Re: bash command substitution problem

[Pollywog]

On Sunday 08 May 2005 05:08 pm, Phil Dyer wrote:

> try this one.
>
> function lsf {
>     for i in *; do
>         if [ -f "$i" ]; then
>             echo "$i"
>         fi;
>     done;
> }
>
> lsf | xargs chmod 600

Thanks, that works.

8)


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