Bash Command Not Found Error with $variable

Question

108   ps --ppid $process | while read -r line ; do
109      #echo $line | awk '{print $1;}
110      child=$($line | awk '{print $1;}')
111      echo $child
113   done

Running this code gives me the following error:

line 111: 3405: command not found

But if I uncomment line 109 it prints the correct value without an error

What are you expecting to happen? without the `echo` `$line` will be treated as a command to run, apparently the command `3405` in your example. — Eric Renouf, Nov 27 '16 at 01:52
This smacks of an XY problem to me, what is it that you're actually trying to do? — Eric Renouf, Nov 27 '16 at 01:54
line 110 looks fishy to me (did you mean to 'echo $line' to awk?) — Jeff Schaller, Nov 27 '16 at 02:00
Post [complete code that reproduces the problem](http://stackoverflow.com/help/mcve) and explain what you want the script to do. — Gilles 'SO- stop being evil', Nov 27 '16 at 21:19

score 1 · Answer 1 · answered Mar 17 '19 at 14:15

The issue is that you're not actually giving the value in $line to awk. Instead you try to execute it as a command.

If all you want to do is output the child processes of a process with a certain PID, then you don't need to loop:

ps --ppid "$process" -o pid=

This would get the list of processes that has $process as their PPID, and for each output their PID.

Also related:

Why is using a shell loop to process text considered bad practice?

score 0 · Answer 2 · answered Nov 27 '16 at 02:00

0

Fix it...

ps --ppid $process | while read -r line ; do
  child=$(echo $line | awk '{print $1}')
done

answered Nov 27 '16 at 02:00

jas-

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Bash Command Not Found Error with $variable

2 Answers2