PHP assignment with a default value

Question

What's a nicer way to do the following, that doesn't call f() twice?

$x = f() ? f() : 'default';

StasM · Accepted Answer · 2011-01-28 19:21:57Z

30

In PHP 5.3, you can also do:

  $a = f() ?: 'default';

See the manual on ?: operator.

answered Jan 28, 2011 at 19:21

StasM

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1 Comment

Andrew Over a year ago

+1 since I haven't seen that used in any of my books that cover PHP 5.3. Very good to know.

dreeves · Accepted Answer · 2011-01-28 19:14:14Z

8

This seems to work fine:

$x = f() or $x = 'default';

answered Jan 28, 2011 at 19:14

dreeves

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Comments

ajreal · Accepted Answer · 2011-01-28 19:17:00Z

2

function f()
{
  // conditions 
  return $if_something ? $if_something : 'default';
}

$x = f();

answered Jan 28, 2011 at 19:17

ajreal

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dreeves Over a year ago

Ah, good answer. I meant for f() to be something we don't have control over and in fact the way this came up for me it wasn't a function at all but $_REQUEST['foo'] so it wasn't an efficiency issue so much as a typing issue.

Mr Griever · Accepted Answer · 2011-01-28 19:23:09Z

1

$x = ($result = foo()) ? $result : 'default';

test

answered Jan 28, 2011 at 19:23

Mr Griever

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1 Comment

dreeves Over a year ago

This works, but it looks like it can be done more cleanly, with no temp variable.

chelmertz · Accepted Answer · 2011-01-28 19:22:23Z

0

You could save it to a variable. Testcase:

function test() {
        echo 'here';
        return 1;
}

$t = test();
$x = $t ? $t : 0;
echo $x;

answered Jan 28, 2011 at 19:22

chelmertz

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