PHP获取mysql数据并使用ajax返回javascript

我用php读取MySql数据库表的值，如下所示：

  <?php

header("Access-Control-Allow-Origin: *");
header("Access-Control-Allow-Methods: PUT, GET, POST");
header("Access-Control-Allow-Headers: Origin, X-Requested-With, Content-Type, Accept");

$servername = "localhost";
$username = "wsl_home";
$password = "wsl";
$dbname = "wsl_home";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

if(isset($_POST['functionname']) == 'getDrinks') {
    $sql = "SELECT ID, Drinkname FROM drinks";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "ID: " . $row["ID"]. " - Name: " . $row["Drinkname"]. "<br>";
        return $row;
    }
} else {
    echo "0 results";
}
}
?>

---

但是我在success函数中没有得到任何输出。有人知道如何完成这项任务吗？

你是从domin获取html文件还是从硬盘上运行它？如果你点击了drink.php，它是否返回信息？为什么要使用“return$row”；这将退出while循环，只需回显该行；您的php函数实际被调用了吗？从中登录以查看您的Ajax是否正在调用它。@Ggg是的