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  • SQL SERVER – Windows File/Folder and Share Permissions – Notes from the Field #029

    - by Pinal Dave
    [Note from Pinal]: This is a 29th episode of Notes from the Field series. Security is the task which we should give it to the experts. If there is a small overlook or misstep, there are good chances that security of the organization is compromised. This is very true, but there are always devils’s advocates who believe everyone should know the security. As a DBA and Administrator, I often see people not taking interest in the Windows Security hiding behind the reason of not expert of Windows Server. We all often miss the important mission statement for the success of any organization – Teamwork. In this blog post Brian tells the story in very interesting lucid language. Read On! In this episode of the Notes from the Field series database expert Brian Kelley explains a very crucial issue DBAs and Developer faces on their production server. Linchpin People are database coaches and wellness experts for a data driven world. Read the experience of Brian in his own words. When I talk security among database professionals, I find that most have at least a working knowledge of how to apply security within a database. When I talk with DBAs in particular, I find that most have at least a working knowledge of security at the server level if we’re speaking of SQL Server. One area I see continually that is weak is in the area of Windows file/folder (NTFS) and share permissions. The typical response is, “I’m a database developer and the Windows system administrator is responsible for that.” That may very well be true – the system administrator may have the primary responsibility and accountability for file/folder and share security for the server. However, if you’re involved in the typical activities surrounding databases and moving data around, you should know these permissions, too. Otherwise, you could be setting yourself up where someone is able to get to data he or she shouldn’t, or you could be opening the door where human error puts bad data in your production system. File/Folder Permission Basics: I wrote about file/folder permissions a few years ago to give the basic permissions that are most often seen. Here’s what you must know as a minimum at the file/folder level: Read - Allows you to read the contents of the file or folder. Having read permissions allows you to copy the file or folder. Write  – Again, as the name implies, it allows you to write to the file or folder. This doesn’t include the ability to delete, however, nothing stops a person with this access from writing an empty file. Delete - Allows the file/folder to be deleted. If you overwrite files, you may need this permission. Modify - Allows read, write, and delete. Full Control - Same as modify + the ability to assign permissions. File/Folder permissions aggregate, unless there is a DENY (where it trumps, just like within SQL Server), meaning if a person is in one group that gives Read and antoher group that gives Write, that person has both Read and Write permissions. As you might expect me to say, always apply the Principle of Least Privilege. This likely means that any additional permission you might add does not need Full Control. Share Permission Basics: At the share level, here are the permissions. Read - Allows you to read the contents on the share. Change - Allows you to read, write, and delete contents on the share. Full control - Change + the ability to modify permissions. Like with file/folder permissions, these permissions aggregate, and DENY trumps. So What Access Does a Person / Process Have? Figuring out what someone or some process has depends on how the location is being accessed: Access comes through the share (\\ServerName\Share) – a combination of permissions is considered. Access is through a drive letter (C:\, E:\, S:\, etc.) – only the file/folder permissions are considered. The only complicated one here is access through the share. Here’s what Windows does: Figures out what the aggregated permissions are at the file/folder level. Figures out what the aggregated permissions are at the share level. Takes the most restrictive of the two sets of permissions. You can test this by granting Full Control over a folder (this is likely already in place for the Users local group) and then setting up a share. Give only Read access through the share, and that includes to Administrators (if you’re creating a share, likely you have membership in the Administrators group). Try to read a file through the share. Now try to modify it. The most restrictive permission is the Share level permissions. It’s set to only allow Read. Therefore, if you come through the share, it’s the most restrictive. Does This Knowledge Really Help Me? In my experience, it does. I’ve seen cases where sensitive files were accessible by every authenticated user through a share. Auditors, as you might expect, have a real problem with that. I’ve also seen cases where files to be imported as part of the nightly processing were overwritten by files intended from development. And I’ve seen cases where a process can’t get to the files it needs for a process because someone changed the permissions. If you know file/folder and share permissions, you can spot and correct these types of security flaws. Given that there are a lot of database professionals that don’t understand these permissions, if you know it, you set yourself apart. And if you’re able to help on critical processes, you begin to set yourself up as a linchpin (link to .pdf) for your organization. If you want to get started with performance tuning and database security with the help of experts, read more over at Fix Your SQL Server. Reference: Pinal Dave (http://blog.sqlauthority.com)Filed under: Notes from the Field, PostADay, SQL, SQL Authority, SQL Query, SQL Security, SQL Server, SQL Tips and Tricks, T SQL

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  • problems with retrieving data that was saved outside of a grails webflow

