CODEWITHSUNDEEP
bash
user1020822
user1020822

Reputation: 33

Bash call a function in a command

I have a question regarding the usage of functions in a command in bash. getRegex is my function, it is defined at the end of the the file. The command that I want to use is the following:

COUNT=`grep -rnE 'getRegex' $HOME/new`

Now I tried a lot of different variants but I cannot make it work, even if I split it in 2. The method works correctly if I call it the following way: getRegex. Any idea what I am missing? TIA

Upvotes: 3

Views: 4304

Answers (2)

ДМИТРИЙ МАЛИКОВ
ДМИТРИЙ МАЛИКОВ

Reputation: 21972

The key words to answer are "bash command substitution", which you could find in man bash or google.

By the way, double quotes are really important here.

#!/bin/bash

function my_func () {
    echo "no"
}

string="no you don't
no you don't
no you don't
no you don't
no you don't"


COUNT="$( echo "${string}" | grep "$( my_func )" -c )"
echo "${COUNT}"

And

$> ./ok.sh 
5

Upvotes: 4

Dan
Dan

Reputation: 10786

If you're trying to call a bash command within another bash command, the inner command (here getRegex) needs to be enclosed in backticks `` or else it will be interpreted as text. Since you here would have backticks inside backticks, you'll need to escape the inner ones. Try this:

COUNT=`grep -rnE '\`getRegex\`' $HOME/new`

But, through the wonders of POSIX, we can use a different syntax. Anywhere you use backticks, you can also use $(). So to avoid backslash emesis, you could write:

COUNT=$(grep -rnE '$(getRegex)' $HOME/new)

Upvotes: 0

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