You caught me away from my reference library, but essentially arguments on the command line of a command that invokes a bash script are referenced as $1, $2, $3, etc. So you could, after your #!/bin/bash line, assign the arguments to your variables like so: VAR1 = $1 VAR2 = $2 I think. You may need to define them as $VAR1 and $VAR2, but I'm not sure, try it both ways and see which works. Also, if you have more than 9 arguments on the command line, you can use the shift command to access those beyond $9. The man page for bash is pretty thorough, as I recall. -- Hugh On Wed, 6 Aug 2003, Gregory Nowak wrote: > Hi all. > > Can someone please tell me how to write a bash script that would accept 2 > arguments from the command line, where the first argument would be an > integer like 1234567, and the second argument would be a text string? > I would then like to be able to access the variables inside the script > via $VAR1 and $VAR2. > > Thanks for any help. > > Greg > > > -- > Free domains: http://www.eu.org/ or mail dns-manager at EU.org > > > _______________________________________________ > Speakup mailing list > Speakup at braille.uwo.ca > http://speech.braille.uwo.ca/mailman/listinfo/speakup >
