Search Results

Search found 474 results on 19 pages for 'filepath'.

Page 18/19 | < Previous Page | 14 15 16 17 18 19  | Next Page >

• How can I pass more than one command line argument via c#

  - by user293392

  I need to pass more than one command line argument via c# for a process called handle.exe: http://www.google.com.mt/search?sourceid=chrome&ie=UTF-8&q=handle.exe First, I need to run the executable file via ADMINISTRATOR permissions. This post has helped me achieve just that: http://stackoverflow.com/questions/667381/programatically-run-cmd-exe-as-adminstrator-in-vista-c But then comes the next problem of calling the actual line arguments such as "-p explore" How can I specify the command line arguments together, or maybe consecutively? Current code is as follows: Process p = new Process(); ProcessStartInfo processStartInfo = new ProcessStartInfo("filePath"); processStartInfo.CreateNoWindow = true; processStartInfo.UseShellExecute = false; processStartInfo.RedirectStandardOutput = true; processStartInfo.RedirectStandardInput = true; processStartInfo.Verb = "runas"; processStartInfo.Arguments = "/env /user:" + "Administrator" + " cmd"; p.StartInfo = processStartInfo; p.Start(); string output = p.StandardOutput.ReadToEnd(); p.WaitForExit(); Console.WriteLine(output); Thanks

  Read the article

• BitmapFrame in another thread

  - by Lasse Lindström

  Hi I am using a WPF BackgroundWorker to create thumbnails. My worker function looks like: private void work(object sender, DoWorkEventArgs e) { try { var paths = e.Argument as string[]; var boxList = new List(); foreach (string path in paths) { if (!string.IsNullOrEmpty(path)) { FileInfo info = new FileInfo(path); if (info.Exists && info.Length 0) { BitmapImage bi = new BitmapImage(); bi.BeginInit(); bi.DecodePixelWidth = 200; bi.CacheOption = BitmapCacheOption.OnLoad; bi.UriSource = new Uri(info.FullName); bi.EndInit(); var item = new BoxItem(); item.FilePath = path; MemoryStream ms = new MemoryStream(); PngBitmapEncoder encoder = new PngBitmapEncoder(); encoder.Frames.Add(BitmapFrame.Create(bi)); encoder.Save(ms); item.ThumbNail = ms.ToArray(); ms.Close(); boxList.Add(item); } } } e.Result = boxList; } catch (Exception ex) { //nerver comes here } } When this fuction is finnished and before the BackgroundWorker "Completed" function is started, I can see on the output window on Vs2008, that a exception is generated. It looks like: A first chance exception of type 'System.NotSupportedException' occurred in PresentationCore.dll The number of exceptions generates equals the number of thumbnails to be generated. Using "trial and error" I have isolated the problem to: BitmapFrame.Create(bi) Removing that line (makes my function useless) also removes the exception. I have not found any explanation to this,,, or a better method to create thumbnails i a background thread. Can anyone help me? //lasse

  Read the article

• how can load images from plist in to UITableView ?

  - by srikanth rongali

  I have stored the videos and the thumbnail images of the videos in Documents folder. And I have given the path in plist. In plist I took an array and I added directories to the array. And in the dictionary I stored the image path /Users/srikanth/Library/Application Support/iPhone Simulator/User/Applications/9E6E8B22-C946-4442-9209-75BB0E924434/Documents/image1 for key imagePath. for video /Users/srikanth/Library/Application Support/iPhone Simulator/User/Applications/9E6E8B22-C946-4442-9209-75BB0E924434/Documents/video1.mp4 for key filePath. I used following code but it is not working. I am trying only for images. I need the images to be loaded in the table in each cell. - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { static NSString *CellIdentifier = @"Cell"; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithFrame:CGRectZero reuseIdentifier:CellIdentifier] autorelease]; [cell setAccessoryType:UITableViewCellAccessoryDetailDisclosureButton]; UIImageView *image2 = [[UIImageView alloc]init]; image2.frame = CGRectMake(0.0f, 0.0f, 80.0f, 80.0f); image2.backgroundColor = [UIColor clearColor]; //image2.image = [UIImage imageNamed:@"snook.png"]; image2.tag = tag7; } NSDictionary *dictOfplist = [cells objectAtIndex:indexPath.row]; [(UIImageView *)[cell viewWithTag:tag7] setImage:[dictOfplist objectForKey:@"imagePath"]]; return cell; } - (void)viewDidLoad { [super viewDidLoad]; self.title = @"Library"; self.navigationItem.rightBarButtonItem = [[UIBarButtonItem alloc] initWithTitle:@"Close" style:UIBarButtonItemStyleBordered target:self action:@selector(close:)]; NSString* plistPath = [[NSBundle mainBundle] pathForResource:@"details" ofType:@"plist"]; contentArray = [NSArray arrayWithContentsOfFile:plistPath]; cells = [[NSMutableArray alloc]initWithCapacity:[contentArray count]]; for(dCount = 0; dCount < [contentArray count]; dCount++) [cells addObject:[contentArray objectAtIndex:dCount]]; } How can I make this work. Thank you.

