PHP JSON Data Selection

Question

How would I go about selecting the data of each title from the following JSON? I have the JSON decoded, but I'm not sure how to select the part I want.

{
    "responseData": {
        "results": [
            {
                "title": "Justin Giesbrecht 749",
                "titleNoFormatting": "Justin Giesbrecht 749",
            },
            {
                "title": "Gopher dunes 09",
                "titleNoFormatting": "Gopher dunes 09",
            },
            {
                "title": "dirtbike Justin",
                "titleNoFormatting": "dirtbike Justin",
            },
            {
                "title": "A Warming",
                "titleNoFormatting": "A Warming",
            }
        ],
        "cursor": {
            "pages": [
                {
                    "start": "0",
                    "label": 1
                },
                {
                    "start": "4",
                    "label": 2
                }
            ],
            "estimatedResultCount": "6",
            "currentPageIndex": 0,
        }
    },
    "responseDetails": null,
    "responseStatus": 200
}

I thought it would be something like this, but I don't get anything:

echo "Response ". $jsonS->responseData->results[1]->title;

When you say you don't get anything, what do you mean? Do you see "Response" or not? Have you tried var_dump($jsonS);? — Josh
– Josh, Commented Oct 14, 2010 at 0:36
Sorry, yes I do see response, but I see nothing of the jsonS, sorry. — Justin
– Justin, Commented Oct 14, 2010 at 15:42

davur · Accepted Answer · 2010-10-14 01:51:55Z

5

Actually you've got the reading of the title part right, it's the JSON that is invalid.

Copying the JSON into a JSON validator/lint e.g. http://www.jsonlint.com/ will show that the you have additional , (commas) after the last object attribute in a few places (5 places to be exact, after each 'titleFormatting' attribute and after 'currentPageIndex').

If you fix those errors and parse it using json_decode e.g.:

$jsonS = json_decode($json_text);

Then your own code:

echo "Response " . $jsonS->responseData->results[1]->title;

Will output the second (index 1 being the second index) results title

Response Gopher dunes 09

answered Oct 14, 2010 at 1:51

davur

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2 Comments

Josh Over a year ago

+1: I answered that it was invalid, but your answer actually tells him why. Nice work!

Justin Over a year ago

Thank you, it was minor, but you were right! appreciate your time! Cheers. [answered]

Karl · Accepted Answer · 2010-10-14 00:27:32Z

0

When you json_decode something it is converted to a regular PHP array, so you can reference it like $decodedObject["nodeName"].

answered Oct 14, 2010 at 0:27

Karl

1

1 Comment

mattbasta Over a year ago

As @Josh mentioned, this only works if you use json_decode($json_data, true)

Josh · Accepted Answer · 2010-10-14 00:43:15Z

When I parse that JSON as you quoted it with PHP's json_decode method, it gives me NULL because PHP doesn't think it's valid JSON. Add a var_dump($jsonS); to see if this is happening to you as well. If it is, you may need to be sure your JSON is valid.

Here's what I did, for reference:

$json_data = <<<END_OF_JSON
{   
    "responseData": {        "results": [
            {
                "title": "Justin Giesbrecht 749",                "titleNoFormatting": "Justin Giesbrecht 749",
            },
            {
                "title": "Gopher dunes 09",
                "titleNoFormatting": "Gopher dunes 09",
            },
            {
                "title": "dirtbike Justin",
                "titleNoFormatting": "dirtbike Justin",
            },
            {
                "title": "A Warming",
                "titleNoFormatting": "A Warming",
            }
        ],
        "cursor": {
            "pages": [                {   
                    "start": "0",
                    "label": 1
                },
                {   
                    "start": "4",
                    "label": 2
                }
            ],
            "estimatedResultCount": "6",
            "currentPageIndex": 0,
        }
    },
    "responseDetails": null,
    "responseStatus": 200
}
END_OF_JSON;
$jsonS = json_decode($json_data);
var_dump($jsonS);
echo "Response ". $jsonS->responseData->results[1]->title;

And the output was:

NULL Response

If you're using different JSON, please edit your question and share it. But if you're actually using the above, it's not valid...

ts. · Accepted Answer · 2010-10-14 03:15:10Z

0

In JSON you are not allowed to leave trailing comma in array/object definition. So, when in PHP is perfectly valid to write:

$a = array(1,2,3,);

(note last comma), in JSON

a : [1,2,3,]

or

a : {x :'y',}

is invalid.

answered Oct 14, 2010 at 3:15

ts.

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