Unable to eval a string array

Question

How can i change this to a JSON Array? im using the eval technique but this is not working. Im getting this kind of response from our upx server:

array(
    'invoice' => array(
        'id' => '4',
        'realid' => '4',
        'address_rev' => NULL,
        'relation_data_id' => '3',
        'contact_set_rev' => '3',
        'business_data_rev' => '4',
        'private_data_rev' => NULL,
        // etc..
    )
); 

var_dump($newdata); // String
eval("\$newdata = \"$newdata\";");
var_dump($newdata); // Still stays as a string as shown above....

Any clue?

Ty Already!

What on earth outputs that? If you want to transmit a literal PHP array, serialize() it... — DaveRandom
– DaveRandom, Commented Dec 20, 2011 at 10:48

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

You'll facepalm really bad when I tell you this..

But the reason it's still a string is because you wrapped it in "" inside your call to eval, and of course wrapping letters inside of quotes will make it.. (drum roll ..) a string.

 eval ('$newdata = ' . $newdata . ';'); // this will do what you want

If you want to turn it into json right away, use the below:

 eval ('$newdata = json_encode (' . $newdata . ');');

PHP: json_encode - Manual

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Dec 20, 2011 at 10:50

Filip Roséen

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2 Comments

Emre Akman Over a year ago

Omg, slap me :(, how can i not see that :O probally because im working for like 13 hours straight now! however, ty for the answers!

Scott C Wilson Over a year ago

For people who arrived here from Google: another possible source of error when doing this is neglecting to add the semicolon to the end of the eval'ed string. Creating a temporary variable which you pass to eval (at least during development) allows you to double check the string for syntactical correctness.

DaveRandom · Accepted Answer · 2011-12-20 10:49:19Z

2

var_dump($newdata); // String
eval("\$newdata = $newdata;");
var_dump($newdata); // Still stays as a string as shown above....
// eval("\$newdata = \"$newdata\";");
//                    ^         ^

Remove the double quotes. Your just putting it back into a string...

Although as I said above, if you want to transmit a PHP array, you should be using serialize()

answered Dec 20, 2011 at 10:49

DaveRandom

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Comments

Shades88 · Accepted Answer · 2011-12-20 10:50:59Z

1

You can try json_encode of php. you can try

$json_array = json_encode($your_array);

This will give you json encoded array. Then you can do json_decode on $json_array to get original contents back

Check if your php has JSON extension

answered Dec 20, 2011 at 10:50

Shades88

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