That did it. Thanks!


>The Process.Start method has another overload that takes command arguments. Try:
>

>string sApp = @"C:\Program Files\Java\jdk1_6_10\bin\javac";
>string sArg = @"c:\myjava\Test.java";
>System.Diagnostics.Process.Start(sApp, sArg);
>

>
>
>>I was just playing around, and I tried this:
>>
>>

>>string sCommand = @"C:\Program Files\Java\jdk1_6_10\bin\javac c:\myjava\Test.java";
>>System.Diagnostics.Process.Start(sCommand);
>>

>>
>>And I get "The system cannot find the specified file"
>>
>>If I leave off the "c:\myjava\Test.java", then the java compiler options are displayed, so that's not the part it can't find. I have verified that the .java file is there.
>>
>>I created a batch file with this in it:
>>

>>"C:\Program Files\Java\jdk1_6_10\bin\javac" "c:\myjava\Test.java"
>>

>>
>>I ran it and it compiled the file fine.
>>
>>I'm guessing it has something to do with the space in the file paths.
>>
>>Anyone know what's wrong?