Develop Reference

This should be so simple, but I can't get it to work

I wrote a simple c# program to automate the process of signing files and I just can't get it to work. Sorry, but I'm no expert in C# (obviously)! Here's my code:
static void Main(string[]
// This progam makes it easy to remember the parameters that need to be passed to signtool and automates the process
if (args.Length < 3)
{
Console.WriteLine("Usage: SignFile FileToSign .pfxfile Password");
}
else
{
try
{
string signTool = "\"C:\\Program Files (x86)\\Microsoft SDKs\\Windows\\v7.0A\\Bin\\signtool.exe\"";
string fileToSign = args[0];
string pfxFile = args[1];
string password = args[2];
string commandLine = "/C " + signTool + " sign /f \"" + pfxFile + "\" /fd SHA256 /p "
+ password + #" /t http://timestamp.verisign.com/scripts/timestamp.dll " + fileToSign;
ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.FileName = "cmd.exe";
System.IO.Directory.SetCurrentDirectory(".");
startInfo.Arguments = commandLine;
Process process = new Process();
startInfo.UseShellExecute = false;
process.StartInfo = startInfo;
process.Start();
}
catch (Exception e)
{
Console.WriteLine("Error: " + e.Message);
throw;
}
}
}
When I run it I get 'C:\Program' is not recognized as an internal or external command.
I also tried using the short version of the directory name:
string signTool = "\"c:\\PROGRA~2\\MICROS~2\\Windows\\v7.0A\\Bin\\signtool.exe\"";
But this I get:
The filename, directory name, or volume label syntax is incorrect.
I could say more, but I think I'll leave this as simple as it is...

Try it with # in front, like this:
string signTool = #"""C:\Program Files (x86)\Microsoft SDKs\Windows\v7.0A\Bin\signtool.exe""";

The error was fixed by comment 2 on my original question:

Command line c# not working as intended

I have the following command that I wish to run (which works when I manually run it in cmd):
"C:\Program Files\APP\APP.exe" -CMV "C:\Program Files\APP\Second\IT" -ID 5 6 7 4 2
This is the code that I have written in C#:
string firstPath = #"""C:\Program Files\APP\APP.exe""";
string secondPath = #"""C:\Program Files\APP\Second\IT""";
string command = firstPath + " -CMV " + secondPath + " -ID 5 6 7 4 2");
I also tried the following piece of code:
int exitCode;
ProcessStartInfo processInfo;
Process process;
processInfo = new ProcessStartInfo("cmd.exe", command);
processInfo.CreateNoWindow = true;
processInfo.UseShellExecute = false;
// *** Redirect the output ***
processInfo.RedirectStandardError = true;
processInfo.RedirectStandardOutput = true;
//processInfo.Arguments
process = Process.Start(processInfo);
process.WaitForExit();
// *** Read the streams ***
// Warning: This approach can lead to deadlocks, see Edit #2
string output = process.StandardOutput.ReadToEnd();
string error = process.StandardError.ReadToEnd();
exitCode = process.ExitCode;
Console.WriteLine("output>>" + (String.IsNullOrEmpty(output) ? "(none)" : output));
Console.WriteLine("error>>" + (String.IsNullOrEmpty(error) ? "(none)" : error));
Console.WriteLine("ExitCode: " + exitCode.ToString(), "ExecuteCommand");
process.Close();
This still doesn't seem to work! Any help is much appreciated.
*********FINAL EDIT*********
This is the string that gets sent in as the command:
""C:\\Program Files\\DEEM\\DEEM.exe" -ENV "C:\\Program Files\\DEEM\\Environments\\IT" -ID 01004698001001 00285209090217 00285209090250 00285209090382 99041250643762"
Which results in the following error:
"'C:\Program' is not recognized as an internal or external
command,\r\noperable program or batch file.\r\n"
All I need to make it work is to take out the first and last quote which are wrapped due to the command being a string. I am not sure how to take the quotes out.

