CODEWITHSUNDEEP
phpdecodejson
user3705546
user3705546

Reputation: 33

PHP json_decode return empty array

I just test this sample from php doc (https://www.php.net/manual/en/function.json-decode.php)

here is my code:

<?php $json = '{"a":1,"b":2,"c":3,"d":4,"e":5}'; echo json_decode($json, true), '<br />';?>

But it just returns an EMPTY array.

Have no idea why...Been searching around but no solution found.

PLEASE help!

Upvotes: 3

Views: 22844

Answers (6)

Aksh
Aksh

Reputation: 654

You can validate at following website: http://jsonlint.com/

You have to use a php "json_decode()" function to decode a json encoded data. Basically json_decode() function converts JSON data to a PHP array.

Syntax: json_decode( data, dataTypeBoolean, depth, options )

data : - The json data that you want to decode in PHP.

dataTypeBoolean(Optional) :- boolean that makes the function return a PHP Associative Array if set to "true", or return a PHP stdClass object if you omit this parameter or set it to "false". Both data types can be accessed like an array and use array based PHP loops for parsing.

depth :- Optional recursion limit. Use an integer as the value for this parameter.

options :- Optional JSON_BIGINT_AS_STRING parameter.

Now Comes to your Code

$json_string = '{"a":1,"b":2,"c":3,"d":4,"e":5}' ;

Assign a valid json data to a variable $json_string within single quot's ('') as json string already have double quots.

// here i am decoding a json string by using a php 'json_decode' function, as mentioned above & passing a true parameter to get a PHP associative array otherwise it will bydefault return a PHP std class objecy array.

$json_decoded_data = json_decode($json_string, true);

// just can check here your encoded array data.
// echo '<pre>';
// print_r($json_decoded_data);

// loop to extract data from an array
foreach ($json_decoded_data as $key => $value) {
    echo "$key | $value <br/>";

}

Upvotes: 4

Lajos Veres
Lajos Veres

Reputation: 13725

It works fine as others mention, but when you print the array it is converted to string, which means only the string "Array" will be printed instead of the real array data. You should use print_r(), var_dump(), var_export() or something similar to debug arrays like this.

If you turn on notices you will see:

PHP Notice:  Array to string conversion in ...

The example you linked uses also var_dump for the same reason.

Upvotes: 2

MH2K9
MH2K9

Reputation: 12039

you should not use echo because it is an array. use print_r or var_dump .it works fine

$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
print_r(json_decode($json, true));

Output:

Array
(
   [a] => 1
   [b] => 2
   [c] => 3
   [d] => 4
   [e] => 5
)

Upvotes: 3

Muhammad Ali
Muhammad Ali

Reputation: 2014

var_dump have pretty print in php5.4

$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}'; 
var_dump( json_decode($json));

Upvotes: 0

Marcin Nabiałek
Marcin Nabiałek

Reputation: 111829

It works fine, but you use wrong method to display array.

To display array you cannot use echo but you need to use var_dump

Upvotes: 2

Chris Hasiński
Chris Hasiński

Reputation: 3085

No, it doesn't return an empty array.

Printing an array with echo just prints a string "Array()".

Use print_r or var_dump to get the structure of the variable.

In newer PHP it will also emit a notice when using echo on an array ("Array to string conversion"), so you shouldn't do it anyway. The manual you've mentioned changed to print_r.

Upvotes: 2

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