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Top Forums Shell Programming and Scripting Reading a string and passing passing arguments to a while loop Post 302590235 by goddevil on Sunday 15th of January 2012 02:40:45 AM
Old 01-15-2012
Quote:
Originally Posted by agama
You had a missing quote on one of the echo statements.

I think this is all that you need. The cat.... statement isn't necessary. Awk reads the param.txt file and thus you don't need to cat it through head/tail to get it into the awk.

Code:
awk -F : -v last=${RESPONSESTOP:-0} -v first=${RESPONSE:-0} '
   /^#/ { next; } 
 
   NR >= first && (last == 0  || NR <= last) { 
        print $1, $2, $3, $0
    }
   ' param.txt | while read x y z i 
   do
        echo "first field $x"
        echo "second field $y"
        echo "third field $z"
        echo "whole record (i) $i"   # <missing close quote was here
 
        if [[ $z == "GOD" ]]
        then
                function1 $x $y
                function2 $y
                function3 $y
        fi
        if [[ $z == "DEVIL" ]]
        then
            function4
        fi
   done

Hi agama,

I have rectified the missing quote. when i run the script, i get the following error
Quote:
awk: syntax error near line 1
awk: bailing out near line 1
---------- Post updated at 07:40 AM ---------- Previous update was at 07:26 AM ----------

Quote:
Originally Posted by Scrutinizer
Or alternatively, use plain shell without pipes, e.g.
Code:
START=2; STOP=7
linenr=0
while IFS=: read x y z ; do
  linenr=$((linenr+1));
  [ $linenr -ge $START ] || continue
  [ $linenr -le $STOP  ] || break
  case $x in 
    \#*) continue
  esac
  echo " Do your stufff with $x $y and $z at $linenr" 
done < infile

Code:
 Do your stufff with RED value2 and Y at 2
 Do your stufff with BLUE value4 and Y at 4
 Do your stufff with RED value5 and Y at 5
 Do your stufff with RED value6 and Y at 6
 Do your stufff with BLUE value7 and Y at 7

In the above code, if i change the values of start and stop to 7 and 18 respectively and try to execute
Code:
echo " Do your stufff with $x $y and $z at $linenr"

i am not getting any output. I can only print the values between 1 and 6. If i try to print from 6 onwards, there is no output (I have over 50 files btw).

Last edited by goddevil; 01-15-2012 at 05:54 AM..
 

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