我正在尝试使用Ajax提交表单,这是我的简化代码我的索引:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script type="text/javascript" src="jquery-1.7.1.min.js"></script>
<script>
$(document).ready(function(){
$("#addemail").click(function(){
var email=$("#email").val();
$.ajax({
type: "POST",
url: "addemail.php",
data: "email="+email,
success: console.log("success!"),
error: console.log("error!")
});
});
});
</script>
</head>
<body>
<div id="wrapper">
<h2>Email:</h2>
<form action="" method="post">
<table>
<tr>
<td><label>Années:</label></td>
<td><input type="text" id="email" name="email" /></td>
<td><input type="submit" id="addemail" value="Ajouter" /></td>
</tr>
</table>
</form>
</div>
</body>
</html>
而这里我的PHP文件
<?php
$connection = mysql_connect('localhost', 'XXX', 'XXX');
$db= mysql_select_db('mydb', $connection);
$email= $_POST["email"];
$query = 'INSERT INTO users(email) VALUES ("'.mysql_real_escape_string($email).'")';
mysql_query($query);
?>
我的问题是它不起作用...但是,如果我在ajax工作之后放置一个断点,它会在db But中写好电子邮件,然后我将创建两个控制台日志(成功!和错误!)...同样如果我将action =“ addemail.php”添加到表单中,它将不会执行Ajax并转到php页面(此位置为空白,但在db中写得很好...)
有人帮我吗?