Argument passing in .sh scripts
Works fune for me.
I don't know what the -N command means, but
#!/bin/bash -l#$ -S /bin/bash#$ -N $2echo $1echo $2when called by sh foo.sh a b promptly echoes
ab
No you can't. The # at the beginning of the line makes it so that the $2 won't be replaced by the argument to the script. The way to do what you're trying to do is
qsub foo.sh -N <name>