    - by callie16
    Hi, this is actually connected to an earlier question of mine here. Anyway, I have 3 domains that look like this: class A { ... static hasMany = [ b : B ] ... } class B { ... static belongsTo = [ a : A ] static hasMany = [ c : C ] ... } class C { ... static belongsTo = [ b : B ] ... } In my GSP page, I call an action in the Controller via a remote function in a javascript block (I'm using Dojo so I'm passing data to be saved this way... it's not a form per se so I use JSON for now to pass the data to the Controller). Let's say, I'm calling something like this: def someAction = { def jsonArr = [parse the JSON here] def tmpA = A.get(params.id) ... def tmpB = new B() b.someParam = jsonArr.someParam ... def tmpC = new C() tmpC.cParam = jsonArr.cParam tmpB.addToC(tmpC) tmpB.save(flush: true) //this may or may not be here but I'm adding it for the sake of completeness tmpA.addToB(tmpB) tmpA.save(flush: true) // NOTE: If I check here via println or whatnot, tmpA has a tmpB which has a tmpC... in other words, the data got saved. It's also in the DB. redirect(action: 'order' ...) } Then comes the fun part. Here's the webflow sample: def orderFlow = { ... someStateIShouldEndUpIn { on("next") { // or on previous... doesn't matter def anId = params.id def currA = A.get(anId) // this does NOT return a null value def testB = currA.b // this DOES return a null value }.to("somePage") ... } ... } Any ideas on why this happens? Moreover, when I dump the data of currA, b=null... instead of b=[] or b=[contents of tmpB]. Any help would be seriously appreciated... been at this for a couple of days now... Thanks!

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  • Can't save my picture

    - by mamii
    I want to save the image that I draw, but I always failure is reported. I have tested and tried but I can correct any errors. Therefore, I appeal to you. This store is for me as a "cancer sore". And what is the drawing application without the possibility shranjevnja? sucks: D Question: What is wrong with my code for storage? or anything else? Posts: 09-12 07:30:34.346: E / Panel (8003): IOEception 09-12 07:30:34.346: E / Panel (8003): java.io.IOException: Parent directory of file does not exist: / sdcard/anppp/2012Sep1273034.png 09-12 07:30:34.346: E / Panel (8003): at java.io.File.createNewFile (File.java: 1263) 09-12 07:30:34.346: E / Panel (8003): at aa.bb.cc.Panel.saveapp (Panel.java: 67) 09-12 07:30:34.346: E / Panel (8003): at aa.bb.cc.AndroidPaint.onOptionsItemSelected (AndroidPaint.java: 94) 09-12 07:30:34.346: E / Panel (8003): at android.app.Activity.onMenuItemSelected (Activity.java: 2170) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.policy.impl.PhoneWindow.onMenuItemSelected (PhoneWindow.java: 730) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.view.menu.MenuItemImpl.invoke (MenuItemImpl.java: 139) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.view.menu.MenuBuilder.performItemAction (MenuBuilder.java: 855) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.view.menu.ExpandedMenuView.invokeItem (ExpandedMenuView.java: 89) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.view.menu.ExpandedMenuView.onItemClick (ExpandedMenuView.java: 93) 09-12 07:30:34.346: E / Panel (8003): at android.widget.AdapterView.performItemClick (AdapterView.java: 284) 09-12 07:30:34.346: E / Panel (8003): at android.widget.ListView.performItemClick (ListView.java: 3285) 09-12 07:30:34.346: E / Panel (8003): at android.widget.AbsListView $ PerformClick.run (AbsListView.java: 1640) 09-12 07:30:34.346: E / Panel (8003): at android.os.Handler.handleCallback (Handler.java: 587) 09-12 07:30:34.346: E / Panel (8003): at android.os.Handler.dispatchMessage (Handler.java: 92) 09-12 07:30:34.346: E / Panel (8003): at android.os.Looper.loop (Looper.java: 123) 09-12 07:30:34.346: E / Panel (8003): at android.app.ActivityThread.main (ActivityThread.java: 4363) 09-12 07:30:34.346: E / Panel (8003): at java.lang.reflect.Method.invokeNative (Native Method) 09-12 07:30:34.346: E / Panel (8003): at java.lang.reflect.Method.invoke (Method.java: 521) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.os.ZygoteInit $ MethodAndArgsCaller.run (ZygoteInit.java: 860) 09-12 07:30:34.346: E / Panel (8003): at com.android.internal.os.ZygoteInit.main (ZygoteInit.java: 618) 09-12 07:30:34.346: E / Panel (8003): at dalvik.system.NativeStart.main (Native Method) There is code: private Bitmap mBitmap; private Canvas mCanvas; private Bitmap tmpBitmap; private Canvas tmpCanvas; private DrawHandler mDrawHandler; private Canvas tCanvas; private String mImagePath = Environment.getExternalStorageDirectory() + "/anppp"; private File file; public void saveapp() { Calendar currentDate = Calendar.getInstance(); SimpleDateFormat formatter= new SimpleDateFormat("yyyyMMMddHmmss"); String dateNow = formatter.format(currentDate.getTime()); file = new File(mImagePath + "/" + dateNow +".png"); FileOutputStream fos; try { file.createNewFile(); fos = new FileOutputStream(file); tmpBitmap.compress(Bitmap.CompressFormat.PNG, 100, fos); fos.close(); } catch (FileNotFoundException e) { Log.e("Panel", "FileNotFoundException", e); } catch (IOException e) { Log.e("Panel", "IOEception", e); } } That's it .. I do not know what could be wrong ;(