  Read the article

• How can I add the "--watch" flag to this TextMate snippet?

  - by Jannis

  I love TextMate as my editor for all things web, and so I'd like to use a snippet to use it with style.less files to automatically take advantage of the .less way of compiling .css files on the fly using the native $ lessc {filepath} --watch as suggested in the less documentation (link) My (thanks to someone who wrote the LESS TM Bundle!) current TextMate snippet works well for writing the currently opened .less file to the .css file but I'd like to take advantage of the --watch parameter so that every change to the .less file gets automatically compiled into the .css file. This works well when using the Terminal command line for it, so I am sure it must be possible to use it in an adapted version of the current LESS Command for TextMate since that only invokes the command to compile the file. So how do I add the --watch flag to this command:? #!/usr/bin/env ruby file = STDIN.read[/lessc: ([^*]+\.less)/, 1] || ENV["TM_FILEPATH"] system("lessc \"#{file}\"") I assume it should be something like: #!/usr/bin/env ruby file = STDIN.read[/lessc: ([^*]+\.less)/, 1] || ENV["TM_FILEPATH"] system("lessc \"#{file}\" --watch") But doing so only crashes the TextMate.app. Any ideas would be much appreciated. Thanks for reading. Jannis

  Read the article

• .NET 4: Process.Start using credentials returns empty output

  - by alexey

  I run an external program from ASP.NET: var process = new Process(); var startInfo = process.StartInfo; startInfo.FileName = filePath; startInfo.Arguments = arguments; startInfo.UseShellExecute = false; startInfo.RedirectStandardOutput = true; startInfo.RedirectStandardError = true; process.Start(); process.WaitForExit(); Console.Write("Output: {0}", process.StandardOutput.ReadToEnd()); Console.Write("Error Output: {0}", process.StandardError.ReadToEnd()); Everything works fine with this code: the external program is executed and process.StandardOutput.ReadToEnd() returns the correct output. But after I add these two lines before process.Start() (to run the program in the context of another user account): startInfo.UserName = userName; startInfo.Password = securePassword; The program is not executed and process.StandardOutput.ReadToEnd() returns an empty string. No exceptions are thrown. userName and securePassword are correct (in case of incorrect credentials an exception is thrown). How to run the program in the context of another user account?

  Read the article

• Access 2007 file picker, replaces all rows with the same choice.

  - by SqlStruggle

  This code is from an Access 2007 project I've been struggling with. The actual mean part is the part where I should put something like "update only current form" DoCmd.RunSQL "Update Korut Set [PikkuKuva]=('" & varFile & "') ;" Could someone please help me with this?` If I use it now, it updates all the tables with the same file picked. Heres the whole code. ' This requires a reference to the Microsoft Office 11.0 Object Library. Dim fDialog As Office.FileDialog Dim varFile As Variant Dim filePath As String ' Set up the File dialog box. Set fDialog = Application.FileDialog(msoFileDialogFilePicker) With fDialog ' Allow the user to make multiple selections in the dialog box. .AllowMultiSelect = False ' Set the title of the dialog box. .Title = "Valitse Tiedosto" ' Clear out the current filters, and then add your own. .Filters.Clear .Filters.Add "All Files", "*.*" ' user picked at least one file. If the .Show method returns ' False, the user clicked Cancel. If .Show = True Then ' Loop through each file that is selected and then add it to the list box. For Each varFile In .SelectedItems DoCmd.SetWarnings True DoCmd.RunSQL "Update Korut Set [PikkuKuva]=('" & varFile & "') ;" Next Else MsgBox "You clicked Cancel in the file dialog box." End If End With

