2017-09-08 32 views
-3

http://example.com/api/orderDetails.php?{“user_id”:1,“itemList”:[{“max_price”:120,“min_price”:100,“trend”:“high”,“c_price”:110,“status”:“1” “类型”: “蔬菜”, “ITEM_ID”:19},{ “MAX_PRICE”:200, “MIN_PRICE”:100, “趋势”: “低”, “c_price”:100, “状态”: “1” ,“type”:“veg”,“item_id”:224}],“table_no”:“1”}如何解码获取我的url中的json数据的json?

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你在哪里接收网址前是一个JSON格式?什么是你的应用程序的类型? –

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我从android.I得到这个url我必须插入每个项目在我的数据库。问题是我如何解码这个JSON,这样我就可以获得数组数据 – Aks

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刚刚添加了答案,但你需要你的URL是一个有效的JSON格式并确保你这样做,你必须使用json内容强制响应头。 –

回答

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如何解码php?

$object_from_json = json_decode('{"user_id":1,"itemList":[{"max_price":120,"min_price":100,"trend":"high","c_price":110,"status":"1","type":"veg","item_id":19},{"max_price":200,"min_price":100,"trend":"low","c_price":100,"status":"1","type":"veg","item_id":224}],"table_no":"1"}'); 

//or 

$array_from_json = json_decode('{"user_id":1,"itemList":[{"max_price":120,"min_price":100,"trend":"high","c_price":110,"status":"1","type":"veg","item_id":19},{"max_price":200,"min_price":100,"trend":"low","c_price":100,"status":"1","type":"veg","item_id":224}],"table_no":"1"}', true); 
//then you got a Array in PHP 
//var_dump($array_from_json); 

//with an URL 
$array_from_json_url = json_decode(file_get_contents($json_url), true); 
//var_dump($array_from_json_url); 

你需要你的网页PHP orderDetails.php上添加响应头需要回声

header("Content-type:application/json"); 
//$output is array or object PHP 
echo json_encode($oupout); 
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如何通过$ _REQUEST方法获取它。我试过了$ array_from_json = json_decode($ _ REQUEST,true);但是获得null在我的回应中 – Aks