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  • Using the HTML5 &lt;input type=&quot;file&quot; multiple=&quot;multiple&quot;&gt; Tag in ASP.NET

    - by Rick Strahl
    Per HTML5 spec the <input type="file" /> tag allows for multiple files to be picked from a single File upload button. This is actually a very subtle change that's very useful as it makes it much easier to send multiple files to the server without using complex uploader controls. Please understand though, that even though you can send multiple files using the <input type="file" /> tag, the process of how those files are sent hasn't really changed - there's still no progress information or other hooks that allow you to automatically make for a nicer upload experience without additional libraries or code. For that you will still need some sort of library (I'll post an example in my next blog post using plUpload). All the new features allow for is to make it easier to select multiple images from disk in one operation. Where you might have required many file upload controls before to upload several files, one File control can potentially do the job. How it works To create a file input box that allows with multiple file support you can simply do:<form method="post" enctype="multipart/form-data"> <label>Upload Images:</label> <input type="file" multiple="multiple" name="File1" id="File1" accept="image/*" /> <hr /> <input type="submit" id="btnUpload" value="Upload Images" /> </form> Now when the file open dialog pops up - depending on the browser and whether the browser supports it - you can pick multiple files. Here I'm using Firefox using the thumbnail preview I can easily pick images to upload on a form: Note that I can select multiple images in the dialog all of which get stored in the file textbox. The UI for this can be different in some browsers. For example Chrome displays 3 files selected as text next to the Browse… button when I choose three rather than showing any files in the textbox. Most other browsers display the standard file input box and display the multiple filenames as a comma delimited list in the textbox. Note that you can also specify the accept attribute in the <input> tag, which specifies a mime-type to specify what type of content to allow.Here I'm only allowing images (image/*) and the browser complies by just showing me image files to display. Likewise I could use text/* for all text formats registered on the machine or text/xml to only show XML files (which would include xml,xst,xsd etc.). Capturing Files on the Server with ASP.NET When you upload files to an ASP.NET server there are a couple of things to be aware of. When multiple files are uploaded from a single file control, they are assigned the same name. In other words if I select 3 files to upload on the File1 control shown above I get three file form variables named File1. This means I can't easily retrieve files by their name:HttpPostedFileBase file = Request.Files["File1"]; because there will be multiple files for a given name. The above only selects the first file. Instead you can only reliably retrieve files by their index. Below is an example I use in app to capture a number of images uploaded and store them into a database using a business object and EF 4.2.for (int i = 0; i < Request.Files.Count; i++) { HttpPostedFileBase file = Request.Files[i]; if (file.ContentLength == 0) continue; if (file.ContentLength > App.Configuration.MaxImageUploadSize) { ErrorDisplay.ShowError("File " + file.FileName + " is too large. Max upload size is: " + App.Configuration.MaxImageUploadSize); return View("UploadClassic",model); } var image = new ClassifiedsBusiness.Image(); var ms = new MemoryStream(16498); file.InputStream.CopyTo(ms); image.Entered = DateTime.Now; image.EntryId = model.Entry.Id; image.ContentType = "image/jpeg"; image.ImageData = ms.ToArray(); ms.Seek(0, SeekOrigin.Begin); // resize image if necessary and turn into jpeg Bitmap bmp = Imaging.ResizeImage(ms.ToArray(), App.Configuration.MaxImageWidth, App.Configuration.MaxImageHeight); ms.Close(); ms = new MemoryStream(); bmp.Save(ms,ImageFormat.Jpeg); image.ImageData = ms.ToArray(); bmp.Dispose(); ms.Close(); model.Entry.Images.Add(image); } This works great and also allows you to capture input from multiple input controls if you are dealing with browsers that don't support multiple file selections in the file upload control. The important thing here is that I iterate over the files by index, rather than using a foreach loop over the Request.Files collection. The files collection returns key name strings, rather than the actual files (who thought that was good idea at Microsoft?), and so that isn't going to work since you end up getting multiple keys with the same name. Instead a plain for loop has to be used to loop over all files. Another Option in ASP.NET MVC If you're using ASP.NET MVC you can use the code above as well, but you have yet another option to capture multiple uploaded files by using a parameter for your post action method.public ActionResult Save(HttpPostedFileBase[] file1) { foreach (var file in file1) { if (file.ContentLength < 0) continue; // do something with the file }} Note that in order for this to work you have to specify each posted file variable individually in the parameter list. This works great if you have a single file upload to deal with. You can also pass this in addition to your main model to separate out a ViewModel and a set of uploaded files:public ActionResult Edit(EntryViewModel model,HttpPostedFileBase[] uploadedFile) You can also make the uploaded files part of the ViewModel itself - just make sure you use the appropriate naming for the variable name in the HTML document (since there's Html.FileFor() extension). Browser Support You knew this was coming, right? The feature is really nice, but unfortunately not supported universally yet. Once again Internet Explorer is the problem: No shipping version of Internet Explorer supports multiple file uploads. IE10 supposedly will, but even IE9 does not. All other major browsers - Chrome, Firefox, Safari and Opera - support multi-file uploads in their latest versions. So how can you handle this? If you need to provide multiple file uploads you can simply add multiple file selection boxes and let people either select multiple files with a single upload file box or use multiples. Alternately you can do some browser detection and if IE is used simply show the extra file upload boxes. It's not ideal, but either one of these approaches makes life easier for folks that use a decent browser and leaves you with a functional interface for those that don't. Here's a UI I recently built as an alternate uploader with multiple file upload buttons: I say this is my 'alternate' uploader - for my primary uploader I continue to use an add-in solution. Specifically I use plUpload and I'll discuss how that's implemented in my next post. Although I think that plUpload (and many of the other packaged JavaScript upload solutions) are a better choice especially for large uploads, for simple one file uploads input boxes work well enough. The advantage of this solution is that it's very easy to handle on the server side. Any of the JavaScript controls require special handling for uploads which I'll also discuss in my next post.© Rick Strahl, West Wind Technologies, 2005-2012Posted in HTML5  ASP.NET  MVC   Tweet !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="//platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() { var po = document.createElement('script'); po.type = 'text/javascript'; po.async = true; po.src = 'https://apis.google.com/js/plusone.js'; var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(po, s); })();

