json_decode does not return array in php

Question

I am facing with a problem, when I am trying to parse json returns from server into array in php. Here is my code ...

<?php
    mb_internal_encoding('UTF-8');  
    $url = 'http://localhost/busexpress/api/v1/mobile_user_register/mobile_user_register/retrieve.json';
    $ch = curl_init($url);
    curl_setopt($ch, CURLOPT_TIMEOUT, 10);
    curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 10);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    $data = curl_exec($ch);
    //$data="'".$data."'";
    echo $data;
    curl_close($ch);

    //$trimspace = preg_replace('/\s+/', '', $data); 
    //echo $trimspace;

    $jdata = json_decode($data, true);
    print_r $jdata; 

?>

This is the json after trimming space. I also want to convert it int array with json_decode() but no result return. I think this json is valid. And suggestion pls. This is my firstly trying to feed web service from server.

Thanks

 '{
"status": "1",
"user": [        
    {
        "id": "27",
        "name": "kktt",
        "phone_no": "1239293",
        "activate_code": "0d08ed",
        "deposit": "0",
        "created": "2015-06-0316:35:08",
        "updated": "1110-11-3000:00:00",
        "status": "0"
    },
    {
        "id": "28",
        "name": "kktt",
        "phone_no": "1239293",
        "activate_code": "fb4876",
        "deposit": "0",
        "created": "2015-06-0316:37:14",
        "updated": "1000-01-0100:00:00",
        "status": "0"
    }
  ]
}'

----------Edit---------

As your suggestion I comment trimming space and correct json format. And echo $data; .....

{
"status": "1",
"user": [        
    {
        "id": "27",
        "name": "kktt",
        "phone_no": "1239293",
        "activate_code": "0d08ed",
        "deposit": "0",
        "created": "2015-06-0316:35:08",
        "updated": "1110-11-3000:00:00",
        "status": "0"
    },
    {
        "id": "28",
        "name": "kktt",
        "phone_no": "1239293",
        "activate_code": "fb4876",
        "deposit": "0",
        "created": "2015-06-0316:37:14",
        "updated": "1000-01-0100:00:00",
        "status": "0"
    }
  ]
}

In decoding array doesn't have any data.

 $jdata = json_decode($data, true);
 print_r $jdata; 
 echo "user status -> ". $jdata["status"];

when I copy that json and hard code in a string, decode it again, it works for me. please see my testing code....

$data =' {"status":"1","mobile_user":[{"id":"1","name":"saa","phone_no":"09978784963","activate_code":"","deposit":"0","created":"2015-05-29 00:00:00","updated":"0000-00-00 00:00:00","status":"1"},{"id":"3","name":"ttr","phone_no":"090930499","activate_code":"","deposit":"0","created":"2015-06-01 00:00:00","updated":"0000-00-00 00:00:00","status":"0"}]}';
$data = json_decode($data,true);
$status = $data['status'];
$mobile_user = $data['mobile_user'];
$id = $mobile_user[0]["id"];
$name = $mobile_user[0]["name"];
echo "id -> ". $id ."<br>";
echo "name -> ". $name;

Any suggestion pls!

venca · Accepted Answer · 2015-06-10 09:16:21Z

3

I think your json is malformed. Remove $data="'".$data."'";

You can check json error if any.

And $trimspace = preg_replace('/\s+/', '', $data); is needless.

answered Jun 10, 2015 at 9:16

venca

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2 Comments

venca Over a year ago

$jdata is null? What say php.net/manual/en/function.json-last-error.php?

SAWJUSTO Over a year ago

thank php.net/manual/en/function.json-last-error.php, this link helps me, the error I got is json syntax error, but still solving.

mariobgr · Accepted Answer · 2015-06-10 09:16:32Z

1

json_decode usually returns an object, so I don't think your code is wrong here.

$arrayObject = new ArrayObject($object);
$array = $arrayObject->getArrayCopy();

This is how you can convert it to an array. It works in PHP 5.3+

answered Jun 10, 2015 at 9:16

mariobgr

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Comments

Lal krishnan S L · Accepted Answer · 2015-06-10 09:17:13Z

1

Try this

  $jdata = json_decode($trimspace, true);
     print_r($jdata);

answered Jun 10, 2015 at 9:17

Lal krishnan S L

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Comments

Ole Spaarmann · Accepted Answer · 2015-06-10 09:21:03Z

1

First of all your json is malformed. Remove the '' from the beginning and the end of your file. The contents of $data should look like this:

{
"status": "1",
"user": [        
    {
        "id": "27",
        "name": "kktt",
        "phone_no": "1239293",
        "activate_code": "0d08ed",
        "deposit": "0",
        "created": "2015-06-0316:35:08",
        "updated": "1110-11-3000:00:00",
        "status": "0"
    },
    {
        "id": "28",
        "name": "kktt",
        "phone_no": "1239293",
        "activate_code": "fb4876",
        "deposit": "0",
        "created": "2015-06-0316:37:14",
        "updated": "1000-01-0100:00:00",
        "status": "0"
    }
  ]
}

Second $jdata is an associative array. You cannot print its contents with echo. Instead do

print_r($jdata);

Third you don't need to remove spaces. Do that in the script that produces the json, otherwise just parse the json with the spaces directly.

answered Jun 10, 2015 at 9:21

Ole Spaarmann

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2 Comments

SAWJUSTO Over a year ago

@ Bjarnstroem, thank for your answer, I edited my question again, could you give me a favor of checking it agin.

Ole Spaarmann Over a year ago

@SAWAUNG Hey, I suggest you read the PHP documentation before posting here... print_r needs () around the parameters. So don't write print_r $jdata; but print_r($jdata);

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