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我建立使用AJAX形式和遇到一个问题:空白值 - Ajax表单
每当我提交表单,存储在数据库中的值变为空白......(仅供参考,当然页不刷新..这是整点)
这里是我的文件的index.php:
<!doctype html>
<html>
<head>
<title>Form Practice</title>
</head>
<body>
<div id="myForm">
Name: <input name="username" type="radio" value="Sagar">Sagar</input>
<input name="username" type="radio" value="Saransh">Saransh</input><br /><br />
Profession: <input name="profession" type="radio" value="Coder">Coder</input>
<input name="profession" type="radio" value="Entrepreneur">Entrepreneur</input>
<input name="profession" type="radio" value="Blogger">Blogger</input><br /><br />
<input type="submit" id="submit" value="Submit"></input>
</div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$.ajax({
url: "data.php",
type: "POST",
async: false,
data:{
"username":$('input[name=username]:checked', '#myForm').val(),
"profession": $('input[name=profession]:checked', '#myForm').val()
}
});
</script>
</body>
</html>
而且我Data.php:
<?php
require 'connect.php';
$db_selected = mysql_select_db(DB_NAME, $link);
if(!$db_selected){
die('cant use'. DB_NAME . ':'. mysql_error());
}
$username = $_POST['username'];
$profession = $_POST['profession'];
$sql = mysql_query("INSERT INTO info (Name, Profession) VALUES('{$username}', '{$profession}')");
if(!mysql_query($sql)){
die('Some Error '. mysql_error());
}
mysql_close();
?>
<!doctype html>
<html>
<head>
<title>Data</title>
</head>
<body>
</body>
</html>
你有回声出的userame和职业,以确保数据被发送通过ajax后? – Matt
页面甚至没有刷新..所以在data.php中回显用户名没有意义。如果你要求在index.php本身回显它,那么这个变量就不存在,因为我们还没有提交表单。所以,无论哪种方式它不会帮助。 –