Convert JSON string to PHP Array

Question

I have the following JSON string, which was an Objective C array then encoded to JSON:

 $jsonString=[\"[email protected]\",\"[email protected]\",\"[email protected]\"]

I want to convert this to a regular PHP array. I've tried many things but none of them seem to work:

$arrayOfEmails=json_decode($jsonString); //doesn't work
$arrayOfEmails=(array)json_decode($jsonString); //doesn't work

Any ideas?

Edit:

I'm still not getting it to work.

$decodeEmails=$_POST["arrayOfEmails"];
sendResponse(200, $decodeEmails);
//the above returns exactly whats after this colon:[\"[email protected]\",\"[email protected]\",\"[email protected]\]

I need to do this: $arrayOfEmails=json_decode($decodeEmails); But I think I need quotes around $decodedEmails for this to work. How can I add quotes around $decodeEmails string?

it seems like you have to escape the quotes inside the string for it to be valid: "[\"[email protected]\",\"[email protected]\",\"[email protected]\"]" — driangle
– driangle, Commented Feb 7, 2012 at 3:35
This is the Objective C version of the string..it gets sent to PHP via POST — Snowman
– Snowman, Commented Feb 7, 2012 at 3:36
You need to define "doesn't work." json_decode() is the proper method for this purpose, so what are you getting when you use it? — JAAulde
– JAAulde, Commented Feb 7, 2012 at 3:38
I edited it now, that's the actual PHP string. Do I have to do anything to it first? — Snowman
– Snowman, Commented Feb 7, 2012 at 3:39

xdazz · Accepted Answer · 2012-02-07 03:37:34Z

10

You should quote your string, it works fine, see here.

$jsonString = '["[email protected]","[email protected]","[email protected]"]';
$arrayOfEmails=json_decode($jsonString);

Or

$jsonString = "[\"[email protected]\",\"[email protected]\",\"[email protected]\"]";
$arrayOfEmails=json_decode($jsonString);

answered Feb 7, 2012 at 3:37

xdazz

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4 Comments

xdazz Over a year ago

@mohabitar If you are already escape the ", then you can use ".

Snowman Over a year ago

var_dump gives this (The actual array has more than 3, I took the extras off from here just so I can show it): string(457) "[\"[email protected]\",\"[email protected]\",\"[email protected]\"]"

xdazz Over a year ago

@mohabitar It's already a string, you can just use json_decode after it .

Snowman Over a year ago

But that's what I'm doing isn't it? $decodeEmails=$_POST["arrayOfEmails"]; $arrayOfEmails=json_decode($decodeEmails);

Mayank Shukla · Accepted Answer · 2017-02-19 06:38:39Z

10

Try this: json_decode($json_string, true);

edited Feb 19, 2017 at 6:38

Mayank Shukla

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answered Feb 7, 2012 at 3:37

the_wizard

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5 Comments

Xeoncross Over a year ago

This is correct, the other answers don't seem to know about the second argument for json_decode() and are only type-casting the first level elements.

Snowman Over a year ago

But this returns an associative array, no? I just want a regular

Xeoncross Over a year ago

@mohabitar, PHP only has one type of array. I think you mean that the array keys are not numeric then? Your input is what needs to be checked or corrected to get the output you desire.

JAAulde Over a year ago

There is no need for the second param in thie case given in the question. The second param is for converting JSON serialized objects to assoc arrays. The OP has a JSON serialized indexed array.

Mohini Over a year ago

When I passed second parameter assoc= true to json_decode() its worked but I have to called it 2 times to achive my output

Botz3000 · Accepted Answer · 2012-05-22 10:07:11Z

3

$data = unserialize($data)

now u can get the $data as array

For example $data have the value like this

$data = "a:2:{s:18:"_1337666504149_149";a:2:{s:8:"fbregexp";s:1:"1";s:5:"value";s:4:"2222";}s:18:"_1337666505594_594";a:2:{s:8:"fbregexp";s:1:"3";s:5:"value";s:5:"45555";}}";

$data = unserialize($data)

now i get value like this

Array ( [fbregexp] => 1 [value] => 2222 ) [_1337666505594_594] => Array ( [fbregexp] => 3 [value] => 45555 ) )

edited May 22, 2012 at 10:07

Botz3000

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answered May 22, 2012 at 6:43

kranthi kumar pulivarhty

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Comments

bortunac · Accepted Answer · 2014-11-29 08:40:29Z

2

    $str=<<<H
    {"a":"AAA","b":"333"}
    H;

     $object = json_decode($str); 
     $array  = json_decode($str , 1 );

    // $arr = get_object_vars( json_decode($str) );

edited Nov 29, 2014 at 8:40

answered Apr 12, 2013 at 20:25

bortunac

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Comments

Will · Accepted Answer · 2012-02-07 03:40:39Z

1

If json_decode isn't working, you could try something like this:

$arr = explode( '\\",\\"', substr( $json, 3, strlen( $json ) - 3 ) );

answered Feb 7, 2012 at 3:40

Will

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1 Comment

Ajeesh Joshy Over a year ago

Worked for me as I was passing JSON string in URL

timgeb · Accepted Answer · 2014-06-17 10:29:21Z

1

You could use json_decode() then print_r() to create a PHP formatted array

<?php
$json = file_get_contents('json/yourscript.json'); // Get the JSON data
$phpObj = json_decode($json,true);  // Convert to PHP Object
print_r($phpObj);  // Convert to PHP Array
?>

edited Jun 17, 2014 at 10:29

timgeb

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answered Jun 17, 2014 at 9:54

mintedsky

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Comments

Prasad Rajapaksha · Accepted Answer · 2012-02-07 03:38:38Z

0

This code works fine.

$jsonString = '["[email protected]","[email protected]","[email protected]"]';
$arrayOfEmails=json_decode($jsonString);
$arrayOfEmails=(array)json_decode($jsonString);
print_r($arrayOfEmails);

answered Feb 7, 2012 at 3:38

Prasad Rajapaksha

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4 Comments

Prasad Rajapaksha Over a year ago

you don't have to escape the quotes in this case.

Xeoncross Over a year ago

Instead of type-casting, you should be using the second argument to convert to an array: json_decode($value, $array = TRUE).

Prasad Rajapaksha Over a year ago

Yes. I just tried to fix his error. Here is my result. Array ( [0] => [email protected] [1] => [email protected] [2] => [email protected] )

JAAulde Over a year ago

There is no need to typecast, or to pass the second param as true

JAAulde · Accepted Answer · 2012-02-07 12:24:12Z

0

Assuming the lack of quotes around your JSON in the question is a transposition error during posting, then the code you're using is fine: http://codepad.org/RWEYM42x

You do need to ensure your string is UTF8 encoded. You can use the built in encoder if it isn't ( http://php.net/manual/en/function.utf8-encode.php ).

For any further help, you need to actually tell us what you are getting with your code.

edited Feb 7, 2012 at 12:24

answered Feb 7, 2012 at 3:53

JAAulde

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Comments

Racil Hilan · Accepted Answer · 2017-02-19 06:40:55Z

-2

$arr = explode( '\\",\\"', substr( $json, 3, strlen( $json ) - 3 ) )

This solution is good but for getting full valid array I'm using strlen( $json ) - 6, so it should be:

$arr = explode( '\\",\\"', substr( $json, 3, strlen( $json ) - 6 ) );

var_dump($arr);

edited Feb 19, 2017 at 6:40

Racil Hilan

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answered Feb 19, 2017 at 6:16

shojol80

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