如何解碼php文件中的json數據

Question

我想解碼來自android的json數據

["{\"User_Registartion_Details\":{\"fullname\":\"ssss\",\"emailid\":\"sumitbwm@gmail.com\",\"mobile\":\"8796485577\",\"location\":\"pune\"}}"]

怎么做....從json獲取數據並使用php腳本嘗試將其保存在我的sql數據庫中，但在我的sql中顯示空值。 這是我的php代碼，它獲取json數據並進行解碼，然后保存在database.I嘗試，但存儲了空值（空白）。 數據以json格式從android頁面獲取並在php腳本中解碼，保存在db..anyboday中對此有所了解，請help..i在android / php中是新的。
這是我的php代碼，它獲取json數據並進行解碼，然后保存在數據庫中

    <?php

    // Include confi.php
    include_once('conection.php');

    if($_SERVER['REQUEST_METHOD'] == "POST" && $_SERVER["CONTENT_TYPE"] == "application/json")
    {

        $request = file_get_contents('php://input');
        $array = json_decode($request,true);
        $name = $array['fullname'];
        $email = $array['emailid'] ;
        $mobileno = $array['mobile'] ;
        $location = $array['location'];

        $sql = "INSERT INTO  usertable (`fullname`, `emailid`, `mobile`, `location`) VALUES ('$array[name]', '$array[email]', '$array[mobileno]', '$array[location]')";
        $qur = mysql_query($sql);
        if($qur)
        {
            $json = array("status" => 1, "msg" => "Done User added!");
        }
        else
        {
            $json = array("status" => 0, "msg" => "Error adding user!");
        }
    }else
    {
        $json = array("status" => 0, "msg" => "Request method not accepted");
    }

    @mysql_close($conn);

    /* Output header */
    header('Content-type: application/json');
    echo json_encode($json);

Answer 1

PHP中的JSON編碼和解碼是使用json_encode和`json_decode``函數完成的。

// To encode
$encodes = json_encode($json);

// To decode 
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
$decoded = json_decode($json);

// To see which variables are inside your decoded json
echo "<pre>";
var_dump($decoded);
echo "</pre>";
exit;

編輯

您的問題是您的json字符串不正確。 我不得不對其解碼兩次，以使其正常工作。

// Incorrect json string
$json = '["{\"User_Registartion_Details\":{\"fullname\":\"ssss\",\"emailid\":\"sumitbwm@g‌​mail.com\",\"mobile\":\"8796485577\",\"location\":\"pune\"}}"]';

// Correct Json string
$json = '{"User_Registartion_Details":{"fullname":"ssss","emailid":"sumitbwm@g‌​mail.com","mobile":"8796485577","location":"pune"}}';

echo "<pre>";
var_dump(json_decode($json));

object(stdClass)#25 (1) {
  ["User_Registartion_Details"]=>
  object(stdClass)#19 (4) {
    ["fullname"]=>
    string(4) "ssss"
    ["emailid"]=>
    string(24) "sumitbwm@g‌​mail.com"
    ["mobile"]=>
    string(10) "8796485577"
    ["location"]=>
    string(4) "pune"
  }
}

您可以通過查看$_POST變量來訪問您的請求

當在請求中使用application / x-www-form-urlencoded或multipart / form-data作為HTTP Content-Type時，通過HTTP POST方法傳遞給當前腳本的變量的關聯數組。

使用var_dump($_POST);

例

// Your android application sends a request
// This request is part of the HTTP protocol
// It is either a POST, GET, PUT or DELETE request

// PHP handles PUT and DELETE as a POST request
// The request data will be stored in the $_POST varaible and the $_REQUEST variable
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    // Use a var_dump to learn what key holds the json data

    var_dump($_POST);

    // let's say its in $_POST['json']
    $json = $_POST['json'];

    var_dump(json_decode($json));
}

資源資源

• json_encode（）-手冊
• json_decode（）-手冊