Question

基本上我是从一个JSON格式的字符串开始的。此字符串的一个示例是：[{"options":[{"valid":true,"content":"Answer"}],"content":"Question1","type":"text"},{"options":[{"valid":false,"content":"Answer"}],"content":"Question2","type":"radios"}]

使用json_decode我将其转换为此示例中的数组：

Array
(
    [0] => stdClass Object
        (
            [options] => Array
                (
                    [0] => stdClass Object
                        (
                            [valid] => 1
                            [content] => Answer
                        )

                )

            [content] => Question1
            [type] => text
        )

    [1] => stdClass Object
        (
            [options] => Array
                (
                    [0] => stdClass Object
                        (
                            [valid] => 
                            [content] => Answer
                        )

                )

            [content] => Question2
            [type] => radios
        )

)

使用此PHP代码将其存储在MySQL中：

$questions = json_decode($_POST['test']);

        $con=mysqli_connect("Localhost","root","","test");

        if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }

        foreach( $questions as $question ){
            $options = $question->options;
            $options_id = array();
            $question_query = "INSERT INTO `question` (Content, Type) VALUES ('{$question->content}', '{$question->type}')";
            echo $question_query;
            if( !mysqli_query($con, $question_query) ){ echo mysqli_error($con);}
            $question_id= mysqli_insert_id($con);
            echo $question_id;
            foreach($options as $option){
            $options_query = "INSERT INTO `option` (Question_id, Content) VALUES ({$question_id}, '{$option->content}')";
            echo $options_query;
            if( !mysqli_query($con, $options_query) ){
            echo mysqli_error($con);
      }
  }

它只是存储问题而且可能是答案。

所以在某些时候我需要向后处理这个过程（从MySQL获取信息并再次构建JSON，以便我可以渲染相应的UI）。有没有办法只需还原此代码并向后执行此过程？也许与json_enconde？

Answer 1

你看起来像下面

$a = '[{"options":[{"valid":true,"content":"Answer"}],"content":"Question1","type":"text"},{"options":[{"valid":false,"content":"Answer"}],"content":"Question2","type":"radios"}]' ;
$b = json_decode($a,true);
//print_r($b);

foreach($b as $key=>$val){
    echo 'Content : '.$val['content'];
    echo '<br />';
    echo 'Type : '.$val['type'];
    echo '<br />';
    foreach($val['options'] as $key1=>$val1){
        echo 'Valid :'.$val1['valid'];
        echo '<br />';
        echo 'Answer : '.$val1['content'];
        echo '<br />';

    }
    echo '<hr><br />';
}

输出看起来像

Content : Question1
Type : text
Valid :1
Answer : Answer
------------------------------------------------------
Content : Question2
Type : radios
Valid :
Answer : Answer

通过还原PHP代码来创建JSON字符串

1 个答案: