How do I convert a string of an array to an array in PHP?

Question

This seems like a simple task, but I haven't been able to find a way to do this. I have an output of data stored as a string like this:

$rawData = "array('first' => '$first', 'middle' => '$middle', 'last' => '$last')";

What I simply need to do is convert it to this array:

$arrData = array('first' => "$first", 'middle' => "$middle", 'last' => "$last");

The closest I have been able to get is this shown below with print_r results:

$Data = explode(',', $rawData);

Array
(
    [0] => 'first' => '$first'
    [1] =>  'middle' => '$middle'
    [2] =>  'last' => '$last'
)

What I need is this:

Array
(
    [first] => $first
    [middle] => $middle
    [last] => $last
)

Must be something very easy I have overlooked. Please help.

I have spent a lot of time looking at similar questions on this site and looking at w3schools and other references. I know I must be overlooking something simple, but can't find it. Something more than just a -1 vote would be greatly appreciated. Thanks — user1028866
– user1028866, Commented Oct 11, 2015 at 7:13
I saw that this fellow had a similar question, but he was starting with the array. I have to start with the string. stackoverflow.com/questions/684553/… — user1028866
– user1028866, Commented Oct 11, 2015 at 7:51

Axel Amthor · Accepted Answer · 2015-10-11 07:38:07Z

2

Depending on where you get the string from, you may use eval. It's highly recommended to not use this function, whenever the contents of that string may be influenced directly or indirectly by a user, see hint at http://php.net/manual/en/function.eval.php. And it must contain valid PHP syntax, thus putting it into a try/catch block is necessary:

$evalArray = false;
$first = 'a';
$middle = 'b';
$last = 'c';

$rawData = "array('first' => '$first', 'middle' => '$middle', 'last' => '$last')";

try {
    $evalArray = eval($rawData);
} catch ( $e )
{
    echo "parsing failed " . $e
}

print_r($evalArray );

answered Oct 11, 2015 at 7:38

Axel Amthor

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3 Comments

user1028866 Over a year ago

this quote at the end was quite revealing! "If eval() is the answer, you're almost certainly asking the wrong question." -- Rasmus Lerdorf, BDFL of PHP Thanks again.

Axel Amthor Over a year ago

Right. Bt in order to re-design that, much more information is needed as disclosed in your question!

user1028866 Over a year ago

I understand. I was appreciating your caution. The whole data string comes from an on-line form. It was cleansed and encrypted before insertion into a sql db. I am pulling it out of the db, de-crypting and then trying to use the information for it's intended purpose. Anyway, the whole string fell into your warning: "whenever the contents of that string may be influenced directly or indirectly by a user". I'm glad you brought that to my attention. Thanks!

Tom · Accepted Answer · 2015-10-11 07:48:02Z

1

This does what you want ( i think ), even if its not pretty

<?php

$arr = array();
$first = 'one';
$middle = 'two';
$last = 'three';

$rawData = "array('first' => '$first', 'middle' => '$middle', 'last' => '$last')";

$arrData = explode( ',', ( str_replace( array( 'array(\'', '=>', '\')', '\'' ), array( '', '%@@%', '', '' ), $rawData ) ) );

foreach( $arrData as $val ) {
  $v = explode( '%@@%', $val );
  $arr[trim($v[0])] = trim($v[1]);
}

echo '<pre>' . print_r( $arr, true ) . '</pre>';

?>

edited Oct 11, 2015 at 7:48

answered Oct 11, 2015 at 7:30

Tom

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4 Comments

user1028866 Over a year ago

Thanks for replying Tom. I get the following error:

user1028866 Over a year ago

Parse error: syntax error, unexpected '[' in: for the $arrData[] = explode( '%@@%', $val )[1];

user1028866 Over a year ago

OK, got the edit and it is working now. Thanks so much for replying. I thought there would be a much simpler way to do this!

user1028866 Over a year ago

Thanks Tom, this is working in my program now. I'm so glad to be unstuck!

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