Question

以下代码不会在浏览器中显示任何内容。所有数据库名称和表名称都是正确的。在while循环中添加行echo json_encode（$ temp）会将结果显示为对象。

 <?php 
     define('DB_HOST', 'localhost');
     define('DB_USER', 'root');
     define('DB_PASS', '');
     define('DB_NAME', 'sample_userdata');

     $conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

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

     $stmt = $conn->prepare("SELECT questions FROM sample_questions;");

     $stmt->execute();

     $stmt->bind_result($post);

     $userdata = array(); 

     while($stmt->fetch()){
       $temp = array();
       $temp['questions'] =$post;


       array_push($userdata,$temp);
     }

     echo json_encode($userdata);
     ?>

Answer 1

当您使用json_decode（）函数时，它将为带有键的数组创建对象

$x['test1'] = 1;
$x['test2'] = 2;
echo json_encode($x);
// Result: {"test1":1,"test2":2}

您可以使用array_values（）删除键

$x['test1'] = 1;
$x['test2'] = 2;
echo json_encode(array_values($x));
// Result: [1,2]

php代码未给出响应

1 个答案: