Question

I want to set a path in my Java program with this Windows command (this path contains some DLL files that are used in a native peripheral of my program):

c:\>path=%path%;"C:\Users\NetBeansProjects\IPTV1.7\3rd_party"

But this causes an exception when the program runs:

java.io.IOException: Cannot run program “path=%path%;C:\Users\NetBeansProjects\IPTV1.7\3rd_party\”: CreateProcess error=2, The system cannot find the file specified

I don’t know why I can set the path with no problem in the command prompt but then get an exception thrown in the code.

String path = "C:\\Users\\NetBeansProjects\\IPTV1.7\\3rd_party\\";
Runtime rt = Runtime.getRuntime();
Process p = rt.exec("path=%path%;"+ path);

Answer 1

Your command

path=%path%;"C:\Users\NetBeansProjects\IPTV1.7\3rd_party"

is not a “real” windows command, but only a variable assignment, interpreted by your shell (cmd.exe), visible only in the same shell session and any commands (other programs) started from there.

When trying to execute this from Java with Runtime.exec(), the windows CreateProcess function tries to find a executable file with this strange name, which obviously does not exist (can’t exist, I think), and you get this error.

Even if you could execute this, for example by calling cmd.exe, it would only influence this same cmd.exe process (and any programs started from there, not your own Java process (and programs started from here).

Depending on what you in fact want, you could, for example:

• give the path with ProcessBuilder directly to the process which in fact needs it (like Aaron showed)
• search by yourself for executable files, if you want this for finding commands in the next exec
• put the variable assignment and other commands together in a .BAT or .CMD file and execute this.