Below is the part of the code that is throwing error.

for (( p=0; p<${var2}; p++ ))

if [[ ${array2[$p]} == *"xyz"* ]]
    awk -F" " '{ array3[p]="123" };1' ${array1[$i]}
    awk -F" " '{ array3[p]="456" };1' ${array1[$i]}

When i run the script in debug mode, i could see that the value of p is not being considered.

+ awk '-F ' '{ array3[p]="456" };1' oct.csv

I am running the script as ./script.sh and /bin/ksh is the shell used. Please help.

  • What is it you're trying to accomplish? – ilkkachu Oct 3 '16 at 19:55
  • 2
    I'm pretty confident you're after awk -v "p=$p". But btw, it is much better to post a working example or describe what you are after. We are just guessing what var2 or array2 may contain. – grochmal Oct 3 '16 at 20:08
  • I am trying to assign array3[p]="123" but since value of p is not getting retrieved, i want to know how to get it where p is a variable. – user2568967 Oct 3 '16 at 20:20
  • @user2568967 You're wanting to assign ${array3[$p]} value to "123" or "456" depending on if ${array2[$p]} is == "xyz", right? But why are you setting a variable inside the awk statement then not using it? Examples of what you want to accomplish with array3 would be helpful. Also as an FYI you don't need to use -F" "with awk as it uses space as the field separator by default. – Zachary Brady Oct 3 '16 at 20:53
  • I am trying to use the variable inside the awk array statement but not sure how to use it since i could not retrieve the value of p. Once it is retrieved, its place will be filled either with 123 or 456. – user2568967 Oct 3 '16 at 20:58

First of all, this is wrong:

awk -F" " '{ array3[p]="123" };1' ${array1[$i]}

${array1[$i]} is not a file, so you can't just throw it as first command line argument to awk. With variables you'd have to use either echo $VARIABLE or here string <<< like so

awk -F" " '{ array3[p]="123" };1' <<< ${array1[$i]}

Second, of all, there is a scope issue. array3[p]="123" only exists within awk code, so you can't assign anything to an outside of it. What you could do, however, simply do


It is also unclear where exactly $i comes from in your script. Overall, your question gives us a bit of code but doesn't provide any clues to fully answer your question

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