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 $resultecho $result['password'];// closes database connectionmysqli_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：19I'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!
2 回答
至尊宝的传说
TA贡献1789条经验 获得超10个赞
您可以使用 不使用事件。clicksubmit
在你的情况下，只要给你的喜欢 -idform
<form validate="" id="submit">
现在，在您的脚本中 -js
$(function() { //shorthand document.ready function
$('#submit').on('submit', function(e) {
e.preventDefault(); //prevent form from submitting
console.log(data);
$.ajax({
type: "POST",
async: true,
url: test.php,
data: $(this).serializeArray(),
success: function (data) {
console.log(data);
}
});
});
});
因此，请检查您的整个代码 -
<!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="" id="submit">
<input type="text" placeholder="Username" id="username" required/><br />
<input type="password" placeholder="Password" id="password" required/><br />
<input type="submit" value="Login" />
</form>
<script type="text/javascript">
// when document is loaded
$(function() { //shorthand document.ready function
$('#submit').on('submit', function(e) {
e.preventDefault(); //prevent form from submitting
console.log(data);
$.ajax({
type: "POST",
async: true,
url: test.php,
data: $(this).serializeArray(),
success: function (data) {
console.log(data);
}
});
});
});
</script>
</body>
</html>
希望这会帮助你。
青春有我
TA贡献1784条经验 获得超8个赞
您需要将一个函数传递给您的 document.ready（） 调用和您的 click（） 调用。
<script type="text/javascript">
$(document).ready(function() {
Your variables here...
$('#submit').click(function() {
... Ajax call here.
});
});
</script>
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