Your first example works if you wait for the process to exit:
static void Main(string[] args)
{
string firstPath = #"""C:\Program Files\APP\APP.exe""";
string secondPath = #"""C:\Program Files\APP\Second\IT""";
string command = firstPath + " -CMV " + secondPath + " -ID 5 6 7 4 2";
System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
//startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.FileName = firstPath; //"cmd.exe";
startInfo.Arguments = command;
process.StartInfo = startInfo;
process.Start();
// **** don't forget to wait for the process to exit ***
process.WaitForExit();
var code = process.ExitCode;
var time = process.ExitTime;
}
In your second example you did use WaitForExit(), but there is some other stuff broken I did not investigate on.

Open Visual Studio command prompt from C# code

I am trying to open Visual studio Command prompt using C# code.
Here is my code
private void Execute(string vsEnvVar)
{
var vsInstallPath = Environment.GetEnvironmentVariable(vsEnvVar);
// vsEnvVar can be VS100COMNTOOLS, VS120COMNTOOLS, VS140COMNTOOLS
if (Directory.Exists(vsInstallPath))
{
var filePath = vsInstallPath + "vsvars32.bat";
if (File.Exists(filePath))
{
//start vs command process
Process proc = new Process();
var command = Environment.GetEnvironmentVariable("ComSpec");
command = #"" + command + #"";
var batfile = #"E:\Test\vstest.bat";
var args = string.Format("/K \"{0}\" \"{1}\"" ,filePath, batfile);
proc.StartInfo.FileName = command;
proc.StartInfo.Arguments = args;
proc.StartInfo.CreateNoWindow = false;
proc.StartInfo.UseShellExecute = false;
proc.Start();
}
else
{
Console.WriteLine("File Does not exists " + filePath);
}
}
}
But the args string is not getting properly formatted.
I am getting below formatted string
"/K \"C:\\Program Files\\Microsoft Visual Studio 10.0\\Common7\\Tools\\vsvars32.bat\" \"E:\\Test\\vstest.bat\""
extra "\" is getting added.
Please point out what I am missing.
Thanks

The string is being formatted as you asked, but you have asked for the wrong thing. "E:\Test\VStest.bat" is being passed as an argument to VCVars.bat, but I suspect you want it to be executed after it.
Try this:
var args = string.Format("/S/K \" \"{0}\" && \"{1}\" \"" ,filePath, batFile);
This should produce:
"/S/K \" \"C:\\Program Files\\Microsoft Visual Studio 10.0\\Common7\\Tools\\vsvars32.bat\" && \"E:\\Test\\vstest.bat\" \" \"
Which as a string is:
/S/K " "C:\Program Files\Microsoft Visual Studio 10.0\Common7\Tools\vsvars32.bat" && "E:\Test\vstest.bat" "

How to start kdiff from Console application?

UPDATED...
I want to call kdiff from Console application. So I'm building two files and want to compare they at the end of executing my program:
string diffCmd = string.Format("{0} {1}", Logging.FileNames[0], Logging.FileNames[1]);
// diffCmd = D:\vdenisenko\DbHelper\DbHelper\bin\Debug\Reports\16_Nov 06_30_46_DiscussionThreads_ORIGIN.txt D:\vdenisenko\DbHelper\DbHelper\bin\Debug\Reports\16_Nov 06_30_46_DiscussionThreads_ORIGIN.txt
System.Diagnostics.Process.Start(#"C:\Program Files (x86)\KDiff3\kdiff3.exe", diffCmd);
//specification is here http://kdiff3.sourceforge.net/doc/documentation.html
It runs kdiff3 tool, but something wrong with filenames or command... Could you please look on screenshot and say what is wrong?

