Php 使用JSON从数据库获取信息

亲爱的各位程序员/黑客/程序员。我需要一些帮助。我使用下面的代码从MySQL数据库中获取信息。这是PHP代码：

<?php
ini_set("display_errors", true);
ini_set("html_errors", false);
require "conn.php";
$query=mysqli_query($conn,"SELECT * FROM UserData");

if ($query){
    while($row=mysqli_fetch_array($query)){
        $flag[] =$row;


    }
    print(json_encode($flag));


}
mysqli_close($conn);
return $flag;
?>

代码将打印以下内容：

{"0":"1","id":"1","1":"netsgets","username":"netsgets","2":"test","password":"test","3":"","likedOne":"","4":"","likedTwo":"","5":"","likedThree":"","6":"","likedFour":"","7":"","likedFive":""}

我希望它是这样的：

{"id":"1","username":"netsgets","password":"test","likedOne":"","likedTwo":"","
likedThree":"","likedFour":"","likedFive":""}

（应该是这样的！！！）

---

请帮助我解决此问题。

假设连接正常，并且它正确地获取数据，您可以使用mysql\u fetch\u assoc来代替：

<?php
ini_set("display_errors", true);
ini_set("html_errors", false);
require "conn.php";
$query=mysqli_query($conn,"SELECT * FROM UserData");

if ($query){
    while($row=mysqli_fetch_assoc($query)){
        $flag[] =$row;


    }
    print(json_encode($flag));


}
mysqli_close($conn);
return $flag;
?>

如果您有选择的话，我个人会转向PDO，因为它允许您进行更好的数据库抽象

<?php
ini_set("display_errors", true);
ini_set("html_errors", false);
require "conn.php";
$query=mysqli_query($conn,"SELECT * FROM UserData");

if ($query){
    while($row=mysqli_fetch_assoc($query)){
        $flag[] =$row;


    }
    print(json_encode($flag));


}
mysqli_close($conn);
return $flag;
?>