how i can pick country_code from this data?

Question

here i got this below data with this

$location = file_get_contents('http://freegeoip.net/json/'.$_SERVER['REMOTE_ADDR']);

and $location contains :

{
      "ip": "77.99.179.98",
      "country_code": "GB",
      "country_name": "United Kingdom",
      "region_code": "H9",
      "region_name": "London, City of",
      "city": "London",
      "zipcode": "",
      "latitude": 51.5142,
      "longitude": -0.0931,
      "metro_code": "",
      "areacode": ""

}

have received this data. but am not getting how to pink country_name and region_name from this?

Did you catch that data in a variable? Then there will be solution with string position function.$location= what a directory or the data after executing your code? — Muhammad Ashikuzzaman, Jul 26 '14 at 11:25
possible duplicate of [How to convert JSON string to array](http://stackoverflow.com/questions/7511821/how-to-convert-json-string-to-array) — Kevin, Jul 26 '14 at 13:32

score 0 · Answer 1 · answered Jul 26 '14 at 11:18

you can user jsone_decode like this

    $output_string = '{
      "ip": "77.99.179.98",
      "country_code": "GB",
      "country_name": "United Kingdom",
      "region_code": "H9",
      "region_name": "London, City of",
      "city": "London",
      "zipcode": "",
      "latitude": 51.5142,
      "longitude": -0.0931,
      "metro_code": "",
      "areacode": ""

}';

$arr = json_decode($output_string, true);
echo "Country  Name :".$arr['country_name'];
echo "\nRegion  Name :".$arr['region_name'];

see Working Demo

score 0 · Accepted Answer · answered Jul 26 '14 at 11:18

0

try this

$obj = json_decode($location, true);

then try to get the object

echo $obj['country_name'];

answered Jul 26 '14 at 11:18

Shah Rukh

3,046
2
19
27

1

True will give u array not a object – bhushya Jul 26 '14 at 11:19

score 0 · Answer 3 · answered Jul 26 '14 at 11:20

0

$locationArray = json_decode($location, true);
echo $locationArray['country_name'];

answered Jul 26 '14 at 11:20

Lorenz Meyer

19,166
22
75
121

IROEGBU · Answer 4 · 2014-07-26T11:33:36.420

This works:

$arr = json_decode('{
      "ip": "77.99.179.98",
      "country_code": "GB",
      "country_name": "United Kingdom",
      "region_code": "H9",
      "region_name": "London, City of",
      "city": "London",
      "zipcode": "",
      "latitude": 51.5142,
      "longitude": -0.0931,
      "metro_code": "",
      "areacode": ""

}');

echo $arr->country_code . ", " . $arr->country_name;

Add a true argument to json_decode($json_string, true) to return an array instead of an object as above.

$arr = json_decode('{
      "ip": "77.99.179.98",
      "country_code": "GB",
      "country_name": "United Kingdom",
      "region_code": "H9",
      "region_name": "London, City of",
      "city": "London",
      "zipcode": "",
      "latitude": 51.5142,
      "longitude": -0.0931,
      "metro_code": "",
      "areacode": ""

}', true);

echo $arr['country_code'] . ", " . $arr['country_name'];

score -2 · Answer 5 · answered Jul 26 '14 at 11:17

-2

Use $location->country_code; it will give you country code.

answered Jul 26 '14 at 11:17

GTS Soft.

187
4

without decoding how do you access the prop.. file_get_contents returns string – bhushya Jul 26 '14 at 11:24
of course, you need to decode it. I am just giving code for accessing element – GTS Soft. Jul 26 '14 at 11:27

how i can pick country_code from this data?

5 Answers5