awk output in a variable

Shell Programming and Scripting
1

Not sure why it is not working the following :

set -- $@
stype ="a"
for shell_args in "$@"
do
$stype=` awk '{print substr ("'"$shell_args"'", 0, 3)}' `
echo $stype
done

Thank you

2

Please put code inside

 tags.



set -- $@

[/quote]

[indent]
Why?

Syntax error; there should be no space before = (and there's no need for the quotes).

stype=` awk '{print substr ("'"$shell_args"'", 0, 3)}' `
3

hi
Thanks for your help. I've made the modifications but still does not work:

code:
[#!/bin/sh
set -x

for shell_args in "$@"
do
stype=` awk '{print substr ("'"$shell_args"'", 0, 3)}' `
echo $stype
done
]

**************************

debug:
[sh mytest.sh es_GELitems1 ab_s1_Abools1 ai_s1_Aints1 as_s1_Astrings1 at_s1_Atests1 ee_ ei_s1_Eints
+ for shell_args in '"$@"'
++ awk '{print substr ("es_GELitems1", 0, 3)}'

]

4 
stype=` awk 'BEGIN {print substr ("'"$shell_args"'", 0, 3)}' `

(It's very inefficient to call awk separately for every argument.)

for arg
do
   stype=${arg%"${arg#???}"}
   printf "%s\n" "$stype"
done
5

Thanks.

Wonderful solution!

6

Wow, I love it! Thanks cfa. All these years I've been doing:

for arg;do
  stype=$(echo $arg|cut -b-3)
  echo $stype
done