  Read the article

• Improve Application Performace

  - by Gtest

  Hello, Want To Improvide Performace Of C#.Net Applicaiton.. In My Application I am using Third Party Interop/Dll To Process .doc Files. It's a Simple Operation, Which Pass Input/Output FilePath to Interop dll ...& dll will execute text form input file. To Improve Performace I have Tried, Execute 2 therad to process 32 files.(each Thread process 16 files) Execute application code by creating 2 new AppDomains(each AppDomain Code process 16 files) Execute Code Using TPL(Task Parellel Library) But all options take around same time (32 sec) to process 32 files.Manually process tooks same 32 sec to process 32 files. Just tried one thing ..when i have created sample exe to process 16 files as input & output for refrence PAth given in TextBox. ..I open 2 exe instance to process. 1 exe has differnt 16 input files & output Created with input file path 2 exe has differnt 16 input files & output Created with input file path When i click on start button of both exe ..it use 100% cpu & Utilize both core significantly & Process Completed within 16 sec for 32 files. Can we provide this kind of explicit prallism to Improve my applicaiton Peformace? Thanks.

  Read the article

• Using A Local file path in a Streamwriter object ASP.Net

  - by Nick LaMarca

  I am trying to create a csv file of some data. I have wrote a function that successfully does this.... Private Sub CreateCSVFile(ByVal dt As DataTable, ByVal strFilePath As String) Dim sw As New StreamWriter(strFilePath, False) ''# First we will write the headers. ''EDataTable dt = m_dsProducts.Tables[0]; Dim iColCount As Integer = dt.Columns.Count For i As Integer = 0 To iColCount - 1 sw.Write(dt.Columns(i)) If i < iColCount - 1 Then sw.Write(",") End If Next sw.Write(sw.NewLine) ''# Now write all the rows. For Each dr As DataRow In dt.Rows For i As Integer = 0 To iColCount - 1 If Not Convert.IsDBNull(dr(i)) Then sw.Write(dr(i).ToString()) End If If i < iColCount - 1 Then sw.Write(",") End If Next sw.Write(sw.NewLine) Next sw.Close() End Sub The problem is I am not using the streamwriter object correctly for what I trying to accomplish. Since this is an asp.net I need the user to pick a local filepath to put the file on. If I pass any path to this function its gonna try to write it to the directory specified on the server where the code is. I would like this to popup and let the user select a place on their local machine to put the file.... Dim exData As Byte() = File.ReadAllBytes(Server.MapPath(eio)) File.Delete(Server.MapPath(eio)) Response.AddHeader("content-disposition", String.Format("attachment; filename={0}", fn)) Response.ContentType = "application/x-msexcel" Response.BinaryWrite(exData) Response.Flush() Response.End() I am calling the first function in code like this... Dim emplTable As DataTable = SiteAccess.DownloadEmployee_H() CreateCSVFile(emplTable, "C:\\EmplTable.csv") Where I dont want to have specify the file loaction (because this will put the file on the server and not on a client machine) but rather let the user select the location on their client machine. Can someone help me put this together? Thanks in advance.

  Read the article

• When downloading a file using FileStream, why does page error message refers to aspx page name, not

  - by StuperUser

  After building a filepath (path, below) in a string (I am aware of Path in System.IO, but am using someone else's code and do not have the opportunity to refactor it to use Path). I am using a FileStream to deliver the file to the user (see below): FileStream myStream = new FileStream(path, FileMode.Open, FileAccess.Read); long fileSize = myStream.Length; byte[] Buffer = new byte[(int)fileSize + 1]; myStream.Read(Buffer, 0, (int)myStream.Length); myStream.Close(); Response.ContentType = "application/csv"; Response.AddHeader("content-disposition", "attachment; filename=" + filename); Response.BinaryWrite(Buffer); Response.Flush(); Response.End(); I have seen from: http://stackoverflow.com/questions/736301/asp-net-how-to-stream-file-to-user reasons to avoid use of Response.End() and Response.Close(). I have also seen several articles about different ways to transmit files and have diagnosed and found a solution to the problem (https and http headers) with a colleague. However, the error message that was being displayed was not about access to the file at path, but the aspx file. Edit: Error message is: Internet Explorer cannot download MyPage.aspx from server.domain.tld Internet Explorer was not able to open this Internet site. The requested site is either unavailable or cannot be found. Please try again later. (page name and address anonymised) Why is this? Is it due to the contents of the file coming from the HTTP response .Flush() method rather than a file being accessed at its address?

  Read the article

• How do I set an absolute include path in PHP?

