shell script: if statement

Question

I'm following the tutorial here: http://bash.cyberciti.biz/guide/If..else..fi#Number_Testing_Script

My script looks like:

lines=`wc -l $var/customize/script.php`
if test $lines -le 10
then
    echo "script has less than 10 lines"
else
    echo "script has more than 10 lines"
fi

but my output looks like:

./boot.sh: line 33: test: too many arguments
script has more than 10 lines

Why does it say I have too many arguments? I fail to see how my script is different from the one in the tutorial.

Answer 1

wc -l file command will print two words. Try this:

lines=`wc -l file | awk '{print $1}'`

---

To debug a bash script (boot.sh), you can:

$ bash -x ./boot.sh

It'll print every line executed.

Answer 2

wc -l file

outputs

1234 file

use

lines=`wc -l < file`

to get just the number of lines. Also, some people prefer this notation instead of backticks:

lines=$(wc -l < file)

Also, since we don't know if $var contains spaces, and if the file exists:

fn="$var/customize/script.php"
if test ! -f "$fn"
then
    echo file not found: $fn
elif test $(wc -l < "$fn") -le 10
then
    echo less than 11 lines
else
    echo more than 10 lines
fi