server, I need to edit the value of CopyVMStorage in the EXP file. This file is an XML file. I wrote the following code in PowerShell to do the update for me. The variable $existing is the existing exp file. $xml = [xml](get-content $existing) $xpath = '//PROPERTY[@NAME ="CopyVmStorage"]' foreach ($node in $xml.SelectNodes($xpath)) {$node.Value = 'TRUE'} $xml.Save($existing) This code makes the correct changes to the XML. However, when I go to import the file on the Hyper-V server, I get an error that says the file format is incorrect. I am wondering if the encoding of the file is incorrect or if there is something else going on. If I edit the file manually in wordpad, it imports without an issue. The filename is a GUID with a .exp extension, and it appears that the file name is too long for notepad to open. Notepad throws an error trying to open the file, which is why I went with WordPad. I have noticed that the file that is updated with PowerShell comes out formatted whereas the raw file is xml all bunched together with no whitespace. Any ideas on what "file format" means in this HyperV error message and how I might be able to use my code to automate this change in the XML without changing the file format?

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  • How to read from database and write into text file with C#?

    - by user147685
    How to read from database and write into text file? I want to write/copy (not sure what to call) the record inside my database into a text file. One row record in database is equal to one line in the text file. I'm having no problem in database. For creating text file, it mentions FileStream and StreamWriter. Which one should I use?

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  • Getting a file pointer from file descriptor

    - by Naga Kiran
    In PHP 5.2.3, "fdopen" was used to read/write to a file descriptor that's opened by another application. fdopen(<fileDescriptorId>,"rw"); //It worked fine with PHP 5.2.3 After upgrading PHP to 5.3.2, it's throwing "undefined reference to 'fdopen' function". Please suggest whats the replacement for this in PHP 5.3.2 or any workaround.

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  • How to compare filename of uploaded file and string

    - by user225269
    I use this code to upload image files in xammp server: <?php if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 100000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file, File must be less than 100Kb in size with .jpg, .jpeg, or .gif file extension"; } ?> What do I do to compare the file name of the uploaded files with the text inputted by the user? My goal is to be able to compare the user input(ID number) and the file name of the image file which should also be an ID number. So that I will be able to display the image that corresponds with the ID Number provided. What do I need to do?Please give me an idea on how can I achieve this. Thanks

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