PHP Convert string into array

Question

How to convert a string $string = 'array(array("a"=>"content"))'; (<-- this is string) into array like this:

array(1) { [0]=> array(1) { ["a"]=> string(7) "content" } }

I am going to build a function to run serialize online:

Input: $string = 'array(array(1))'; echo serialize($string);

Result: a:1:{i:0;a:1:{i:0;i:1;}} NOT s:16:"array(array(1)))";

You call: $returnValue = serialize('array(array(1))');

My question would be why you have such a string in the first place? — Felix Kling
– Felix Kling, Commented Aug 10, 2013 at 14:24
Please provide more context. You had two people answer eval when you already knew that eval is insecure and not a solution. Don't waste people's time suggesting things that you rule out right from the start. Note the section "Be specific" in stackoverflow.com/questions/ask-advice — Gordon
– Gordon, Commented Aug 10, 2013 at 14:48

Amal · Accepted Answer · 2013-08-10 14:24:19Z

4

You /could/ use eval():

$string = 'array(array("a"=>"content"))';
eval("\$array = $string;");
print_r($array);

Output:

Array
(
    [0] => Array
        (
            [a] => content
        )

)

But if you're accepting user inputs, you should not use eval().

Also, consider using json_encode() / json_decode() instead.

Alternatively, to store arrays in strings, you could use serialize and unserialize.

answered Aug 10, 2013 at 14:24

Amal

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6 Comments

Truongnq Over a year ago

if $string get data from $_POST like: $string = $_POST['string']; Like this: $string = 'array(array("a"=>"content")); die("HACK: add more functions before this")'; eval("\$array = $string;"); print_r($array);

Truongnq Over a year ago

I still hack if using eval()

$string = 'array(array("a"=>"content")); die("HACK: add more functions before this")'; eval("\$array = $string;"); print_r($array);

Output: HACK: add more functions before this

Sylwester Over a year ago

@Truongnq If you are gettig the string from outside you should write a parser. It might be better to agree on something simpler, like JSON.

Amal Over a year ago

@Truongnq: As I mentioned in the answer, you shouldn't use eval if it involves accepting input from user. Would've been kind of nice if you'd mentioned in your question that you don't want to use eval. Anyway I've suggested alternatives - you might want to check it out. :-)

Truongnq Over a year ago

@AmalMurali: I can see your updated answer. Can you explain more. Thanks!

|

Raymond Nijland · Accepted Answer · 2013-08-10 14:19:56Z

0

eval? check http://php.net/manual/en/function.eval.php

take note on the Caution the manual shows...

answered Aug 10, 2013 at 14:19

Raymond Nijland

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2 Comments

Truongnq Over a year ago

eval is not security when I public this kind of function only. I tried, and hacked it when use eval.

Truongnq Over a year ago

I tried: $reg_exp = "/array(*.*)/"; $subject ="array(array(1=>'123'))"; $result = preg_split($reg_exp,$subject); var_dump($result); Output: array(2) { [0]=> string(0) "" [1]=> string(0) "" } but this is NOT correct. Please help!

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