JSON decode in PHP from a url

Question

Hello i am new with Json in php. I have a web service that gives me data in json format. I take this data making decode put when i try to use this data i cant

Here is my code:

 $url = "http://www.webinsurer.gr/....;

    $json = json_decode(@file_get_contents($url), true);

and if i make debug i see the data i take :

[file] => C:\xampp\htdocs\development.insurancemarket.gr\mvc\protected\models\Ratingsmail.php
[line] => 18
[data] => Array
    (
        [0] => Array
            (
                [POL_EXPIREDATE] => 2014-05-19 12:00:00
                [INCO_IWCODE] => 41
                [INCO_DESC] => MAPFRE ASISTENCIA
                [PACK_IWCODE] => 0
                [PACK_DESC] => 
                [OFFERCODE] => 
                [PAYMENTCODE] => 
            )

        [1] => Array
            (.....

But i dont now how to use that data. when i try this :

$b= $json->{1}->{'INCO_IWCODE'};

    Debug::debuger($b);

the result is nothing

what is wrong? sorry for long post.

show us console.log($json), it will output to the firebug/developer tools console (f12 in any browser) — Bartłomiej Wach
– Bartłomiej Wach, Commented May 15, 2014 at 8:28
"if I make debug" with what code? Is this print_r($json)? If so you need to use $json['data'][0]['INCO_IWCODE']. — h2ooooooo
– h2ooooooo, Commented May 15, 2014 at 8:28
@BartłomiejWach What? You cannot debug PHP variables on a javascript console. — EJTH
– EJTH, Commented May 15, 2014 at 8:29

EJTH · Accepted Answer · 2014-05-15 08:28:45Z

1

When setting the second argument on json_decode to true, you are actively asking for the data to be returned in an associative array and not objects. Thats why your code didn't work.

answered May 15, 2014 at 8:28

EJTH

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3 Comments

theodosis Over a year ago

i removed the true and again in the debug the data was empty

Daniel W. Over a year ago

@theodosis $json->data[1]->INCO_IWCODE - you are using {...} which is wrong and has a different meaning in PHP

EJTH Over a year ago

Also what Dan said :-) To elaborate: {...} is for dynamic referencing, eg. if the variable you want depends on a computed result, or when the variable has a naming which is incompatible with PHP syntax: $obj->{$myvar} or $obj->{'invalid-variable-name'} are examples of valid uses. It does not work with arrays of any kind, only objects.

Hüseyin BABAL · Accepted Answer · 2014-05-15 08:35:19Z

0

Demo

You are converting json to associative array. You need to use;

$b = $json["data"][1]["INCO_IWCODE"];

edited May 15, 2014 at 8:35

answered May 15, 2014 at 8:28

Hüseyin BABAL

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3 Comments

theodosis Over a year ago

its not working the result is : Array ( [file] => C:\xampp\htdocs\development.insurancemarket.gr\mvc\protected\models\Ratingsmail.php [line] => 30 [data] => )

Hüseyin BABAL Over a year ago

@user3438570 it is not working with what? Your formatting a bit hard to read. Please see my demo in answer

theodosis Over a year ago

I took your code but again in the debug the data was empty. i dont now why.

theodosis · Accepted Answer · 2014-05-15 08:58:52Z

0

$a = $json[0]->INCO_IWCODE;

That worked for my guys. thanks you all!!!

answered May 15, 2014 at 8:58

theodosis

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