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  • Making file in user's homedir accessible from web/webserver

    - by evident
    Hi everybody, I have a txt-file one of my user's homedir which is regularly updated there by a script. I now want to be able to access (read) this file from the web. /home/user/folder/file.txt So what I tried now is to log in as root, go into my webservers httpdocs folder /var/www/path/to/domain/httpdocs and there I tried to create a symbolic link with ln -s /home/user/foler/file.txt /var/www/path/to/domain/httpdocs/file.txt But this didn't work... I already tried changing the chmod of the symlink (which changes the ones from the original file of course) and also a chown to the user from webserver, but no matter what I tried I cannot open the file from the web or from a php-script (which is what I want to do) Can anybody help me and tell me what I need to do? What rights do I need to give? Or is there another way of achieving this?

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  • Uploading a file automatically for speed test?

    - by Abhi
    I am building a Web UI for a device for internet connection and one of the requirements in it is a speed test. I know the basic concept of how speed test works. A file is downloaded for a limited time then the same file is uploaded again and the speed is tracked at regular intervals. Downloading the file is not an issue, but how am I supposed to upload the file without the client knowing that the file is getting uploaded? I've read through a lot of documentation, but I'm still not able to get the answer to how I will upload the file from clients machine without asking him to select the file.

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  • Edit 100MB+ file

    - by Majid Fouladpour
    I have captured some traffic with Wireshark and saved the result as a file. The file has 3 sections now: request headers response headers response body The response body is to become an flv file, but now everything is saved as a single file. So I need a way to delete the first two sections from the file, but the problem is that the file is very big (over a thousand mega bytes). I have tried to open it with gedit, but no matter how long I wait, gedit hangs and remains unresponsive until I kill it. What tool can I use to edit this big file easily?

