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that was it.

thank you very much for your help

I also tryied that: VAR = "3"; echo $VAR

with this result: bash: VAR: command not found

thankx eitherway

Hello,
I have a (basic I guess) problem with bash scripting.
In the command line, this piece of code returns nothing (a blank line), and if I'm not worng it should return 3.


set VAR = "3"...