Use JSON in PHP

Question

I use some website that give me information about IP but the information that that website return in JSON and I don't know the JSON. I want to use this to check the user if it is from IR do something but I dont know how to use JSON in php,
Here is the JSON that the website return:

{"address":"0.0.0.0.0","country":"IR","stateprov":"somewhere ","city":"Tehr\somewhere (somewhere)"}

I want to save the country in a variable and add this code to my website:

<?php
if($country == 'IR'){
//Do somethong  
}

$country is the country name that return from the website,

possible duplicate of [PHP- Decode JSON](http://stackoverflow.com/questions/19609354/php-decode-json) — Mark Miller, Jul 28 '14 at 07:54

Darren · Accepted Answer · 2014-07-28T08:02:08.267

4

You'll need to use json_decode().

$s = '{"address":"0.0.0.0.0","country":"IR","stateprov":"somewhere ","city":"Tehrsomewhere (somewhere)"}';

$d = json_decode($s);

Which returns:

stdClass Object
(
    [address] => 0.0.0.0.0
    [country] => IR
    [stateprov] => somewhere 
    [city] => Tehrsomewhere (somewhere)
)

That would allow you to check the country/other fields like this:

if($d->country == 'IR') {
    // do something
}

NOTE: you had an error (invalid json) in your "city" field, the \ makes it invalid.

Example

You can ensure that your json is valid by checking it at JSON Lint.

edited Jul 28 '14 at 08:02

answered Jul 28 '14 at 07:53

Darren

13,050
4
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A question How can I use [country] in my php if? – King_Far Jul 28 '14 at 08:03
@King_Far Look at updated answer. You'd use `$d->country` from my example I posted. (*using it as an object, not as an array*) – Darren Jul 28 '14 at 08:04

score 0 · Answer 2 · answered Jul 28 '14 at 07:50

0

I think you are looking for the function json_decode.It decodes the JSON string

See the documentation here

answered Jul 28 '14 at 07:50

Avinash Babu

6,171
3
21
26

score 0 · Answer 3 · edited Jul 28 '14 at 08:21

First of all i would like to inform you that given json is not valid. "city" : "Tehr\somewhere (somewhere)" is not valid because of "\".

So change it into below given format.

$jsonEncode = { "address": "0.0.0.0.0","country": "IR","stateprov": "somewhere ","city": "There somewhere (somewhere)"}

$jsonDecode = json_decode($jsonEncode,true);

Now you will get the value in array format.

Array(
    [address] => 0.0.0.0.0
    [country] => IR
    [stateprov] => somewhere 
    [city] => There somewhere (somewhere)

);

print_r($jsonDecode['city']); will give you city name or details

score 0 · Answer 4 · answered Jul 28 '14 at 08:07

0

You have to first decode this json string.

$data = '{"address":"0.0.0.0.0","country":"IR","stateprov":"somewhere ","city":"Tehr\somewhere (somewhere)"}';

$decodeData = json_decode($data);

Then use this decode json string in php like this.

if($decodeData->country == 'IR'){
//Do somethong  
}

answered Jul 28 '14 at 08:07

Anshul Lonkar

83
9

And what's the difference in your and @Darrens answer? Except the variable names? – Vucko Jul 28 '14 at 08:09

Use JSON in PHP

4 Answers4