  - by Nathan Long

  In HTML, I can find a file starting from the web server's root folder by beginning the filepath with "/". Like: /images/some_image.jpg I can put that path in any file in any subdirectory, and it will point to the right image. With PHP, I tried something similar: include("/includes/header.php"); ...but that doesn't work. I think that that this page is saying that I can set include_path once and after that, it will be assumed. But I don't quite get the syntax. Both examples start with a period, and it says: Using a . in the include path allows for relative includes as it means the current directory. Relative includes are exactly what I don't want. How do I make sure that all my includes point to the root/includes folder? (Bonus: what if I want to place that folder outside the public directory?) Clarification My development files are currently being served by XAMPP/Apache. Does that affect the absolute path? (I'm not sure yet what the production server will be.)

  Read the article

• How to stream pdf document from servlet?

  - by Kumar

  Hi,I am creating pdf document using jasper report and i need to stream that pdf document from servlet.Can anyone help me where i did mistake.This is the code snippet which i am using in my application. ServletOutputStream servletOutputStream = response.getOutputStream(); String fileName="test.pdf"; response.setContentType("application/pdf"); response.setHeader("Content-Disposition","attachment; filename=\"" + fileName + "\""); response.setHeader("Cache-Control", "no-cache"); try { Map parameters = new HashMap(); parameters.put("SUBREPORT_DIR", JasperReportFilepath); parameters.put("TestId", testID); JasperPrint jprint=JasperFillManager.fillReport(filePath, parameters, conn); byte[] output=JasperExportManager.exportReportToPdf(jprint); System.out.println("Size====>"+output.length); servletOutputStream.write(output); servletOutputStream.flush(); servletOutputStream.close(); System.out.println("===============>Streaming perfectly"); } catch(Exception e) { System.out.println("===============>+JasperException"+e.getMessage()); } and i could not get any error message also.Everything is working fine but document is not streaming. Please help me to sort out the problem.

  Read the article

• Getting method arguments with Roslyn

  - by Kevin Burton

  I can get a list from the solution of all calls to a particuliar method using the following code: var createCommandList = new List<MethodSymbol>(); INamedTypeSymbol interfaceSymbol = (from p in solution.Projects select p.GetCompilation().GetTypeByMetadataName("BuySeasons.BsiServices.DataResource.IBsiDataConnection")).FirstOrDefault(); foreach (ISymbol symbol in interfaceSymbol.GetMembers("CreateCommand")) { if (symbol.Kind == CommonSymbolKind.Method && symbol is MethodSymbol) { createCommandList.Add(symbol as MethodSymbol); } } foreach (MethodSymbol methodSymbol in createCommandList) { foreach (ReferencedSymbol referenceSymbol in methodSymbol.FindReferences(solution)) { foreach (ReferenceLocation referenceLocation in from l in referenceSymbol.Locations orderby l.Document.FilePath select l) { if (referenceLocation.Location.GetLineSpan(false).StartLinePosition.Line == referenceLocation.Location.GetLineSpan(false).EndLinePosition.Line) { Debug.WriteLine(string.Format("{0} {1} at {2} {3}/{4} - {5}", methodSymbol.Name, "(" + string.Join(",", (from p in methodSymbol.Parameters select p.Type.Name + " " + p.Name).ToArray()) + ")", Path.GetFileName(referenceLocation.Location.GetLineSpan(false).Path), referenceLocation.Location.GetLineSpan(false).StartLinePosition.Line, referenceLocation.Location.GetLineSpan(false).StartLinePosition.Character, referenceLocation.Location.GetLineSpan(false).EndLinePosition.Character)); } else { throw new ApplicationException("Call spans multiple lines"); } } } } But this gives me a list of ReferencedSymbol's. Although this gives me the file and line number that the method is called from I would also like to get the specific arguments that the method is called with. How can I either convert what I have or get the same information with Roslyn? (notice the I first load the solution with the Solution.Load method and then loop through to find out where the method is defined/declared (createCommandList)). Thank you.

  Read the article

• Add xml-stylesheet and get standalone = yes.

  - by tumba25

  The code at the bottom is what I have. I removed the creation of all tags. At the top in the xml file I get.<?xml version="1.0" encoding="UTF-8" standalone="no"?> Note that standalone is no, even thou I have it set to yes. The first question: How do I get standalone = yes? I would like to add <?xml-stylesheet type="text/xsl" href="my.stylesheet.xsl"?> at line two in the xml file. Second question: How do I do that? Some useful links? Anything? DocumentBuilderFactory dbfac = DocumentBuilderFactory.newInstance(); DocumentBuilder docBuilder = dbfac.newDocumentBuilder(); Document doc = docBuilder.newDocument(); <cut> TransformerFactory transfac = TransformerFactory.newInstance(); transfac.setAttribute("indent-number", new Integer(2)); Transformer trans = transfac.newTransformer(); trans.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "no"); trans.setOutputProperty(OutputKeys.STANDALONE, "yes"); trans.setOutputProperty(OutputKeys.INDENT, "yes"); trans.setOutputProperty(OutputKeys.CDATA_SECTION_ELEMENTS, "name"); FileOutputStream fout = new FileOutputStream(filepath); BufferedOutputStream bout= new BufferedOutputStream(fout); trans.transform(new DOMSource(doc), new StreamResult(new OutputStreamWriter(bout, "utf-8")));

