Problems assigning variable to output of command

Question

I am using wc -l < songs.txt to find out how many lines are in songs.txt

Then I want to store that as a variable. I've tried these:

OUT = $(wc -l < songs.txt)
OUT = "$(wc -l < songs.txt)"

But I still get command not found error for OUT

wc -l < songs.txt works on its own

What simple thing am I missing here?

Tip: [shellcheck](http://shellcheck.net) automatically identifies this and other problems. — that other guy, Jan 28 '17 at 19:07

score 0 · Answer 1 · answered Jan 28 '17 at 19:07

0

it's easy first of all don't use space after and before equal sign

OUT=$(wc -l < songs.txt)

second for call it you can write something like this :

echo $OUT

Good luck

answered Jan 28 '17 at 19:07

Freeman

You should double-quote variable references, e.g. `echo "$OUT"`. – Gordon Davisson Jan 29 '17 at 20:47
@GordonDavisson, thank you , is it a rule ? because it's working without "" too! – Freeman Jan 31 '17 at 09:49
1

Not a strict rule, and it'll *often* work without double-quotes, but there are enough cases where *not* double-quoting variables [will](http://stackoverflow.com/questions/26616849/how-to-set-shell-variable-to-result-of-evaluation-in-os-x) [cause](http://stackoverflow.com/questions/41793270/grep-on-a-variable-containing-a-string-with-a-tab/41793296) [trouble](http://superuser.com/questions/655926/why-does-checking-whether-a-folder-exists-using-d-on-an-empty-string-return-tru) that it's safest to treat it as a rule. – Gordon Davisson Jan 31 '17 at 16:10
good to know, thank you brother , 1++ :D – Freeman Jan 31 '17 at 18:13

Grzegorz Górkiewicz · Accepted Answer · 2017-01-28T20:17:47.293

0

Spaces should be avoided.

out=$(wc -l < songs.txt)

edited Jan 28 '17 at 20:17

answered Jan 28 '17 at 19:07

1

The command is correct, but shouldn't it be "*can't have* spaces" instead of "are missing"? – Benjamin W. Jan 28 '17 at 20:16
Yes, you are right. – Grzegorz Górkiewicz Jan 28 '17 at 20:17

2 Answers2