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  • Entity Framework: Connecting to a mdf user database file via localDB during script execution

    - by Marko Apfel
    Problem If you run the “Generate database from model” wizard and execute the generated script the destination database could be the wrong one (for instance master of the SQL Server). Solution To use an own mdf attachable user database some connection information must specified during script execution. Execute your script opens the dialog “Connect to Server”. Press “Options” and go to the second tab “Connection Properties”. Select “Browse server” in the “Connect to database” dropdown box: Confirm the information dialog with “Yes”. In the following dialog you could choose your user database. Now the schema is created in the user database.

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  • File Upload with HttpWebRequest doesn't post the file

    - by Sri Kumar
    Hello All, Here is my code to post the file. I use asp fileupload control to get the file stream. HttpWebRequest requestToSender = (HttpWebRequest)WebRequest.Create("http://localhost:2518/Web/CrossPage.aspx"); requestToSender.Method = "POST"; requestToSender.ContentType = "multipart/form-data"; requestToSender.KeepAlive = true; requestToSender.Credentials = System.Net.CredentialCache.DefaultCredentials; requestToSender.ContentLength = BtnUpload.PostedFile.ContentLength; BinaryReader binaryReader = new BinaryReader(BtnUpload.PostedFile.InputStream); byte[] binData = binaryReader.ReadBytes(BtnUpload.PostedFile.ContentLength); Stream requestStream = requestToSender.GetRequestStream(); requestStream.Write(binData, 0, binData.Length); requestStream.Close(); HttpWebResponse responseFromSender = (HttpWebResponse)requestToSender.GetResponse(); string fromSender = string.Empty; using (StreamReader responseReader = new StreamReader(responseFromSender.GetResponseStream())) { fromSender = responseReader.ReadToEnd(); } XMLString.Text = fromSender; In the page load of CrossPage.aspx i have the following code NameValueCollection postPageCollection = Request.Form; foreach (string name in postPageCollection.AllKeys) { Response.Write(name + " " + postPageCollection[name]); } HttpFileCollection postCollection = Request.Files; foreach (string name in postCollection.AllKeys) { HttpPostedFile aFile = postCollection[name]; aFile.SaveAs(Server.MapPath(".") + "/" + Path.GetFileName(aFile.FileName)); } string strxml = "sample"; Response.Clear(); Response.Write(strxml); I don't get the file in Request.Files. The byte array is created. What was wrong with my HttpWebRequest?

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  • Trouble converting an MP3 file to a WAV file using Naudio

    - by WebDevHobo
    Naudio Library: http://naudio.codeplex.com/ I'm trying to convert an MP3 file to a WAV file, but I've run in to a small error. I know what's going wrong, but I don't really know how to go about fixing it. Here's the piece of code I'm running: private void button1_Click(object sender, EventArgs e) { using(Mp3FileReader reader = new Mp3FileReader(@"path\to\MP3")) { using(WaveFileWriter writer = new WaveFileWriter(@"C:\test.wav", new WaveFormat())) { int counter = 0; while(reader.Read(test, counter, test.Length + counter) != 0) { writer.WriteData(test, counter, test.Length + counter); counter += 512; } } } } reader.Read() goes into the Mp3FileReader class, and the method looks like this: public override int Read(byte[] sampleBuffer, int offset, int numBytes) { if (numBytes % waveFormat.BlockAlign != 0) //throw new ApplicationException("Must read complete blocks"); numBytes -= (numBytes % waveFormat.BlockAlign); return mp3Stream.Read(sampleBuffer, offset, numBytes); } mp3Stream is an object of the Stream class. The problem is: I'm getting an ArgumentException. MSDN says that this is because the sum of offset and numBytes is greater than the length of sampleBuffer. Documentation: http://msdn.microsoft.com/en-us/library/system.io.stream.read.aspx This happens because I increase the counter every time, but the size of the byte array test remains the same. What I've been wondering is: do I need to increase the size of the array dynamically, or do I need to find out the needed size at the beginning and set it right away? And also, instead of 512, the method in Mp3FileReader returns 365 the first time. Which is the size of a whole block. But I'm writing the full 512. I'm basically just using the read to check if I'm not at the end of the file yet. Do I need to catch the return value and do something with that, or am I good here?

