Hi, here is another try, that sometimes works: Brian Hatch wrote: > To emulate this, let's set it ourselves in a normal shell: > > bash$ export VAR='`cat /etc/passwd`' > bash$ echo $VAR > `cat /etc/passwd` bash:~# export VAR='-e test\ntest' bash:~# echo $VAR test test bash:~# echo "$VAR" -e test\ntest bash:~# well, the "$VAR" means, that $VAR is the first argument for echo. IMHO there is nothing you can do to execute code in this statement. bye ralf
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