PHPUnit - Bash Script Output to Variable

Question

I'm trying to capture the output of my test suite from PHPUnit to determine whether a failure occurred. However, when I attempt to store the output in a bash variable, the variable is always empty:

PHPUNIT_RESULT=`vendor/bin/phpunit`

if [ -z "$PHPUNIT_RESULT" ]; then
        echo "something there!
fi

However, the variable always seems to be empty.

EDIT: Sample output

PHPUnit 3.4.5 by Sebastian Bergmann.

......F.......F

Time: 0 seconds, Memory: 8.00Mb

There was 1 failure:

1) MyTest::testTemp
Failed asserting that <boolean:false> is true.

/path/to/myTest.php:68

FAILURES!
Tests: 4, Assertions: 5, Failures: 1, Incomplete: 1.

To start with, generally speaking, using backticks to perform command substitution in bash is discouraged. Use the $(...) syntax instead. In either case, what's the expected output of vendor/bin/phpunit? — Sebastian Lenartowicz
– Sebastian Lenartowicz, Commented Jul 24, 2016 at 16:32
@SebastianLenartowicz thanks for the info. I've also tried it with $() and had the same results. I've updated the question with a sample output. It is multi-line output. — djt
– djt, Commented Jul 24, 2016 at 16:35
Don't forget as well that the -z operator checks for an empty string - so, if you print "Something there!", it actually means you're getting output! — Sebastian Lenartowicz
– Sebastian Lenartowicz, Commented Jul 24, 2016 at 16:48
@Matteo thanks for that link, it lead to some useful results — djt
– djt, Commented Jul 24, 2016 at 17:18

Jeff Puckett · Accepted Answer · 2016-07-24 17:10:02Z

6

if there's any test failure, phpunit will exit with a non-zero status. you can check this with the $? variable.

./vendor/bin/phpunit /path/to/myTest.php

if [ $? -ne 0 ]; then
        echo "failed test"
fi

answered Jul 24, 2016 at 17:10

Jeff Puckett

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1 Comment

djt Over a year ago

Thanks for this. Thats exactly what I ended up doing

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