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  • Pentaho - Reporting Tool - Is the .prpt file (report template file) contains datasource information

    - by Yatendra Goel
    I am new Pentaho Reporting Tool. I have the following question: When I created a report using Pentaho Report Designer, it output a report file having .prpt extension. After that I found an example on internet where the following code were used to display the report in html format:| protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { ResourceManager manager = new ResourceManager(); manager.registerDefaults(); String reportPath = "file:" + this.getServletContext().getRealPath("sampleReport.prpt"); try { Resource res = manager.createDirectly(new URL(reportPath), MasterReport.class); MasterReport report = (MasterReport) res.getResource(); HtmlReportUtil.createStreamHTML(report, response.getOutputStream()); } catch (Exception e) { e.printStackTrace(); } } And the report got printed successfully. So as we haven't specified any datasource information here, I think that the .prpt file contains that information in it. If that's true than Isn't Jasper is better Reporting tool than Pentaho because when we display Jasper reports, we have to provide datasource details also so in that way our report is flexible and is not bound to any particular database.

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  • How get file names using OpenFileDialog in .NET (1000+ file multiselect)

    - by Cole
    Maybe some of you have come across this before.... I am opening files for parsing. I'm using OpenFileDialog, of course, but i'm limited to a buffer of 2048 on the .FileNames string. Thus, I can only select a few hundred files. This is OK for most cases. However, fore example, I have in one case 1400 files to open. Do you know a way to do this with the open file dialog. I just want the string array of .FileNames, I pass that to parser class. I was also thinking of offering a FolderBrowserDialog option and then I'd use some other method to just loop through all the files in a directory, like the DirectoryInfo class. I'd do this as a last resort if I can't have an all in one solution.

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  • Open File Dialog Asp.Net

    - by Nick LaMarca
    I am creating an excel report in vb.net using the office interop. When the report is completed I am saving the excel file on the C drive. The users have asked to save file anywhere they want not just the c drive. Can someone give me some code to popup an opend file dialog in asp.net? I want the dialog to popup in a saveAs in ASP.NET. I know how to do it in win forms, but I am creating an excel report in asp.net and calling the worksheet objects SaveAs property that excepts a fileName. So right now I just hardcode a file name in there. The users want to choose a file location

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  • Python: saving objects and using pickle. Error using pickle.dump

    - by Peterstone
    Hello I have an Error and I don´t the reason: >>> class Fruits:pass ... >>> banana = Fruits() >>> banana.color = 'yellow' >>> banana.value = 30 >>> import pickle >>> filehandler = open("Fruits.obj",'w') >>> pickle.dump(banana,filehandler) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "C:\Python31\lib\pickle.py", line 1354, in dump Pickler(file, protocol, fix_imports=fix_imports).dump(obj) TypeError: must be str, not bytes >>> I don´t know how to solve this error because I don´t understand it. Thank you so much.

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  • Django/jQuery - read file and pass to browser as file download prompt

    - by danspants
    I've previously asked a question regarding passing files to the browser so a user receives a download prompt. However these files were really just strings creatd at the end of a function and it was simple to pass them to an iframe's src attribute for the desired effect. Now I have a more ambitious requirement, I need to pass pre existing files of any format to the browser. I have attempted this using the following code: def return_file(request): try: bob=open(urllib.unquote(request.POST["file"]),"rb") response=HttpResponse(content=bob,mimetype="application/x-unknown") response["Content-Disposition"] = "attachment; filename=nothing.xls" return HttpResponse(response) except: return HttpResponse(sys.exc_info()) With my original setup the following jQuery was sufficient to give the desired download prompt: jQuery('#download').attr("src","/return_file/"); However this won't work anymore as I need to pass POST values to the function. my attempt to rectify that is below, but instead of a download prompt I get the file displayed as text. jQuery.get("/return_file/",{"file":"c:/filename.xls"},function(data) { jQuery(thisButton).children("iframe").attr("src",data); }); Any ideas as to where I'm going wrong? Thanks!

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  • Need a way to determine if a file is done being written to.

    - by Khorkrak
    The situation I'm in is this - there's a process that's writing to a file, sometimes the file is rather large say 400 - 500MB. I need to know when it's done writing. How can I determine this? If I look in the directory I'll see it there but it might not be done being written. Plus this needs to be done remotely - as in on the same internal LAN but not on the same computer and typically the process that wants to know when the file writing is done is running on a Linux box with a the process that's writing the file and the file itself on a windows box. No samba isn't an option. xmlrpc communication to a service on that windows box is an option as well as using snmp to check if that's viable. Ideally Works on either Linux or Windows - meaning the solution is OS independent. Works for any type of file. Good enough: Works just on windows but can be done through some library or whatever that can be accessed with Python. Works only for PDF files. Current best idea is to periodically open the file in question from some process on the windows box and look at the last bytes checking for the PDF end tag and accounting for the eol differences because the file may have been created on Linux or Windows.

