我正在尝试创建一个表单,用户正在选择一个选项,假设他将选择自己的名字,我使用ajax调用另一个页面,在这个页面中,我创建连接并插入答案-问题是插入null密码index.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>MyDataBase </title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"> </script>
$(document).ready(function(){
$("#mysubmit").click( function (){
$.ajax({
url:"Final.php",
type:"POST",
success: function(success_array)
{
alert(" Well Done ");
},
error: function(xhr, ajaxOptions, thrownError)
{
alert( " Didn't work ! ");
}
});
});
});
</script>
</head>
<form method="post" id = "myform">
<input type="radio" name="kname" value="X1" />X1<br />
<input type="radio" name="kname" value="Y1" />Y1<br />
<input type="radio" name="kname" value="A1" />A1<br />
<input type="radio" name="kname" value="G1" />G1<br />
<input type="submit" onclick="save()" id="mysubmit"/>
</form>
<body>
</body>
</html>
Final.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Thanks</title>
</head>
<body>
<?php
$con = mysqli_connect("localhost","root","xyz","test");
mysqli_query($con,"INSERT INTO name values ('$_POST[kname]')");
?>
</body>
</html>
您必须通过$_POST['kname']
引用kname。
但是请!总是逃避你的价值观!
通过ajax提交/发布时,需要添加数据。最简单的是CCD_ 2。
var form = $('#myform');
$.ajax({
url:"Final.php",
type:"POST",
data: form.serialize(),
...
或者你可以得到一个单独的字段-
data: {kname: $('#id_of_your_radio').val()},
只需确保在插入数据库之前对用户数据进行清理即可。