我是jQuery和AJAX的新手，我正在将登录页面作为项目进行处理，我需要使用AJAX从数据库中检索数据。我不是100%流利的英语，所以我会尽我所能解释这个问题（在谷歌翻译的帮助下）。以下是我正在使用的代码：
索引.html
<!DOCTYPE html>
<html>
<head>
<title>Login</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
</head>
<body>
<form validate="">
<input type="text" placeholder="Username" id="username" required/><br />
<input type="password" placeholder="Password" id="password" required/><br />
<input type="submit" id="submit" value="Login" />
</form>
<script type="text/javascript">
// when document is loaded
$(document).ready (
// when submit is clicked
$("#submit").click (
// sets test to null
var test = null;
// sets username to value of username input
var username = document.getElementById("username").value;
// AJAX request
$.ajax({
type: "POST",
async: true,
url: test.php,
data: {username: username},
success: function (data) {
test = data;
console.log(test);
return test;
}
});
);
);
</script>
</body>
</html>
测试.php
<?php
// connects to database
$conn = mysqli_connect('server', 'username', 'password', 'database');
// sets var username to POST username value
$username = $_POST['username'];
// SQL Query
$sql = "SELECT * FROM users WHERE username='" . $username . "'";
$result = mysqli_query($conn, $sql);
// sets result to mysqli_fetch_assoc()
$result = mysqli_fetch_assoc( $result );
// echos $result
echo $result['password'];
// closes database connection
mysqli_close( $conn );
?>
控制台日志
控制台输出：“”[DOM] 输入元素应具有自动完成属性（建议：“当前密码”）：（更多信息：https://www.googlesite.com）
未捕获的语法错误：意外的令牌 var ajax.html：19
I've looked at the code and I can't seem to find an error.
Thanks in advance! ;)
>P.S.
>It's probably going to end up being some stupid typo.
>Other than that, have a great day!
至尊宝的传说
青春有我