php - 将 JSON 字符串解析为数据库

Question

我在将 JSON 字符串解析到我的 MYSQL 数据库时遇到问题。这是发送到服务器的 JSON 字符串:

    [{"Description":"Detta är mitt quiz!","Title":"Mitt Quiz","Category":"Music","Language":"Swedish","Difficulty":1},{"QuestionNr1":{"WrongAnswer3":"Visby","WrongAnswer1":"Stockholm","RightAnswer":"Uppsala","WrongAnswer2":"Umeå","Question":"Vilken stad bor jag i?"},"QuestionNr2":{"WrongAnswer3":"Visby","WrongAnswer1":"Stockholm","RightAnswer":"Uppsala","WrongAnswer2":"Umeå","Question":"Vilken stad bor jag inte i?"}}]

这是调用[mArray JSonrepresentation]之前的数据；在上面

(
    {
    Category = Music;
    Description = "Detta \U00e4r mitt quiz!";
    Difficulty = 1;
    Language = Swedish;
    Title = "Mitt Quiz";
},
    {
    QuestionNr1 =         {
        Question = "Vilken stad bor jag i?";
        RightAnswer = Uppsala;
        WrongAnswer1 = Stockholm;
        WrongAnswer2 = "Ume\U00e5";
        WrongAnswer3 = Visby;
    };
    QuestionNr2 =         {
        Question = "Vilken stad bor jag inte i?";
        RightAnswer = Uppsala;
        WrongAnswer1 = Stockholm;
        WrongAnswer2 = "Ume\U00e5";
        WrongAnswer3 = Visby;
    };
}

)

我正在使用 ASIHTTPRequest 的 POST 方法，但不知道如何在服务器端接收它并使用 PHP 将其解析到我的数据库。

谁能指出我正确的方向，我会很高兴!

//谢谢!

Answer 1

http://php.net/manual/en/function.json-decode.php

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

遍历 $json 并构建 sql 查询