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POHow to specify output directory for .exe in C#
 

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  1. POHow to specify output directory for .exe in C#
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    <p>I am developing wpf application in C#. I have one button on which I am browsing file system through Microsoft.Win32.OpenFileDialog. There is one submit button on which I am calling the Process.Start() to run .exe on grib file. The exe generate the .csv files for me successfully. First I browse the file system, select the file and then I click on submit button. My application execution path is D:\Projects\ApiRouting\ApiRouting\bin\Debug. There is one folder in my application at D:\Projects\ApiRouting\ApiRouting\Files. When I select the file from path D:\Projects\ApiRouting\ApiRouting\Files and click on submit button then the .csv files get generated at D:\Projects\ApiRouting\ApiRouting\Files which is correct. When I select the file from D:\Documents and click on submit button the .csv files are generated at D:\Documents. My code to run .exe is as follows</p> <pre><code> public static void GenerateCsvFile(string fileName) { System.Diagnostics.Process process = new System.Diagnostics.Process(); System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo(); startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden; startInfo.FileName = @"C:\ndfd\degrib\bin\degrib.exe"; startInfo.Arguments = @"" + fileName + "" +" -C -msg 1 -Csv"; startInfo.UseShellExecute = true; process.StartInfo = startInfo; process.Start(); process.WaitForExit(); process.Close(); System.Diagnostics.Process process1 = new System.Diagnostics.Process(); System.Diagnostics.ProcessStartInfo startInfo1 = new System.Diagnostics.ProcessStartInfo(); startInfo1.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden; startInfo1.FileName = @"C:\ndfd\degrib\bin\degrib.exe"; startInfo1.Arguments = @"" + fileName + "" + " -C -msg all -nMet -Csv"; startInfo1.UseShellExecute = true; process1.StartInfo = startInfo1; process1.Start(); process1.WaitForExit(); process1.Close(); } private void BrowseButton_Click(object sender, RoutedEventArgs e) { safeFileName = string.Empty; // Create OpenFileDialog Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog(); // Set filter for file extension and default file extension //dlg.DefaultExt = ".txt"; //dlg.Filter = "Zip Files|*.zip*"; dlg.Multiselect = false; // Display OpenFileDialog by calling ShowDialog method Nullable&lt;bool&gt; result = dlg.ShowDialog(); // Get the selected file name and display in a TextBox if (result == true) { FileNameTextBox.Text = string.Empty; // Open document string fileName = dlg.FileName; safeFileName = dlg.SafeFileName; App.ZipFileSafeName = safeFileName; FileNameTextBox.Text = fileName; App.ZipFileName = fileName; } //dlg.InitialDirectory = @"D:\Projects\ApiRouting\ApiRouting\bin\Debug"; //dlg.FileName = @"D:\Projects\ApiRouting\ApiRouting\bin\Debug\Pacificwind.grb"; //dlg.Reset(); } </code></pre> <p>When the user selects the file from any location of file system I copy that file at D:\Projects\ApiRouting\ApiRouting\Files and then run the .exe. So GenerateCsvFile method always has the fileName parameter value D:\Projects\ApiRouting\ApiRouting\Files\xyz.grb. So why my application is generating the .csv file at D:\Documents when I select the grib file from D:\Documents and why it is generating the .csv file at D:\Projects\ApiRouting\ApiRouting\Files when I select the .csv file from D:\Projects\ApiRouting\ApiRouting\Files ?</p>
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      1. PTAnswer
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    1. PTQuestion
    1. USShailesh Jaiswal
    1. USShailesh Jaiswal
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      1. POHow to specify output directory for .exe in C#
      1. This table or related slice is empty.
      1. VTDownMod
    2. VO
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      1. POHow to specify output directory for .exe in C#
      1. This table or related slice is empty.
      1. VTDownMod
    3. VO
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      1. POHow to specify output directory for .exe in C#
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      1. VTDownMod
    1. COyour escaping seems "odd"... another point: the code you show doesn't have anything to do with the problem you describe (OpenFileDialog)... what is the value of the parameter `fileName` when the above code is executed (check via Debugger)?
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      1. POHow to specify output directory for .exe in C#
      1. USYahia
    2. CO@Yahia, i think he have problems with understanding how degrib.exe works and where it creates result, not OpenFileDialog problems
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      1. POHow to specify output directory for .exe in C#
      1. USDmitry Dovgopoly
    3. COfilename always has value D:\Projects\ApiRouting\ApiRouting\Files\xyz.grb
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      1. POHow to specify output directory for .exe in C#
      1. USShailesh Jaiswal
 

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