Count number of objects in JSON PHP?

Question

I have a JSON array and in that I have 3 JSON objects. I want to count the number of objects that is 3. but it is giving me 1. If I do not add the key "like", then it works. but after adding it, it is not working.

 $JSON = '{"like":['
            . '{"username":"suraj","password":"abc"},'
            . '{"username":"don","password":"abc"},'
            . '{"username":"rana","password":"abc"}'
            . ']}';


    $jsonInPHP = json_decode($JSON);
    echo count($jsonInPHP);

$jsonInPHP = json_decode($JSON,true); echo count($jsonInPHP['like']); — JYoThI
– JYoThI, Commented Aug 27, 2016 at 11:30
Note: If you set the 2nd param as true it makes it an associative array. — Ekin
– Ekin, Commented Aug 27, 2016 at 11:30

JYoThI · Accepted Answer · 2016-08-27 11:33:15Z

4

pass the second parameter true like this

 $jsonInPHP = json_decode($JSON,true); 
  echo count($jsonInPHP['like']);

answered Aug 27, 2016 at 11:33

JYoThI

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1 Comment

Suraj Roy Over a year ago

adding true gives error,, though i got solution,, count($jsonInPHP->like);

R.K.Saini · Accepted Answer · 2020-01-24 19:23:55Z

3

Your JSON represents an object, not an array. In your object, you have like property which is an array so you need to write like this

count($jsonInPHP->like);

edited Jan 24, 2020 at 19:23

answered Aug 27, 2016 at 11:31

R.K.Saini

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Comments

u_mulder · Accepted Answer · 2016-08-27 11:32:10Z

1

This happens because after json decoding your string you get object with one property (like) (or array with one element with key like).

In both ways you want to count size of this property (or key) which is:

// if $jsonInPHP is array
echo count($jsonInPHP['like']); 
// if $jsonInPHP is object
echo count($jsonInPHP->like);

answered Aug 27, 2016 at 11:32

u_mulder

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