  Read the article

• Secure Copy File from remote server via scp and os module in Python

  - by user1063572

  I'm pretty new to Python and programming. I'm trying to copy a file between two computers via a python script. However the code os.system("ssh " + hostname + " scp " + filepath + " " + user + "@" + localhost + ":" cwd) won't work. I think it needs a password, as descriped in How do I copy a file to a remote server in python using scp or ssh?. I didn't get any error logs, the file just won't show in my current working directory. However every other command with os.system("ssh " + hostname + "command") or os.popen("ssh " + hostname + "command") does work. - command = e.g. ls When I try ssh hostname scp file user@local:directory in the commandline it works without entering a password. I tried to combine os.popen commands with getpass and pxssh module to establish a ssh connection to the remote server and use it to send commands directly (I only tested it for an easy command): import pxssh import getpass ssh = pxssh.pxssh() ssh.force_password = True hostname = raw_input("Hostname: ") user = raw_input("Username: ") password = getpass.getpass("Password: ") ssh.login(hostname, user, password) test = os.popen("hostname") print test But I'm not able to put commands through to the remote server (print test shows, that hostname = local and not the remote server), however I'm sure, the conection is established. I thought it would be easier to establish a connection than always use "ssh " + hostname in the bash commands. I also tried some of the workarounds in How do I copy a file to a remote server in python using scp or ssh?, but I must admit due to lack of expirience I didn't get them to work. Thanks a lot for helping me.

  Read the article

• How do I fix a "Request format is unrecognized for URL..." error in a web service running in IIS?

  - by Hary

  I am get the following error while running web service in IIS: Server Error in '/Inbox Sevice' Application. Request format is unrecognized for URL unexpectedly ending in '/GetMailsInfo'. Description: An unhandled exception occurred during the execution of the current web request. Please review the stack trace for more information about the error and where it originated in the code. Exception Details: System.InvalidOperationException: Request format is unrecognized for URL unexpectedly ending in '/GetMailsInfo'. Source Error: An unhandled exception was generated during the execution of the current web request. Information regarding the origin and location of the exception can be identified using the exception stack trace below. Stack Trace: [InvalidOperationException: Request format is unrecognized for URL unexpectedly ending in '/GetMailsInfo'.] System.Web.Services.Protocols.WebServiceHandlerFactory.CoreGetHandler(Type type, HttpContext context, HttpRequest request, HttpResponse response) +490982 System.Web.Services.Protocols.WebServiceHandlerFactory.GetHandler(HttpContext context, String verb, String url, String filePath) +104 System.Web.Script.Services.ScriptHandlerFactory.GetHandler(HttpContext context, String requestType, String url, String pathTranslated) +127 System.Web.HttpApplication.MapHttpHandler(HttpContext context, String requestType, VirtualPath path, String pathTranslated, Boolean useAppConfig) +175 System.Web.MapHandlerExecutionStep.System.Web.HttpApplication.IExecutionStep.Execute() +120 System.Web.HttpApplication.ExecuteStep(IExecutionStep step, Boolean& completedSynchronously) +155 Version Information: Microsoft .NET Framework Version:2.0.50727.42; ASP.NET Version:2.0.50727.42 Does anyone know why I am seeing this error and if there is any way to fix it?

