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awk output in a variable

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# 1  
Old 04-01-2009
awk output in a variable
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  
Old 04-01-2009
Quote:
Originally Posted by andaluzia
Not sure why it is not working the following :

Please put code inside [code] tags.
Quote:
Code:
set -- $@


Why?
Quote:
Code:
stype ="a"


Syntax error; there should be no space before = (and there's no need for the quotes).
Quote:
Code:
for shell_args in "$@"
do
$stype=` awk '{print substr ("'"$shell_args"'", 0, 3)}' `

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

Quote:
Code:
echo $stype
done

# 3  
Old 04-01-2009
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  
Old 04-01-2009

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

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

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

# 5  
Old 04-01-2009
Thanks.

Wonderful solution!
# 6  
Old 04-01-2009
Quote:
Originally Posted by cfajohnson

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

Wow, I love it! Thanks cfa. All these years I've been doing:
Code:
for arg;do
  stype=$(echo $arg|cut -b-3)
  echo $stype
done

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