Piping output to cut

Question

I am trying to get the name of the shell executing a script.

Why does

echo $(ps | grep $PPID) | cut -d" " -f4

work while

echo ps | grep $PPID | cut -d" " -f4

does not?

Answer 1

The reason is that

echo ps

just prints out the string ps; it doesn't run the program ps. The corrected version of your command would be:

ps | grep $PPID | cut -d" " -f4

Edited to add: paxdiablo points out that ps | grep $PPID includes a lot of whitespace that will get collapsed by echo $(ps | grep $PPID) (since the result of $(...), when it's not in double-quotes, is split by whitespace into separate arguments, and then echo outputs all of its arguments separated by spaces). To address this, you can use tr to "squeeze" repeated spaces:

ps | grep $PPID | tr -s ' ' | cut -d' ' -f5

or you can just stick with what you had to begin with. :-)

Answer 2

The first line:

echo $(ps | grep $PPID) | cut -d" " -f4

says:

• Execute ps | grep $PPID in a sub-shell

• Return the output - which will be something like this:

  3559 pts/1 00:00:00 bash
  

  and then use that output as the first parameter of echo - which, in practice, means just echo the output

• Then run cut -d" " -f4 on that - which gives you the command name in this case

The second command:

echo ps | grep $PPID | cut -d" " -f4

says:

• Echo string ps
• Grep that string for $PPID - this will never return anything, as $PPID contains a number, so it will never be ps. Thus, grep returns nothing
• Execute cut -d" " -f4 with input of the previous command - which is nothing, so you get nothing