parsing object output data in PHP

Question

$location = '{ latitude: (location tatitude), longitude: (location longitude), description: (location description) }';

$json = json_decode($location,true);

$locationlat = $json['latitude'];

I want to pull the latitude data in php. Blank data comes when I do it this way. How can I pull latitude and long data?

It's not JSON, but you can do something like that : `preg_match('/{ latitude: (.*), longitude: (.*), description: (.*)}/', $location, $matches);`. Results in `$matches`. — RadicalDingos, May 27 '21 at 17:56

Pat Freeze · Answer 1 · 2021-05-27T17:58:32.193

-1

In your $location is not a valid JSON So json_decode return null.

try

$location = '{ latitude: "location tatitude", longitude: "location longitude", description: "location description" }';

edited May 27 '21 at 17:58

answered May 27 '21 at 17:39

Pat Freeze

1
4

The data is coming directly baby, the way I wrote it, I have no chance to fix it – SH Yazılım Geliştirme May 27 '21 at 17:40
Then you need to do something else, it is not JSON. – AbraCadaver May 27 '21 at 17:41
So if you want to use JSON you will need to parse it first. – Pat Freeze May 27 '21 at 17:53

score -1 · Answer 2 · edited May 27 '21 at 17:46

-1

First, the JSON code is wrong. JSON should look like this:

{
     "fooString": "barString",
     "fooObject": {
          "barString": "Hello World"
     }
}

Now... When you do json_decode(), it makes an object and you shall therefore retrieve a value like this:

$json = json_decode($location);
$locationlat = $json->latitude;

edited May 27 '21 at 17:46

Dharman

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answered May 27 '21 at 17:40

ThorBilsby

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parsing object output data in PHP

2 Answers2