How to correct syntax in php using json_encode

Question

I am trying to get json_encode output as: {"match_all":{}}. The sample program which I have written is:

<?php
 $qarray=Array("match_all"=>Array());
 print_r(json_encode($qarray));
?>

But above is giving me below o/p: {"match_all":[]}

Answer 1

What you want is an object not an array. This will generate the output you want.

$qarray = array("match_all" => new stdClass());
print_r(json_encode($qarray));

https://www.w3schools.com/js/js_json_objects.asp

Answer 2

The result you get is expected because [] denotes a JSON array and not a JSON object. If you need an object you need to explicitly provide one e.g.

<?php
$qarray = Array("match_all"=>(object)Array());
print_r(json_encode($qarray));

However this has little practical importance because Json encode will correctly create objects when you give it associative arrays e.g.

<?php
$qarray = [ "match_all"=> [ "assoc" => true ] ];
print_r(json_encode($qarray));

Prints:

{"match_all":{"assoc":true}}

For the purposes of JSON encoding associative arrays are any arrays that don't have sequential numbers as indices.

Also, short array syntax is a bit easier on the eyes in my opinion but is not different to using Array()

• mickmackusa: apokryfos wins on speed and explanation! +1 (explanation is critical on low-hanging fruit)

Answer 3

PHP json_encode function has option flag as 2nd parameter. You can set it to JSON_FORCE_OBJECT, then will get result you want.

Try this code.

<?php
 $qarray=Array("match_all"=>Array());
 print_r(json_encode($qarray, JSON_FORCE_OBJECT));
?>

Answer 4

You should specify this is an object, and not an array :

$qarray=Array("match_all"=>(object) Array());
print_r(json_encode($qarray));

Answer 5

As an addon to the other provided answers, if you have multiple values it would be pretty annoying to cast them all as objects each time, so you could pretty easily just add a loop after the declaration to set all of the values to objects.

$qarray=Array("match_all"=>Array());
foreach($qarray as $key => $val) {
 if(!is_array($val)) continue;
 $qarray[$key] = (object)$val;
}
print_r(json_encode($qarray));

This makes the code more dynamic as no matter how many keys, their values will all be set as an object before the encoding as long as the value is an array.

Answer 6

In PHP an array can have strings as the key (aka an associative array) and when JSON encoded this is converted into a JSON object ({}). If however the array only has numeric keys, then the array is encoded as a JSON array - a list ([]).

To create an empty JSON object, you need to create an empty object in PHP instead of an empty array.

$qarray = ['match_all' => (object)[]];
print_r(json_encode($qarray));

• GrumpyCrouton: Probably because your answer is very similar to ones already posted.
• Jim Wright: Thanks, that was a typo :)