How to access object array in php

Question

I am trying to filter it using array_filter

$time = (time() -10);

$data = json_decode($json, TRUE);

$datafilter = array_filter($data, function($obj)
        {
        return $obj->response->trade_offers_received->time_created > $time;
        });

echo $datafilter;

I am getting this notice: Trying to get property of non-object. So am doing something wrong. How can i filter this array to get only object where time_created > $time ?

This is my JSON data

Looks like you need to iterate through the trade_offers_received object first, then access the time_created attribute inside each iteration. — Defiant
– Defiant, Commented Feb 2, 2016 at 10:42
If you want an object, remove the TRUE param from the json_decode() — RiggsFolly
– RiggsFolly, Commented Feb 2, 2016 at 10:43

RiggsFolly · Accepted Answer · 2016-02-02 11:15:12Z

2

As I am not sure what you will get if you do the json_decode() without the TRUE parameter that converts objects into arrays, we will stick with the array notation.

Also if you are only interested in data that exists at the $data['response']['trade_offers_received'] level of this data structure, start your array_filter at that level.

But mainly as they are arrays, use the array notation and not the object notation and things will probably work

More Importantly: You will also need to amend the closure code to pass the $time variable into the closure.

$time = (time() -10);

$data = json_decode($json, TRUE);

$datafilter = array_filter($data['response']['trade_offers_received'],
                              function($arr) use($time)
                              {
                                  return $arr['time_created'] > $time;
                              }
                           );

print_r( $datafilter );

edited Feb 2, 2016 at 11:15

answered Feb 2, 2016 at 10:49

RiggsFolly

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5 Comments

MHB2011 Over a year ago

Thanks , I ll try . Basicly what am trying to do is found how many offers are there with time_created > time()-10. I am trying first to filter the array and than count it using count($datafilter['response']['trade_offers_received']) . Currently am getting this errors Undefined variable: time and : Array to string conversion when am trying to echo $datafilter

RiggsFolly Over a year ago

To pass $time to the closure, you need to use the USE($time) syntax. Refresh the page as I missed that first time round

MHB2011 Over a year ago

Thanks . I am still getting this Notice: Array to string conversion in C:\xampp\htdocs\offer\funkcija.php on line 15. Line 15 is echo $datafilter;

RiggsFolly Over a year ago

Ok change that to a print_r( $datafilter ); as its an array of course. I am really not paying attention today. Sorry

MHB2011 Over a year ago

Nice , U are helped me a lot. Its working perfectly now. Thanks

Nazmul Alam · Accepted Answer · 2016-02-02 10:48:58Z

-1

When you use TRUE as the second parameter of json_decode, it becomes an Array. If you want to get an object in return, remove the second parameter.

    $data = json_decode($json); //returns an object
    $arr  = json_decode($json); //returns an array

answered Feb 2, 2016 at 10:48

Nazmul Alam

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2 Comments

RiggsFolly Over a year ago

You might want to check that!

RiggsFolly Over a year ago

Its not particularly relevant to the question anyway as he wants an array so he can feed it into the array_filter() function. Not always a good idea to jump on comments and try to vampire them into answers!

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