JSON array parsing in PHP

Question

please help me how to parse following JSON in PHP.

I got following JSON using PIWIK reporting api. How do I get PageTitle from following json in PHP. I tried following code for JSON parsing.

$json = '[
{
    "idSite": "1",
    "idVisit": "84",
    "visitorId": "f08dc1f2a3e1f839",
    "visitorType": "returning",
    "visitorTypeIcon": "plugins/Live/images/returningVisitor.gif",
    "visitConverted": "0",
    "visitConvertedIcon": null,
    "visitEcommerceStatus": "none",
    "visitEcommerceStatusIcon": null,
    "searches": "0",
    "events": "4",
    "actions": "9",
    "actionDetails": [
        {
            "type": "action",
            "url": "http://mywwebsiteurl.com",
            "pageTitle": "PageTitle",
            "pageIdAction": "110"
        }
    ]
}
]';

$visits = json_decode($json, true);
foreach ($visits->actionDetails as $data) { 
      echo $data->pageTitle;
}

I got following notice

Notice: Trying to get property of non-object

How to get pageTitle from above JSON.

$visits->actionDetails['pageTitle'] should access the data. As you put true as the second parameter it returns an array not a stdClass. — Matt Burrow, Jul 11 '14 at 09:17

score 1 · Accepted Answer · answered Jul 11 '14 at 09:21

1

It should be :-

$visits = json_decode($json, true);
foreach ($visits[0]["actionDetails"] as $data) { 
      echo $data["pageTitle"];
}

answered Jul 11 '14 at 09:21

Khushboo

1,819
1
11
17

Thanks for your response. I got pageTitle. – sathish Jul 11 '14 at 09:37

score 0 · Answer 2 · answered Jul 11 '14 at 09:21

0

You set second attribute to TRUE - so according to the manual. "When TRUE, returned objects will be converted into associative arrays."

Try using the array aproach

foreach ($visits['actionDetails'] as $data){
    echo $data['pageTitle'];
}

answered Jul 11 '14 at 09:21

Axel

61
1
9

Thank you Alex for your response. – sathish Jul 11 '14 at 09:38

score 0 · Answer 3 · edited May 23 '17 at 12:29

0

As documented in the manual, and mentioned by Matt, the second parameter of json_decode() controls the return type. If it is omitted or set to false (the default) then an object is returned. If it is set to true an array is returned.

Your code, json_decode($json, true);, will return an array but you then try to use the array as an object.

edited May 23 '17 at 12:29

Community

1
1

answered Jul 11 '14 at 09:24

timclutton

12,682
3
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JSON array parsing in PHP

3 Answers3