I am trying to run ".sh" file from c# core application.But it doesn't seem to be running properly.Here is my scenario.
I am working on .Net core project which is hosted on Linux environment.We are trying to create "PDF" in our project for which we have used "Apache FOP". Here i have created one "shell script" file "transform.sh" which internally calls "fop" with required parameters.Since developement is being done on windows machine we tested the same usinf "batch" file i.e. "transform.bat",but since we cannot use the "batch" file on linux enviornment we have created shell script file "transform.sh"
Following is the code from"transform.sh"
./fop -xml $1 -xsl $2 -pdf $3
Following is C# code from which i am calling the "shell script file
var process = new Process
{
StartInfo = new ProcessStartInfo
{
UseShellExecute = false,
RedirectStandardOutput = true,
Arguments = string.Format("{0} {1} {2}", XML_filename, XSL_filename, output)
}
};
process.StartInfo.FileName = "Path to shell script file";
process.Start();
process.WaitForExit();
Above code doesnot give any error but it also does not create the pdf file.If i directly run the shell script file from "Terminal" it works fine and create pdf file.
./transform.sh "/home/ubuntu/psa//PdfGeneration/ApacheFolder/XMLFolder/test.xml" "/home/ubuntu/psa/PdfGeneration/ApacheFolder/XSLTFolder/Certificate.xsl" "/home/ubuntu/psa/PdfGeneration/ApacheFolder/PDFFolder/t444t.pdf"
Please let me know if there is something wrong i am doing?How can i make the sheel script run on linux through C# core application. Thanks.
touch \some\filein the script to see if it's even executing./home/ubuntu/my folder/myfile.xmlWhen you pass arguments to a shell script in linux it's important to encapsulate those arguments in quotes so they aren't split into multiple arguments. Perhaps, to see what's coming out of the script, change the scripts to:echo "$1" "$2" "$3" > /home/ubuntu/scripttest.txt. Then run your program that executes the script and see what gets kicked into that file.Process.on your lineProcess.StartInfo.FileName="Path"Then use that ProcessStartInfo in the next line likeProcess.Start(StartInfo). Check out that example in the link in my first comment also here as it seem very similar to what you are doing.