try using an absolute path. Also not that there is an output and error stream which you are not aquireing that may have more detailed information.
-rickb -----Original Message----- From: Patrick Kosiol [mailto:[EMAIL PROTECTED]] Sent: Monday, January 13, 2003 10:58 AM To: Tomcat Users List Subject: >>> Starting an .exe though out of a servlet Hi, I'm runnung TomCat 4.0.4 and I want to start an external .exe-File. My Code: String[] runString = {relativeDataPath + "data.exe", parameter0, parameter1, parameter2, parameter3}; Process p = Runtime.getRuntime().exec(runString); System.out.println(runString); if(p.waitFor() == 0){ ........ The 'data.exe'-File is placed in $CATALINA_HOME/webapps/ROOT/relativeDataPath so it should be no problem to access this File. The Number of Parameters etc. is OK. But I get the following Error-Message: java.io.IOException: CreateProcess: 'relativeDataPath'/bfpldata.exe parameter0 parameter1 parameter2 parameter3 error=3 Does someone know what 'error=3' mean? And how can I access this 'data.exe' and execute this File with the Parameters? The starting-routine seems to be correct. Thx Patrick -- To unsubscribe, e-mail: <mailto:[EMAIL PROTECTED]> For additional commands, e-mail: <mailto:[EMAIL PROTECTED]> -- To unsubscribe, e-mail: <mailto:[EMAIL PROTECTED]> For additional commands, e-mail: <mailto:[EMAIL PROTECTED]>
