PHP - Assign variable and use it to assign value to array all in one step

Question

I've this config.php file that contains this example array:

return [
  myarray => [
    'foo' => 'bar_'.str_random(5).'.log'
  ]
]

I'd like to assign the str_random(5) into a $var variable and use it all in one step; for example, something like (this example doesn't work):

return [
  myarray = [
    'foo' => 'bar_'.(($var=str_random(5)) && $var).'.log'
  ]
]

The output should be:

echo $var;
uoEUu

print_r($myarray);
[
  'foo' => 'bar_uoEUu.log'
]

Thank you

EDIT

Ok, I found this solution, but It doesn't like me:

return [
  myarray = [
    'foo' => 'bar_'.$var=str_random(5).''.$var.'.log'
  ]
]

any cleaner suggestions?

Why? You could set $var before creating the array and just use it in the array. — Jay Blanchard
– Jay Blanchard, Commented Jul 17, 2019 at 12:16
This content is in a config file; and this array is returned: return [ myarray = [ 'foo' => 'bar_'.(($var=str_random(5)) && $var).'.log' ] ] — vlauciani
– vlauciani, Commented Jul 17, 2019 at 12:18
There is no reason I can see why you'd want to do this all in one step, and depending on what str_random is / does, the && $var may be unneeded, and is the bit that means that $var is a boolean, so you're concat-ing a bool to a string which appears to not be what you want. — Jonnix
– Jonnix, Commented Jul 17, 2019 at 12:21

IMSoP · Accepted Answer · 2019-07-17 13:22:12Z

1

The solution is actually much simpler than you think:

$myarray = [
  'foo' => 'bar_'.($var=str_random(5)).'.log'
];

The reason this works is that any assignment can be used as an expression whose value is the value assigned. This isn't often necessary, but has a few uses, like setting multiple variables to the same value:

$foo = $bar = 42;

In this case, you're concatenating a fixed string, an expression (which happens to be an assignment), and another fixed string; the extra parentheses just make sure that the assignment expression is evaluated before the concatenation.

answered Jul 17, 2019 at 13:22

IMSoP

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Comments

Caconde · Accepted Answer · 2019-07-17 13:22:29Z

1

IMO, you are overthinking the problem. If you need to use $var after assigning it to foo you can just do the following:

$var = str_random(5);
$myarray = [
  'foo' => "bar_$var.log"
];

answered Jul 17, 2019 at 13:22

Caconde

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1 Comment

IMSoP Over a year ago

As clarified in comments, this is a simplified example, which comes in the middle of a larger block of code in a config file. While unusual, I can imagine scenarios where this is useful.

delboy1978uk · Accepted Answer · 2019-07-17 13:28:26Z

1

Just wrap it in brackets:

$myarray = [
  'foo' => 'bar_'.($var=uniqid()).'.log'
]

var_dump($var, $myarray);

Which gives you :

string(13) "5d2f1f1e01338" 
array(1) { ["foo"]=> string(34) "bar_5d2f1f1e01338.log" }

Check it out here: https://3v4l.org/JfZf6

edited Jul 17, 2019 at 13:28

answered Jul 17, 2019 at 13:15

delboy1978uk

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3 Comments

IMSoP Over a year ago

This doesn't quite match the desired output, as the random string gets inserted twice.

delboy1978uk Over a year ago

His code wanted 2? (based on his updated question) 'bar_'.($var=uniqid()).''.$var.'.log' the only bit i changed was uniqid(), as the function he uses is a custom one

IMSoP Over a year ago

The desired output shows just one instance of the random string: bar_uoEUu.log. The duplication seems to just be a mistake in trying to get it working.

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