  Read the article

• Nested model form with collection in Rails 2.3

  - by kristian nissen

  How can I make this work in Rails 2.3? class Magazine < ActiveRecord::Base has_many :magazinepages end class Magazinepage < ActiveRecord::Base belongs_to :magazine end and then in the controller: def new @magazine = Magazine.new @magazinepages = @magazine.magazinepages.build end and then the form: <% form_for(@magazine) do |f| %> <%= error_messages_for :magazine %> <%= error_messages_for :magazinepages %> <fieldset> <legend><%= t('new_magazine') %></legend> <p> <%= f.label :title %> <%= f.text_field :title %> </p> <fieldset> <legend><%= t('new_magazine_pages') %> <% f.fields_for :magazinepages do |p| %> <p> <%= p.label :name %> <%= p.text_field :name %> </p> <p> <%= p.file_field :filepath %> </p> <% end %> </fieldset> <p> <%= f.submit :save %> </p> </fieldset> <% end %> problem is, if I want to submit a collection of magazinepages, activerecord complaints because it's expected a model and not an array. create action: def create @magazine = Magazine.new params[:magazine] @magazine.save ? redirect_to(@magazine) : render(:action => 'new') end

  Read the article

• BufferedImage & ColorModel in Java

  - by spol

  I am using a image processing library in java to manipulate images.The first step I do is I read an image and create a java.awt.Image.BufferedImage object. I do it in this way, BufferedImage sourceImage = ImageIO.read( new File( filePath ) ); The above code creates a BufferedImage ojbect with a DirectColorModel: rmask=ff0000 gmask=ff00 bmask=ff amask=0. This is what happens when I run the above code on my macbook. But when I run this same code on a linux machine (hosted server), this creates a BufferedImage object with ColorModel: #pixelBits = 24 numComponents = 3 color space = java.awt.color.ICC_ColorSpace@c39a20 transparency = 1 has alpha = false isAlphaPre = false. And I use the same jpg image in both the cases. I don't know why the color model on the same image is different when run on mac and linux. The colormodel for mac has 4 components and the colormodel for linux has 3 components.There is a problem arising because of this, the image processing library that I use always assumes that there are always 4 components in the colormodel of the image passed, and it throws array out of bounds exception when run on linux box. But on macbook, it runs fine. I am not sure if I am doing something wrong or there is a problem with the library. Please let me know your thoughts. Also ask me any questions if I am not making sense!

  Read the article

• Howto open a file and remove the last line?

  - by sologhost

  I am looking to open up a file, grab the last line in the file where the line = "?", which is the closing tag for a php document. Than I am wanting to append data into it and add back in the "?" to the very last line. I've been trying a few approaches, but I'm not having any luck. Here's what I got so far, as I am reading from a zip file. Though I know this is all wrong, just needing some help with this please... // Open for reading is all we can do with zips and is all we need. if (zip_entry_open($zipOpen, $zipFile, "r")) { $fstream = zip_entry_read($zipFile, zip_entry_filesize($zipFile)); $fp = fopen($curr_lang_file, 'r+b'); while (!feof($fp)) { $output = fgets($fp, 16384); if (trim($output) == '?>') break; fwrite($fp, $output); } fclose($fp); file_put_contents($curr_lang_file, $fstream, FILE_APPEND); } $curr_lang_file is a filepath string to the actual file that needs to have the fstream appended to it, but after we remove the last line that equals '?'

  Read the article

• OpenCV compare two images and get different pixels

  - by Richard Knop

  For some reason the code bellow is not working. I have two 640*480 images which are very similar but not the same (at least few hundred/thousand pixels should be different). This is how I am comparing them and counting different pixels: unsigned char* row; unsigned char* row2; int count = 0; // this happens in a loop // fIplImageHeader is current image // lastFIplImageHeader is image from previous iteration if ( NULL != lastFIplImageHeader->imageData ) { for( int y = 0; y < fIplImageHeader->height; y++ ) { row = &CV_IMAGE_ELEM( fIplImageHeader, unsigned char, y, 0 ); row2 = &CV_IMAGE_ELEM( lastFIplImageHeader, unsigned char, y, 0 ); for( int x = 0; x < fIplImageHeader->width*fIplImageHeader->nChannels; x += fIplImageHeader->nChannels ) { if (row[x] == row2[x]) // the pixel in the first channel (usually G) { count++; } if (row[x+1] == row2[x+1]) // ... second channel (usually B) { count++; } if (row[x+2] == row2[x+2]) // ... third channel (usually R) { count++; } } } } Now at the end I get number 3626 which would seem alright. But, I tried opening one of the images in MS Paint and drawing thick red lines all over it which should increase the number of different pixels substantially. I got the same number again: 3626. Obviously I am doing something wrong here. I am comparing these images in a loop. This line is before the loop: IplImage* lastFIplImageHeader = cvCreateImageHeader(cvSize(640, 480), 8, 3); Then inside the loop I load images like this: IplImage* fIplImageHeader = cvLoadImage( filePath.c_str() ); // here I compare the pixels (the first code snippet) lastFIplImageHeader->imageData = fIplImageHeader->imageData; So lastFIplImageHeader is storing the image from the previous iteration and fIplImageHeader is storing the current image.

