how to convert this json data in php

Question

I am using an api which is giving me a very strange json format.. i'm posting its data which i am getting thru it..

info = { "title" : "Asian Dad: B Again!? (you die)", "image" : "http://i.ytimg.com/vi/IN7o2Iy89WQ/default.jpg", "length" : "2", "status" : "serving", "progress_speed" : "", "progress" : "", "ads" : "", "pf" : "", "h" : "f53762dab34022e9d851ab71e0bf166f" };

I'm trying to print this data in php but i'm not able to do that..nothing is showing on my webpage,.... My code are..

First i tried,

<?php
$url="http://www.website-name.com/a/itemInfo/?video_id=IN7o2Iy89WQ&ac=www";
$info=file_get_contents($url);
$info=json_decode($info,true);
echo $info;
?>

My second attempt was,

<?php
$url="http://www.website-name.com/a/itemInfo/?video_id=IN7o2Iy89WQ&ac=www";
$info=file_get_contents($url);
$info=json_decode($info,true);
$info->h;
?>

My last attempt was,

<?php
$url="http://www.website-name.com/a/itemInfo/?video_id=IN7o2Iy89WQ&ac=www";
$info=file_get_contents($url);
$info=json_decode($info,true);
$info['h'];
?>

Nothing is happening..

please somebody help me

API Url converted...

$url="http://www.website-name.com/itemInfo/?video_id=IN7o2Iy89WQ&ac=www"; IS this sending a JSON? Are you sure? — Sasanka Panguluri
– Sasanka Panguluri, Commented Sep 26, 2013 at 18:13
Can you var_dump($info); after your second and third line and show us the output? — TheWolf
– TheWolf, Commented Sep 26, 2013 at 18:13
@TheWolf ...my webpage is showing NULL if i use var_dump($info); — Kapil
– Kapil, Commented Sep 26, 2013 at 18:17

Kapil · Accepted Answer · 2013-09-26 18:41:24Z

2

The page is sending "info = { "title" : "Asian Dad: B Again!? (you die)", "image" : "http://i.ytimg.com/vi/IN7o2Iy89WQ/default.jpg", "length" : "2", "status" : "serving", "progress_speed" : "", "progress" : "", "ads" : "", "pf" : "", "h" : "058ce93db26fce4a9f1cb41ae2e7c1bb" };"

You cannot use json_decode on this because the info = and the ; at the end are not json. You have to strip the info = and the ;.

   $url="http://www.website-name.com/a/itemInfo/?video_id=IN7o2Iy89WQ&ac=www";
   $info = file_get_contents($url);
   $info = trim($info, "info = ");
   $info = rtrim($info, ";");
   $json = json_decode($info, true);
   echo $json['status'];

edited Sep 26, 2013 at 18:41

Kapil

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answered Sep 26, 2013 at 18:23

Nico

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7 Comments

Kapil Over a year ago

getting error...Fatal error: Cannot use object of type stdClass as array in /home/lpmqcrax/public_html/download/info.php on line 8

Nico Over a year ago

@Kapil Sorry, missing the second argument to json_decode. It works now :-)

Kapil Over a year ago

but it is not working properly...I want to print $json['h'] value which is f53762dab34022e9d851ab71e0bf166f but it is giving me 0cedafe839fc2337f52355ba959576bb which is unfair...do you why is that happening ?

Kapil Over a year ago

it is giving all the information correct except the "h" value..for which i'm doing all of this..any way to get the right value.

Nico Over a year ago

@Kapil To get the right value, you have to fix the server code.

|

JAL · Accepted Answer · 2013-09-26 18:20:30Z

The data I got from the URL in your example is

info = { "title" : "Asian Dad: B Again!? (you die)", "image" : "http://i.ytimg.com/vi/IN7o2Iy89WQ/default.jpg", "length" : "2", "status" : "serving", "progress_speed" : "", "progress" : "", "ads" : "", "pf" : "", "h" : "5cddd4d1667f24aa9a0f5a6cc21e24e3" };

That's an executable JavaScript snippet, not actually JSON. The reason your php is failing is due to the 'info =' part... json_encode returns null on decoding failure.

While this is an assignment of a variable to a JavaScript object, that would work as JSON too if you removed the 'info =' and semicolon. Assuming the responses are predictable, you could do this with php str_replace, but finding an API that returns JSON for your source would be a more reliable and clean solution.

Aristona · Accepted Answer · 2013-09-26 18:20:48Z

0

If you're getting NULL when you var_dump($info) on the second line as you mentioned in comments, then file_get_contents() doesn't retrieve the data.

It's most likely because allow_url_fopen is set to false in PHP.ini

answered Sep 26, 2013 at 18:20

Aristona

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1 Comment

Kapil Over a year ago

getting error Warning: Invalid argument supplied for foreach() in /home/lpmqcrax/public_html/download/info.php on line 4

Martijn Pieters · Accepted Answer · 2020-09-03 20:58:44Z

0

From the json_decode() documentation:

Takes a JSON encoded string and converts it into a PHP variable.

If 2nd parameter is set to true it returns an array !

SO your first and second attempt is wrong.! It should work third attempt. Just check if you retrieve the data.

edited Sep 3, 2020 at 20:58

Martijn Pieters

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answered Sep 26, 2013 at 18:23

Arun Kumar M

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