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I send this Javascript array into PHP page using form submit {"1":"2","2":"2","3":"2","4":"2"}

Now, I want to convert this array into PHP array, like this

$cars = array("Volvo", "BMW", "Toyota");

So, this is what I tried:

$phparray = str_replace(':', ',', $_POST["questionandanswers"]); // Remove : and replace it with ,
$phparray = str_replace('}', '', $phparray); // Remove }
$phparray = str_replace('{', '', $phparray); // Remove {
echo '<br/>';
echo $phparray; // Output of this is: "1","2","2","2","3","2","4","2"



$questionandanswers = array($phparray); // Now convert it into PHP array

But it is not working. Looks like i cannot put $phparray variable here array($phparray)

But, If instead of putting $phparray variable in array($phparray), If i put the output of $phparray manually, then, it works like: array("1","2","2","2","3","2","4","2")

What's the solution?

Josh Poor
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2 Answers2

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It seems that your form submits a json object to the server so use json_decode with second parameter set to true to convert it to an array.

$php_array=  json_decode($_POST["questionandanswers"],true)
Sara Yusri
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Osama
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  • Thanks, but the thing here is, the object name is not always in a serial order: It can also be like `Array ( [3] => 0 [6] => 0 [7] => 0 [11] => 0 ) `. So, how to access the object, without knowing it's name? – Josh Poor Oct 07 '17 at 08:23
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    Just use foreach loop with if statment according to your condition. – Osama Oct 07 '17 at 08:28
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    Btw, you misspelled json_decode – Josh Poor Oct 07 '17 at 09:21
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Use json_decode like json_decode($json). It will return an object. If you want array, use json_decode($json, true). See: http://sg2.php.net/manual/en/function.json-decode.php

Arka Roy
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