json_encode, json_decode, array? [duplicate]

Question

I'm trying to get into just data from

{
    "data": [{
        "media_count": 3045,
        "name": "snow",
    },
    {
        "media_count": 79,
        "name": "snowman",
    },
    {
        "media_count": 40,
        "name": "snowday",
    },
    {
        "media_count": 29,
        "name": "snowy",
    }]
}

I've been trying, using:

$obj = json_decode($res[0], true);
echo $obj['data']; //this returns an array

I also tried this:

$obj = json_encode($res[0], true);
echo $obj; // this returns json, but not inside `data`

"data": [{
"media_count":54373,
"name":"test"
}]

I just want to get inside data. How would I do so?

Thanks in advance!

UPDATE: Sorry to mention, I would like this in json format please

eventually, I would like to only see

{
    "media_count":54373,
    "name":"test"
}

Something like thiat

Btw, your original JSON, as it stands, is not valid due to the trailing commas. — Ayush
– Ayush, Commented Jun 7, 2012 at 5:58

flowfree · Accepted Answer · 2012-06-07 05:55:41Z

2

Use json_encode() to get what you want:

$obj = json_decode($res[0], true);
echo json_encode($obj['data']);

answered Jun 7, 2012 at 5:55

flowfree

16.5k12 gold badges55 silver badges79 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Ayush · Accepted Answer · 2012-06-07 05:50:22Z

1

In your first example, $obj['data'] returns an array because that's how the JSON is set up. According to the JSON, data is a collection of elements.

To access within the array, you can do this:

foreach($obj['data'] as $object) {
    print_r($object);
}

You can also index into it as you want:

print_r($obj['data'][0]);

EDIT

If I'm getting you correct, you want to convert the first JSON to this:

"data": [{
"media_count":54373,
"name":"test"
}]

If so, that is not possible since the second fragment is not valid JSON. (use http://jsonlint.com)

answered Jun 7, 2012 at 5:50

Ayush

42.5k52 gold badges169 silver badges243 bronze badges

2 Comments

hellomello Over a year ago

I would like to only index into data so essentially, when you echo the new json, data shouldn't be there.

Ayush Over a year ago

In that case, after json_decode do $obj = $obj['data']. $obj now holds what you want.

Shiven · Accepted Answer · 2012-06-07 05:51:43Z

0

firstly access the data item of the json array;

    $obj1=$obj[item];
    $sizeobj=sizeof($obj1);

    for($i=0;$i<$sizeobj;$i++)
{
// your code to access the data items
// for eg $obj[item][$i][media_count}
}

answered Jun 7, 2012 at 5:51

Shiven

2061 gold badge9 silver badges17 bronze badges

2 Comments

hellomello Over a year ago

So if I try $obj1 = $obj['data'], I get this "

Shiven Over a year ago

$obj1 is not access it is just to find the size of the 'data'

Collectives™ on Stack Overflow

json_encode, json_decode, array? [duplicate]

3 Answers 3

Comments

2 Comments

2 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

2 Comments

Linked

Related