UNIX for Dummies Questions & Answers

What's wrong with the solution provided and where exactly are you stuck?

well, as I said I need to have the result of $(echo "$x" | cut -d"~" -f1) stored into a variable, then to echo this one. but quoting the input for echo seems not to work inside $(...); i.e echo "$x" prints ok, but in $(echo "$x" | cut -d"~" -f1), the * in x is still considered a special char and I get the list of files in current directory, just like echo $x will do.

Hmm.... I must be missing something....
Have you tried what I posted last (exactly as did)?

Code:

#!/bin/ksh

x="hello * world~something"
y=$(echo "$x" | cut -d'~' -f1)
echo "y->[${y}]"

If the last 'echo' doesn't work, I'll let others to chime in as I don't understand the problem/explanation.

I guess my explanations were not great, 'cause ur code works great. 10x and sorry for the trouble

no problem - glad to be of help.