Return one result from Json String in PHP

Question

hi im trying to return 1 set of array from a json string in my code from searching nas i would like to show the data associated with that name. Any suggestions would be great .. thanks

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);

//echo $_GET['urlname'];

$json_string =    file_get_contents("http://192.168.1.104/testing_JSON/newjson.json");
$parsed_json = json_decode($json_string, true);
$parsed_json = $parsed_json['contacts'];

for(var i = 0; i < $parsed_json.length; i++)
{
if($parsed_json[i].$parsed_json.name == 'nas')
{
 return $parsed_json[i].$parsed_json.name;
}
}
?

im getting a white screen with the above code.

thanks

You didn't showed any JSON string in your question. `192.168.1.104` is something different for most of us. Keep in mind, that we're not in the same local area network as you are. also try to create a self-contained example. — hakre, May 25 '14 at 19:18
you are trying to use javascript syntax here. rtfm http://www.php.net/manual/en/language.types.array.php http://www.php.net/count — chiliNUT, May 25 '14 at 19:20

score 2 · Accepted Answer · answered May 25 '14 at 19:16

2

Use echo instead of return. Return doesn't print anything.

answered May 25 '14 at 19:16

Alex

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thanks for reply .. didnt work still white screen – Nasir Shiraz May 25 '14 at 19:18
1

Thanks for pointing out the white screen. Error referenced. – hakre May 25 '14 at 19:19
I think you're confusing PHP with Javascript. `length` is not a function. Use `count($parsed_json)` instead. Also I'm not sure what you're trying to achieve with this `$parsed_json[i].$parsed_json.name`. – Alex May 25 '14 at 19:19
php is new to me im just learning.. any suggestions. the count($parsed) didnt work – Nasir Shiraz May 25 '14 at 19:27

Return one result from Json String in PHP

1 Answers1