You need to use Process.Start():
string kdiffPath = #"c:\Program Files\Kdiff3.exe"; // here is full path to kdiff utility
string fileName = #"d:\file1.txt";
string fileName2 = #"d:\file2.txt";
Process.Start(kdiffPath,String.Format("\"{0}\" \"{1}\"",fileName,fileName2));
Arguments as described in the docs: kdiff3 file1 file2

var args = String.Format("{0} {1}", fileName, fileName2);
Process.Start(kdiffPath, args);

string kdiffPath = #"c:\Program Files\Kdiff3.exe"; // here is full path to kdiff utility
string fileName = #"d:\file1.txt";
string fileName2 = #"d:\file2.txt";
ProcessStartInfo psi = new ProcessStartInfo(kdiffPath);
psi.RedirectStandardOutput = true;
psi.WindowStyle = ProcessWindowStyle.Hidden;
psi.UseShellExecute = false;
psi.Arguments = fileName + " " + fileName2;
Process app = Process.Start(psi);
StreamReader reader = app.StandardOutput;
//get reponse from console app in your app
do
{
string line = reader.ReadLine();
}
while(!reader.EndOfStream);
app.WaitForExit();

This will run the program from your console app
Process p = new Process();
p.StartInfo.FileName = kdiffPath;
p.StartInfo.Arguments = "\"" + fileName + "\" \"" + fileName2 + "\"";
p.Start();
Unless you are trying to do something else, in which case you need to provide more details.

ffmpeg c# asp.net help on ProcessStartInfo

I wrote the code to convert the file in c#asp.net using ffmpeg. While executing it shows error as "The filename, directory name, or volume label syntax is incorrect" but my paths are correct. So what is the solution for this?
ProcessStartInfo info = new ProcessStartInfo("e:\ffmpeg\bin\ffmpeg.exe", "-i cars1.flv -same_quant intermediate1.mpg");
process.Start();
I tried another way as shown below. But that is also won't works.
ProcessStartInfo info = new ProcessStartInfo("ffmpeg.exe", "-i cars1.flv -same_quant intermediate1.mpg");
Please help me with sample code for converting one video file format to another using ffmpeg in c# asp.net. Thanks in advance.

There is wrapper library available
http://www.ffmpeg-csharp.com/
Also refer
http://www.codeproject.com/Articles/24995/FFMPEG-using-ASP-NET
https://stackoverflow.com/questions/tagged/asp.net+c%23+video
string apppath = Request.PhysicalApplicationPath;
string outf = Server.MapPath(".");
string fileargs = "-i" + " " + "\"C:/file.mp4 \"" + " " + "\"C:/files/test.flv \"";
Process myProcess = new Process();
myProcess.StartInfo.FileName = Server.MapPath(".")+"//ffmpeg.exe";
myProcess.StartInfo.Arguments = fileargs;
myProcess.StartInfo.UseShellExecute = false;
myProcess.StartInfo.CreateNoWindow = false;
myProcess.StartInfo.RedirectStandardOutput = false;
myProcess.Start();
myProcess.WaitForExit(50 * 1000);
Thanks
Deepu

The backslashes (\) in your path are considered to start escape sequences. To prevent this, either use double backslashes (\\), or prefix the string with #.
ProcessStartInfo info = new ProcessStartInfo("e:\\ffmpeg\\bin\\ffmpeg.exe",
"-i cars1.flv -same_quant intermediate1.mpg");
Or:
ProcessStartInfo info = new ProcessStartInfo(#"e:\ffmpeg\bin\ffmpeg.exe",
"-i cars1.flv -same_quant intermediate1.mpg");

You need to have backslashes in your path see below this is how I create videos using ffmpeg.exe
string sConverterPath = #"C:\ffmpeg.exe";
string sConverterArguments = #" -i ";
sConverterArguments += "\"" + sAVIFile + "\"";
sConverterArguments += " -s " + iVideoWidth.ToString() + "x" + iVideoHeight.ToString() + " ";
sConverterArguments += " -b " + iVideoBitRate.ToString() + "k -ab " + iAudioBitRate.ToString() + "k -ac 1 -ar 44100 -r 24 -absf remove_extra ";
sConverterArguments += "\"" + sOutputFile + "\"";
Process oProcess = new Process();
oProcess.StartInfo.UseShellExecute = true;
oProcess.StartInfo.Arguments = sConverterArguments;
oProcess.StartInfo.FileName = sConverterPath;