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function parameter issue

 
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# 1  
Old 02-02-2011
function parameter issue
hi,
i have a function say "mask". i'm passing parameters to the function like file_name \$1 |.
inside the function i'm naming the second parameter as position (i.e value of position will be "$1").
i'm passing the position's value like " $position=a[$position]"
But it is nor picking the value "$1". Instead, while executing it just goes as $position itself.
Please help.
# 2  
Old 02-02-2011
Please tell us what shell/language you are using. Also please show an excerpt of your code since there are things unclear from your description (at least for me), thanks.
When doing so, use [code] and [/code] tags to enhance readability and preserve formatting, thanks.
# 3  
Old 02-02-2011
i'm using ksh.

the below is my code. i'm not able to retrieve the "position" value.

Code:
function mask
{
   set -x
   file_name=$1
   position=$2
   separator=$3
 
   if [ -e $file_name ];then
      awk '
         BEGIN {FS="[,\|]"}
         FNR==NR{a[$1]=$2;next}
         {if ($position in a) {$position=a[$position];print}
            else print}
      ' OFS="$separator" $DIR_FEEDBASE/GEID_lookup.txt $file_name > $file_name_tmp.dat
      mv $file_name_tmp.dat $file_name
   fi
}

The parameters which will be passed to the function will be like " mask <file_name> \$1 |


Moderator's Comments:
Mod Comment I asked you kindly to use code tags - instead you just formatted it with fonts. It is awful to remove all those font formattings to be able to add indention that you also did not use. You get a reminder for this and a PM with an instruction how and when to use code tags.
Last edited by zaxxon; 02-02-2011 at 10:08 AM.. Reason: code tags again
# 4  
Old 02-02-2011
  • Not escaping the | when handing it over as parameter of your function will not work, as it is a special symbol in the shell to pipe output from stdout to stdin of the next command. You should try escaping it or putting double quotes around it. You did escape the $ of $1 so why not do the same with the pipe?! Smilie
  • $position is a variable not known/defined in awk. You might want to use something like awk -v position=$position ... to pass it to awk.

After correcting those, try again and let's see if it get's stuck again.
# 5  
Old 02-03-2011
hi zaxxon,
no luck.even now it is not picking the value.
# 6  
Old 02-03-2011
Without having the input files available and the actual code of your function, it is not easy to recreate your problem.
Did you just try to print out the values inside awk? You should try to check out, at which point the values are not the ones you expect.
Last edited by zaxxon; 02-03-2011 at 10:21 AM.. Reason: typo
 
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