bash if [ -n "$output of a variable" ]

Question

line=$(grep "# rvm line" ~/.bashrc)

if [ ! -n "$line" ]; then
  echo "found"
else
  echo "not found"
fi

What's wrong with my quotes in the first line?

EDIT: The problem is set -o errexit, which I use in my script. I suppose -n is treated as an error, exiting the following processes. How can I overcome this, keeping the error check? (Alternatives to -n could work too).

wrong kind of red-hue? Change the shebang to #!/bin/bash -x or use set -vx and examine the output... — Fredrik Pihl
– Fredrik Pihl, Commented Sep 29, 2011 at 13:37
you are using -n wrongly, means True if string is not empty and then you negate that using ! — Fredrik Pihl
– Fredrik Pihl, Commented Sep 29, 2011 at 13:41

Kusalananda · Accepted Answer · 2011-09-29 14:00:07Z

3

If you don't plan to use $line for anything else than testing if it's empty or not, then you might as well do this:

if grep -q '# rvm line' ~/.bashrc; then
  echo "found"
else
  echo "not found"
fi

answered Sep 29, 2011 at 14:00

Kusalananda

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1 Comment

octosquidopus Over a year ago

That's perfect. It's shorter and doesn't exit the script when used with set -o errexit

ring bearer · Accepted Answer · 2011-09-29 13:54:27Z

1

Nothing

Nothing is apparently wrong with the quotes? Are you not getting expected result ?

ok,your if condition is wrong - if it was intended to check that something was returned by the grep statement. I will leave that as a clue for yourself to figure out. Goodluck

edited Sep 29, 2011 at 13:54

answered Sep 29, 2011 at 13:37

ring bearer

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1 Comment

octosquidopus Over a year ago

hmm.. you're right. It's working by itself, but when it's within my script, it just locks.

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