  Read the article

• How can I determine whether or not to add project items using IWizard?

  - by taarskog

  Hi, I am generating entity wrappers in VS2010 based on dynamic objects in a CRM system. In addition to the entity code, I want to add an EntityBase of which all entities inherit from. If the file exists in the project from before, it should not be added. I am using an IWizard implementation to give the generator the object names etc. Is it possible in the IWizard implementation to determine whether or not to add an item if it exists in the project from before? How do I get a hold of the project handle and its items in or before the ShouldAddProjectItem method? My code so far (not completed): public class EntityWizardImplementation : IWizard { public void BeforeOpeningFile(ProjectItem projectItem) { //Note: Nothing here. } public void ProjectFinishedGenerating(Project project) { //Note: Nothing here. } public void ProjectItemFinishedGenerating(ProjectItem projectItem) { //Note: Nothing here. } public void RunFinished() { //Note: Nothing here. } public void RunStarted(object automationObject, Dictionary<string, string> replacementsDictionary, WizardRunKind runKind, object[] customParams) { try { var window = new WizardWindow(); // Replace parameters gathered from the wizard replacementsDictionary.Add("$crmEntity$", window.CrmEntity); replacementsDictionary.Add("$crmOrganization$", window.CrmOrganization); replacementsDictionary.Add("$crmMetadataServiceUrl$", window.CrmMetadataUrl); window.Close(); } catch (SoapException se) { MessageBox.Show(se.ToString()); } catch (Exception e) { MessageBox.Show(e.ToString()); } } public bool ShouldAddProjectItem(string filePath) { // This is where I assume it is correct to handle the preexisting file. return true; } }

  Read the article

• How to open a file and remove the last line?

  - by sologhost

  I am looking to open up a file, grab the last line in the file where the line = "?", which is the closing tag for a php document. Than I am wanting to append data into it and add back in the "?" to the very last line. I've been trying a few approaches, but I'm not having any luck. Here's what I got so far, as I am reading from a zip file. Though I know this is all wrong, just needing some help with this please... // Open for reading is all we can do with zips and is all we need. if (zip_entry_open($zipOpen, $zipFile, "r")) { $fstream = zip_entry_read($zipFile, zip_entry_filesize($zipFile)); $fp = fopen($curr_lang_file, 'r+b'); while (!feof($fp)) { $output = fgets($fp, 16384); if (trim($output) == '?>') break; fwrite($fp, $output); } fclose($fp); file_put_contents($curr_lang_file, $fstream, FILE_APPEND); } $curr_lang_file is a filepath string to the actual file that needs to have the fstream appended to it, but after we remove the last line that equals '?'

  Read the article

• How to make buttons in python/pygame?

  - by user1334014

  I'm making a game in pygame and on the first screen I want there to be buttons that you can press to (i) start the game, (ii) load a new screen with instructions, and (iii) exit the program. I've found this code online for button making, but I don't really understand it (I'm not that good at object oriented programming). If I could get some explanation as to what it's doing that would be great. Also, when I use it and try to open a file on my computer using the file path, I get the error sh: filepath :Permission denied, which I don't know how to solve. #load_image is used in most pygame programs for loading images def load_image(name, colorkey=None): fullname = os.path.join('data', name) try: image = pygame.image.load(fullname) except pygame.error, message: print 'Cannot load image:', fullname raise SystemExit, message image = image.convert() if colorkey is not None: if colorkey is -1: colorkey = image.get_at((0,0)) image.set_colorkey(colorkey, RLEACCEL) return image, image.get_rect() class Button(pygame.sprite.Sprite): """Class used to create a button, use setCords to set position of topleft corner. Method pressed() returns a boolean and should be called inside the input loop.""" def __init__(self): pygame.sprite.Sprite.__init__(self) self.image, self.rect = load_image('button.png', -1) def setCords(self,x,y): self.rect.topleft = x,y def pressed(self,mouse): if mouse[0] > self.rect.topleft[0]: if mouse[1] > self.rect.topleft[1]: if mouse[0] < self.rect.bottomright[0]: if mouse[1] < self.rect.bottomright[1]: return True else: return False else: return False else: return False else: return False def main(): button = Button() #Button class is created button.setCords(200,200) #Button is displayed at 200,200 while 1: for event in pygame.event.get(): if event.type == MOUSEBUTTONDOWN: mouse = pygame.mouse.get_pos() if button.pressed(mouse): #Button's pressed method is called print ('button hit') if __name__ == '__main__': main() Thank you to anyone who can help me.

