How to pass variable array structure to array

Question

I've the following variable that is dinamically created:

$var = "'a'=>'123', 'b'=>'456'";

I use it to populate an array:

$array=array($var);

I can't do $array=array('a'=>'123', 'b'=>'456') because $var is always different.

So it shows me:

Array
(
    [0] => 'a'=>'123', 'b'=>'456'
)

This is wrong, 'cause I need to get:

Array
(
    [a] => 123
    [b] => 456
)

What is wrong on my code? Thanks in advance.

If for whatever reason you must generate a string as $var in the first place. Consider generating it as a JSON string - {"a":"123", "b":"456"} as you can then use standard JSON methods to convert it. — Nigel Ren
– Nigel Ren, Commented Aug 17, 2019 at 15:43
If you have control over the creation of the dynamic string, you should use JSON encoding to pass the values. Then it’s as simple as json_encode() and json_decode() — Tim Morton
– Tim Morton, Commented Aug 17, 2019 at 15:57

Tim Biegeleisen · Accepted Answer · 2019-08-17 15:47:27Z

1

Ideally you should just leverage PHP's syntax to populate an associative array, something like this:

$array = [];
$array['a'] = '123';
$array['b'] = '456';

However, you could actually write a script which parses your input to generate an associate array:

$var = "'a'=>'123', 'b'=>'456'";
preg_match_all ("/'([^']+)'=>'([^']+)'/", $var, $matches);
$array = [];
for ($i=0; $i < count($matches[0]); $i++) {
    $array[$matches[1][$i]] = $matches[2][$i];
}
print_r($array);

This prints:

Array
(
    [a] => 123
    [b] => 456
)

answered Aug 17, 2019 at 15:47

Tim Biegeleisen

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