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  • Deploy an hta file in an exe or msi file

    - by rjmunro
    I've written an HTA file frontend for our web app that allows the web app to run without web browser status bars etc, and allows it to access the local system for certain tasks. I need a way to deploy this to customers. I need an installer to supply an hta file, an ico file, and add a link to them in the start menu and on the users desktop. I looked at building an installer with NSIS, but I couldn't figure out how to assign the icon to the shortcuts - The icon had to be a standard HTA one. Can this be done with NSIS, or should I be using another installer? P.s. I've got no particular preference for NSIS, it's just something I once used a very long time ago. When I download stuff, I think I prefer them to be msi files that launch with windows installer (it feels more like downloading a .rpm or .deb on linux which I am used to) but I know nothing about how those are created. I'm a web/linux guy who knows very little about windows programming.

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  • Rails - Permission denied when try to save uploaded file in windows

    - by logoin
    I'm writing my own file upload in rails. I saw some related questions but it doesn't answer my question. I use File.open ("#{RAILS_ROOT}/public/docs/attachments/#{@file_name}", "wb") {|f| f.write(@temp_file.read)} to write the file on my local machine (OS: Windows XP) instead of saving it in database. I got a Permission denied error on the File.open method. Since I have cygwin installed, I chmod 777 the folder that files should write to and also make sure the file I upload can be read. But I'm still getting the same error. Any ideas? Thanks!

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  • Reading a binary file in perl: Bad File Descriptor

    - by Magicked
    I'm trying to read a binary file 40 bytes at a time, then check to see if all those bytes are 0x00, and if so ignore them. If not, it will write them back out to another file (basically just cutting out large blocks of null bytes). This may not be the most efficient way to do this, but I'm not worried about that. However, right now I'm getting a "Bad File Descriptor" error and I cannot figure out why. my $comp = "\x00" * 40; my $byte_count = 0; my $infile = "/home/magicked/image1"; my $outfile = "/home/magicked/image1_short"; open IN, "<$infile"; open OUT, ">$outfile"; binmode IN; binmode OUT; my ($buf, $data, $n); while (read (IN, $buf, 40)) { ### Problem is here ### $boo = 1; for ($i = 0; $i < 40; $i++) { if ($comp[$i] != $buf[$i]) { $i = 40; print OUT $buf; $byte_count += 40; } } } die "Problems! $!\n" if $!; close OUT; close IN; I marked with a comment where it is breaking. Thanks for any help!

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  • Java - Reading a csv file line by line - stuck with weird non-existent characters being read!

    - by rockit
    hello fellow java developers. I'm having a very strange issue. I'm trying to read a csv file line by line. Im at the point where Im just testing out the reading of the lines. ONly each time that I read a line, the line contains square characters between each character of text. I even saved the file as a txt file in wordpad and notepad with no change. Thus I must be doing something stupid... I have a csv file, standard csv file, yes a text file with commas in it. I try to read a line of text, but the text is all f-ed up and cannot find the phrase within the text. Any advice? code below. //open csv File filReadMe = new File(strRoot + "data2.csv"); BufferedReader brReadMe = new BufferedReader(new InputStreamReader(new FileInputStream(filReadMe))); String strLine = brReadMe.readLine(); //for all lines while (strLine != null){ //if line contains "(see also" if (strLine.toLowerCase().contains("(see also")){ //write line from "(see also" to ")" int iBegin = strLine.toLowerCase().indexOf("(see also"); String strTemp = strLine.substring(iBegin); int iLittleEnd = strTemp.indexOf(")"); System.out.println(strLine.substring(iBegin, iBegin + iLittleEnd)); } //update line strLine = brReadMe.readLine(); } //end for brReadMe.close();

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  • Short file names versus long file names in Windows

