Bash oneliner with pipes and if condition giving error

Question

I am trying to find the number of a particular process in bash using if condition as

if ps auwx | grep -v grep | grep -ic python -le 2; then echo error; else echo no_error; fi

and I am getting output as

grep: python: No such file or directory

no_error

The one-liner seems to break if I use pipe, and no error is thrown if I omit pipe, and it doesn't matter if I use the absolute path to grep either.I cannot get the required result without the pipe. What am I doing wrong here? I can get this done in a script file, by breaking it into variables and then doing comparing it, but I was using this as an exercise to learn bash. Any help is greatly appreciated.

Answer 1

First of all, the syntax of if command is:

if cmd; then
    # cmd exited with status 0 (success)
else
    # cmd exited with status >0 (fail)
fi

The cmd above is the so-called list - a sequence of pipelines. Each pipeline is a sequence of commands separated with |.

The -le operator is interpreted only by the test command (also known as [, or [[), not by the if command.

So, when you say:

if ps auwx | grep -v grep | grep -ic python -le 2; then ... fi

you actually call grep with arguments:

grep -ic python -le 2

And since -e is used to specify the search pattern, the argument python is interpreted as a filename of the file to search for pattern 2. That's why grep tells you it can't find file named python.

To test the output of a command pipeline in if, you can use the command substitution inside the [[/[/test (as the other answer suggests):

if [[ $(ps auwx | grep -v grep | grep -ic python) -le 2 ]]; then ... fi

or, within (( .. )), with implicit arithmetic comparisons:

if (( $(ps auwx | grep -v grep | grep -ic python) <= 2 )); then ... fi

Answer 2

using a command substitution in a condition

if [[ $(ps ...) -le 2 ]]; then