Correctly count number of lines a bash variable

Question

I need to count the number of lines of a given variable. For example I need to find how many lines VAR has, where VAR=$(git log -n 10 --format="%s").

I tried with echo "$VAR" | wc -l), which indeed works, but if VAR is empty, is prints 1, which is wrong. Is there a workaround for this? Something better than using an if clause to check whether the variable is empty...(maybe add a line and subtract 1 from the returned value?).

Answer 1

The wc counts the number of newline chars. You can use grep -c '^' for counting lines. You can see the difference with:

#!/bin/bash

count_it() {
    echo "Variablie contains $2: ==>$1<=="
    echo -n 'grep:'; echo -n "$1" | grep -c '^'
    echo -n 'wc  :'; echo -n "$1" | wc -l
    echo
}

VAR=''
count_it "$VAR" "empty variable"

VAR='one line'
count_it "$VAR" "one line without \n at the end"

VAR='line1
'
count_it "$VAR" "one line with \n at the end"

VAR='line1
line2'
count_it "$VAR" "two lines without \n at the end"

VAR='line1
line2
'
count_it "$VAR" "two lines with \n at the end"

what produces:

Variablie contains empty variable: ==><==
grep:0
wc  :       0

Variablie contains one line without \n at the end: ==>one line<==
grep:1
wc  :       0

Variablie contains one line with \n at the end: ==>line1
<==
grep:1
wc  :       1

Variablie contains two lines without \n at the end: ==>line1
line2<==
grep:2
wc  :       1

Variablie contains two lines with \n at the end: ==>line1
line2
<==
grep:2
wc  :       2