How to PHP decode JSON from MYSQL

Question

On MySQL I have this JSON value

{"length":1847,"data":[0,0,0,0,-46,37]}

On PHP I need to string in array the "data" objects, so I tried:

$json = $row['wave'];
$json_array = json_decode($json);
$json_wave = $json_array["data"];

It gives me internal server error 500, so I tried:

var_dump(json_decode($json));

What I got is:

object(stdClass)#7 (5) {
  ["length"]=>
  int(1847)
  ["data"]=>
  array(3694) {
    [0]=>
    int(0)
    [1]=>
    int(0)
    [2]=>
    int(0)
    [3]=>
    int(0)
    [4]=>
    int(-46)
    [5]=>
    int(37)
  }
}

I need to output this string 0,0,0,0,-46,37, why is my code not working?

add second parameter TRUE into $json_array = json_decode($json,TRUE); to make it associative array then use $json_wave = $json_array["data"]; — Saty
– Saty, Commented Aug 9, 2016 at 13:39

Adrian Pietka · Accepted Answer · 2016-08-09 13:39:35Z

4

If you want decode JSON to array you shoud use json_decode function with additional parameter ($assoc as true):

$json_array = json_decode($json, true);

You can look to documentation here: http://php.net/manual/en/function.json-decode.php

answered Aug 9, 2016 at 13:39

Adrian Pietka

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Alok Patel · Accepted Answer · 2016-08-09 13:41:28Z

2

json_decode() by default returns an Object instead of an array. If you want to decode the JSON String in Array json_decode accepts second parameter as boolean.

If the second parameter is true json_decode() will return the value in forms of array.

So your code will look something like this,

$json = $row['wave'];
$json_array = json_decode($json,true);
$json_wave = $json_array["data"];

Second parameter's by default value is false.

To get the data from Object you can access it like this,

print_r($json_array->data);

answered Aug 9, 2016 at 13:41

Alok Patel

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Machavity · Accepted Answer · 2016-08-09 13:41:51Z

2

So assuming you want that last variable to be the string you need this

$json_array = json_decode($json);
$json_wave = implode(',', $json_array->data);

Because your JSON contains an object, PHP puts it into an object by default. That's what var_dump means when it says stdClass

answered Aug 9, 2016 at 13:41

Machavity♦

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Comments

BeetleJuice · Accepted Answer · 2016-08-09 13:40:27Z

0

Your var_dump shows that the row is decoded as an object, so you cannot access its fields using array notation.

Change

$json_wave = $json_array["data"];

To

$json_wave = $json_array->data;

Alternatively, if you want to work with arrays rather than objects, you can set the 2nd parameter to json_decode as below:

$json_array = json_decode($json, true);

answered Aug 9, 2016 at 13:40

BeetleJuice

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user5222939 · Accepted Answer · 2016-08-09 13:46:48Z

0

It doesn't work because you're trying to treat object like an array. json_decode wihtout second parameter converts json_string to an object. use json_decode($json, true) to say 'hey, convert my json string to an array, not object'. On the other hand you can just use $json_array->data instead $json_array["data"].

answered Aug 9, 2016 at 13:46

user5222939

Comments

Aman Maurya · Accepted Answer · 2016-08-09 13:50:47Z

0

Try This :

$json = '{"length":1847,"data":[0,0,0,0,-46,37]}';
$a = json_decode($json);
print_r($a->data);

Value after json_decode in variable $a is stored in the object form and you can not get the object value like $a['data'] so you have to use in this manner $a->data

answered Aug 9, 2016 at 13:50

Aman Maurya

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