    - by normski
    I have some code which gets the short name from a file path, using GetShortNameW(), and then later retrieves the long name view GetLongNameA(). The original file is of the form "C:/ProgramData/My Folder/File.ext" However, following conversion to short, then back to long, the filename becomes "C:/Program Files/My Folder/Filename.ext". The short name is of the form "C:/PROGRA~2/MY_FOL~1/FIL~1.EXT" The short name is being incorrectly resolved. The code compiles using VS 2005 on Windows 7 (I cannot upgrade the project to VS2008) Does anybody have any idea why this might be happening? DWORD pathLengthNeeded = ::GetShortPathNameW(aRef->GetFilePath().c_str(), NULL, 0); if(pathLengthNeeded != 0) { WCHAR* shortPath = new WCHAR[pathLengthNeeded]; DWORD newPathNameLength = ::GetShortPathNameW(aRef->GetFilePath().c_str(), shortPath, pathLengthNeeded); if(newPathNameLength != 0) { UI_STRING unicodePath(shortPath); std::string asciiPath = StringFromUserString(unicodePath); pathLengthNeeded = ::GetLongPathNameA(asciiPath.c_str(),NULL, 0); if(pathLengthNeeded != 0) {// convert it back to a long path if possible. For goodness sake can't we use Unicode throughout?F char* longPath = new char[pathLengthNeeded]; DWORD newPathNameLength = ::GetLongPathNameA(asciiPath.c_str(), longPath, pathLengthNeeded); if(newPathNameLength != 0) { std::string longPathString(longPath, newPathNameLength); asciiPath = longPathString; } delete [] longPath; } SetFullPathName(asciiPath); } delete [] shortPath; }

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  • Unable to delete a file using bash script

    - by user3719091
    I'm having problems removing a file in a bash script. I saw the other post with the same problem but none of those solutions solved my problem. The bash script is an OP5 surveillance check and it calls an Expect process that saves a temporary file to the local drive which the bash script reads from. Once it has read the file and checked its status I would like to remove the temporary file. I'm pretty new to scripting so my script may not be as optimal as it can be. Either way it does the job except removing the file once it's done. I will post the entire code below: #!/bin/bash #GET FLAGS while getopts H:c:w: option do case "${option}" in H) HOSTADDRESS=${OPTARG};; c) CRITICAL=${OPTARG};; w) WARNING=${OPTARG};; esac done ./expect.vpn.check.sh $HOSTADDRESS #VARIABLES VPNCount=$(grep -o '[0-9]\+' $HOSTADDRESS.op5.vpn.results) # Check if the temporary results file exists if [ -f $HOSTADDRESS.op5.vpn.results ] then # If the file exist, Print "File Found" message echo Temporary results file exist. Analyze results. else # If the file does NOT exist, print "File NOT Found" message and send message to OP5 echo Temporary results file does NOT exist. Unable to analyze. # Exit with status Critical (exit code 2) exit 2 fi if [[ "$VPNCount" > $CRITICAL ]] then # If the amount of tunnels exceeds the critical threshold, echo out a warning message and current threshold and send warning to OP5 echo "The amount of VPN tunnels exceeds the critical threshold - ($VPNCount)" # Exit with status Critical (exit code 2) exit 2 elif [[ "$VPNCount" > $WARNING ]] then # If the amount of tunnels exceeds the warning threshold, echo out a warning message and current threshold and send warning to OP5 echo "The amount of VPN tunnels exceeds the warning threshold - ($VPNCount)" # Exit with status Warning (exit code 1) exit 1 else # The amount of tunnels do not exceed the warning threshold. # Print an OK message echo OK - $VPNCount # Exit with status OK exit 0 fi #Clean up temporary files. rm -f $HOSTADDRESS.op5.vpn.results I have tried the following solutions: Create a separate variable called TempFile that specifies the file. And specify that in the rm command. I tried creating another if statement similar to the one I use to verify that file exist and then rm the filename. I tried adding the complete name of the file (no variables, just plain text of the file) I can: Remove the file using the full name in both a separate script and directly in the CLI. Is there something in my script that locks the file that prevents me from removing it? I'm not sure what to try next. Thanks in advance!

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  • How to copy file preserving directory path in Linux?

    - by gasan
    I have Eclipse projects and ".project" file in them, the directory structure looks like 'myProject/.project'. I want to copy these '.project' files to another directory, but I want the enclosing directory name to be preserved. Let's say I have 'a/myProject/.project', I want to copy 'myProject/.project' to 'b', so it be 'b/myProject/.project', but 'b/myProject' doesn't exist. When I try in a: cp -r ./myProject/.project ../b it copies only '.project' file itself, without 'myProject' directory. Please advise.

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  • File association error while trying to browse to a network share in explorer or from run?

    - by ChrisFletcher
    I'm getting the below error message while trying to browse to a local share on a Windows Server 2003 machine: Windows cannot find ### this file does not have a program associated with it for performing this action. Create an Association in the folder options control panel The server is on the network, has an assigned IP address, can access the internet and is otherwise functioning normally. I realise this message can occur when opening a document or file which has no application associated with it, but I'm trying to access a share. What's going on here?

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