  Read the article

• Move and rename file in android

  - by Andre Fróes

  I am trying to copy a file to another folder in the android, but so far, i got no success. I manage to do so with a selected image and when taking a photo, but not with files. I've read and tried several solutions passed by the community (searched over the forum and the internet), but none of it was able to solve my problem when copying. First things first. I added the permissions to my manifest: <uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" /> <uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" /> after that, before copying a file, i print its filepath and the directory file path: 06-10 11:11:11.700: I/System.out(1442): /mimetype/storage/sdcard/Misc/Javascript erros for Submit and Plan buttons in IE.doc 06-10 11:11:11.710: I/System.out(1442): /storage/sdcard/mywfm/checklist-files both exists: to copy the file to the expected folder I used the FileUtils: try { FileUtils.copyFile(selectedFile, dir); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } The problem is: I get no exception and the file isn't there. I tried this solution either: How to move/rename file from internal app storage to external storage on Android? same thing, no exception, but no file either.

  Read the article

• Java variables across methods

  - by NardCake

  I'm making a basic text editor, and I have 2 methods the first one is triggered when a user click 'Open' and it prompts the user to pick a file and it opens the file fine. I just want to access the same file path which is in a variable in the method that is triggered when the user clicks save. My methods are public, Iv'e tried accessing it through a class, still no. Please help! Code: public void open(){ try{ //Open file JFileChooser fc = new JFileChooser(); fc.showOpenDialog(null); File file = fc.getSelectedFile(); String haha = file.getPath(); BufferedReader br = new BufferedReader(new FileReader(file.getPath())); String line; while((line = br.readLine()) != null){ text.append(line + "\n"); } } catch (FileNotFoundException e){ e.printStackTrace(); }catch (IOException e){ } } public void save(){ try { BufferedWriter bw = new BufferedWriter(new FileWriter(file.filePath)); bw.write(text.getText()); bw.close(); } catch (IOException e) { e.printStackTrace(); } }

  Read the article

• What is a good method to solve cabal install problems?

  - by sp3ctum

  I've used the cabal package manager for Haskell programs to install libraries and new projects that I've cloned from some repositories. More often than not, I keep running into problems. Most projects make installing them seem super easy, but in my case that's not always true - sometimes they are very hard to get running. Some are so hard, in fact, that I've lost interest in the project solely because of not being able to install it. So instead of complaining, I'd like to ask what I should do to better this situation. I'd like to use my most recent problem as an example. I'm interested in trying out the Gitit project. It's a promising looking personal wiki that runs on various version control systems. So here's what I've done: Clone from Github run cabal install in the project directory like I'm told on the project install page: mika@eka:~/git/gitit$ ls BLUETRIP-LICENSE CHANGES HCAR-gitit.tex LICENSE Network README.markdown RELANN-0.6.1 Setup.lhs TANGOICONS YUI-LICENSE data expireGititCache.hs gitit.cabal gitit.hs plugins mika@eka:~/git/gitit$ cabal install Resolving dependencies... cabal: cannot configure happstack-server-7.0.7. It requires base64-bytestring ==1.0.* For the dependency on base64-bytestring ==1.0.* there are these packages: base64-bytestring-1.0.0.0. However none of them are available. base64-bytestring-1.0.0.0 was excluded because gitit-0.10 requires base64-bytestring ==0.1.* mika@eka:~/git/gitit$ So now I'm thinking: well, I'll install happstack-server on its own, maybe that will work: mika@eka:~/git/gitit$ cabal install happstack-server Resolving dependencies... Warning: happstack-server.cabal: Ignoring unknown section type: test-suite Configuring happstack-server-7.0.7... cabal: At least the following dependencies are missing: blaze-html ==0.5.*, hslogger >=1.0.2, monad-control ==0.3.*, network >=2.2.3, sendfile >=0.7.1 && <0.8, system-filepath >=0.3.1, text >=0.10 && <0.12, threads >=0.5, transformers-base ==0.4.* cabal: Error: some packages failed to install: happstack-server-7.0.7 failed during the configure step. The exception was: ExitFailure 1 So looks like there are some dependencies missing. But isn't installing these dependencies the whole point of using cabal in the first place? What should I do? File bug reports (to which project?), install the dependencies manually or something else? Bonus points for explaining what causes these kinds of problems.

  Read the article

< Previous Page | 14 15 16 17 18 19  | Next Page >