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  • Explained: EF 6 and “Could not determine storage version; a valid storage connection or a version hint is required.”

    - by Ken Cox [MVP]
    I have a legacy ASP.NET 3.5 web site that I’ve upgraded to a .NET 4 web application. At the same time, I upgraded to Entity Framework 6. Suddenly one of the pages returned the following error: [ArgumentException: Could not determine storage version; a valid storage connection or a version hint is required.]    System.Data.SqlClient.SqlVersionUtils.GetSqlVersion(String versionHint) +11372412    System.Data.SqlClient.SqlProviderServices.GetDbProviderManifest(String versionHint) +91    System.Data.Common.DbProviderServices.GetProviderManifest(String manifestToken) +92 [ProviderIncompatibleException: The provider did not return a ProviderManifest instance.]    System.Data.Common.DbProviderServices.GetProviderManifest(String manifestToken) +11431433    System.Data.Metadata.Edm.Loader.InitializeProviderManifest(Action`3 addError) +11370982    System.Data.EntityModel.SchemaObjectModel.Schema.HandleAttribute(XmlReader reader) +216 A search of the error message didn’t turn up anything helpful except that someone mentioned that the error messages was bogus in his case. The page in question uses the ASP.NET EntityDataSource control, consumed by a Telerik RadGrid. This is a fabulous combination for putting a huge amount of functionality on a page in a very short time. Unfortunately, the 6.0.1 release of EF6 doesn’t support EntityDataSource. According to the people in charge, support is planned but there’s no timeline for an EntityDataSource build that works with EF6.  I’m not sure what to do in the meantime. Should I back out EF6 or manually wire up the RadGrid? The upshot is that you might want to rethink plans to upgrade to Entity Framework 6 for Web forms projects if they rely on that handy control. It might also help to spend a User voice vote here:  http://data.uservoice.com/forums/72025-entity-framework-feature-suggestions/suggestions/3702890-support-for-asp-net-entitydatasource-and-dynamicda

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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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  • Architecture for a business objects / database access layer

    - by gregmac
    For various reasons, we are writing a new business objects/data storage library. One of the requirements of this layer is to separate the logic of the business rules, and the actual data storage layer. It is possible to have multiple data storage layers that implement access to the same object - for example, a main "database" data storage source that implements most objects, and another "ldap" source that implements a User object. In this scenario, User can optionally come from an LDAP source, perhaps with slightly different functionality (eg, not possible to save/update the User object), but otherwise it is used by the application the same way. Another data storage type might be a web service, or an external database. There are two main ways we are looking at implementing this, and me and a co-worker disagree on a fundamental level which is correct. I'd like some advice on which one is the best to use. I'll try to keep my descriptions of each as neutral as possible, as I'm looking for some objective view points here. Business objects are base classes, and data storage objects inherit business objects. Client code deals with data storage objects. In this case, common business rules are inherited by each data storage object, and it is the data storage objects that are directly used by the client code. This has the implication that client code determines which data storage method to use for a given object, because it has to explicitly declare an instance to that type of object. Client code needs to explicitly know connection information for each data storage type it is using. If a data storage layer implements different functionality for a given object, client code explicitly knows about it at compile time because the object looks different. If the data storage method is changed, client code has to be updated. Business objects encapsulate data storage objects. In this case, business objects are directly used by client application. Client application passes along base connection information to business layer. Decision about which data storage method a given object uses is made by business object code. Connection information would be a chunk of data taken from a config file (client app does not really know/care about details of it), which may be a single connection string for a database, or several pieces connection strings for various data storage types. Additional data storage connection types could also be read from another spot - eg, a configuration table in a database that specifies URLs to various web services. The benefit here is that if a new data storage method is added to an existing object, a configuration setting can be set at runtime to determine which method to use, and it is completely transparent to the client applications. Client apps do not need to be modified if data storage method for a given object changes. Business objects are base classes, data source objects inherit from business objects. Client code deals primarily with base classes. This is similar to the first method, but client code declares variables of the base business object types, and Load()/Create()/etc static methods on the business objects return the appropriate data source-typed objects. The architecture of this solution is similar to the first method, but the main difference is the decision about which data storage object to use for a given business object is made by the business layer, not the client code. I know there are already existing ORM libraries that provide some of this functionality, but please discount those for now (there is the possibility that a data storage layer is implemented with one of these ORM libraries) - also note I'm deliberately not telling you what language is being used here, other than that it is strongly typed. I'm looking for some general advice here on which method is better to use (or feel free to suggest something else), and why.

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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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  • Uncompress OpenOffice files for better storage in version control

    - by Craig McQueen
    I've heard discussion about how OpenOffice (ODF) files are compressed zip files of XML and other data. So making a tiny change to the file can potentially totally change the data, so delta compression doesn't work well in version control systems. I've done basic testing on an OpenOffice file, unzipping it and then rezipping it with zero compression. I used the Linux zip utility for my testing. OpenOffice will still happily open it. So I'm wondering if it's worth developing a small utility to run on ODF files each time just before I commit to version control. Any thoughts on this idea? Possible better alternatives? Secondly, what would be a good and robust way to implement this little utility? Bash shell that calls zip (probably Linux only)? Python? Any gotchas you can think of? Obviously I don't want to accidentally mangle a file, and there are several ways that could happen. Possible gotchas I can think of: Insufficient disk space Some other permissions issue that prevents writing the file or temporary files ODF document is encrypted (probably should just leave these alone; the encryption probably also causes large file changes and thus prevents efficient delta compression)

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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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  • Program crashes after trying to use a recently created file. C#

    - by Jason T.
    So here is my code if (!File.Exists(pathName)) { File.Create(pathName); } StreamWriter outputFile = new StreamWriter(pathName,true); But whenever I run the program the first time the path with file gets created. However once I get to the StreamWriter line my program crashes because it says my fie is in use by another process. Is there something I'm missing between the File.Create and the StreamWriter statements?

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