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Smokey
Smokey

Reputation: 1897

Extracting data from JSON failed in PHP

{
    "firstName":"sunny",
    "religion":  {"holly":"1",
    "bolly":"colly",
    "nolly":"only"
},
    "lonely":"money",
    "ronney":"leone",
    "honey":"bunny"
}

This is my JSON. I want to get all the data from this and to be stored into some php variables or an array.

I Used the following code to extract data from my JSON. I decoded it first and then stored it in an array..

    $val_array = json_decode($jsondata, true);
    echo $jsondata;
    $AAA    = $val_array->firstName;
    $BBB    = $val_array->religion;
    $CCC_id = $val_array->bolly;
    $DDD    = $val_array->nolly;
    $CCC_id = $val_array->lonely;
    $DDD    = $val_array->ronney;

But it didn't give me any output. Then I used this.

foreach($data['val_array'] as $item)
{
     echo  $item[0];
    }
}

No output. Help??

Upvotes: 0

Views: 102

Answers (5)

Smokey
Smokey

Reputation: 1897

$val_array = json_decode($jsondata, true);

$m1=$val_array['firstName'];
$m2=$val_array['lonely'];
$m3=$val_array['ronney'];
$m4=$val_array['honey'];

$m4=$val_array['religion']['holly'];
$m5=$val_array['religion']['bolly'];
$m6=$val_array['religion']['nolly'];

BY using this, we don't have to use foreach loops or inner loops for accessing data. Viola!

Upvotes: 0

SidOfc
SidOfc

Reputation: 4584

json_decode by default returns an object yet since you are setting the second parameter to true, you are given an associative array with the information instead.

It basically comes down to the fact that either you do not need to fill in the second parameter and get the object you want, or you work with arrays when you set the parameter to true.

A little reading on PHP.net will do you good for further reference since their documentation is well presented, usually commented by others with helpful suggestions and quite clean as well!

Upvotes: 0

Igoooor
Igoooor

Reputation: 407

You have an error in your JSON : 

"nolly":"only",

"honey":"bunny",

remove the ',' at the end of these 2 lines, then json_decode() will return you an array. And if you want an object, do not pass second argument to json_decode()

Upvotes: 0

Bartek
Bartek

Reputation: 1359

Your JSON is not valid. Remove commas after last elements:

{
    "firstName" : "sunny",
    "religion" : {
        "holly" : "1",
        "bolly" : "colly",
        "nolly" : "only" # Here
    },
    "lonely"    : "money",
    "ronney"    : "leone",
    "honey"     : "bunny" # And here
}

Upvotes: 0

Daniel W.
Daniel W.

Reputation: 32350

You get this second param wrong:

$val_array = json_decode($jsondata, true);

$AAA = $val_array['firstName'];

OR

$val_array = json_decode($jsondata, false);

$AAA=$val_array->firstName;

Upvotes: 4

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