Php 使用JQuery显示未定义的POST
我是jQuery的新手,在提交表单时,我一直将Php 使用JQuery显示未定义的POST,php,jquery,html,Php,Jquery,Html,我是jQuery的新手,在提交表单时,我一直将未定义的作为数据。我在网上查看并尝试了serialize()、$.ajax函数、$.post函数以及其他方法,但均无效 我正在尝试在不刷新页面的情况下完成表单提交。如果我注释掉脚本以在表单提交后保留在页面上,代码工作正常 index2.php <script> function newlock() { $("#new-lock").load("new-lock.php"); } </script> <h3>
未定义的作为数据。我在网上查看并尝试了serialize()
、$.ajax
函数、$.post
函数以及其他方法,但均无效
我正在尝试在不刷新页面的情况下完成表单提交。如果我注释掉脚本以在表单提交后保留在页面上,代码工作正常
index2.php
<script>
function newlock() {
$("#new-lock").load("new-lock.php");
}
</script>
<h3>Lock Settings</h3>
<div>
<a href="#" onclick="newlock();">New Lock</a>
</div>
//code in between
<div id="tabs-1">
<div id="new-lock"></div>
</div>
require "connect.php";
$name = $_POST['name'];
echo $name;
$IP = $_POST['IP'];
echo $name;
$sql="INSERT INTO door_lock (lock_IP, lock_name)
VALUES
('$name','$IP')";
if (!mysql_query($sql,$conn))
{
die('Error: ' . mysql_error());
}
mysql_close($conn);
?>
<!doctype html>
<html lang="us">
<h2 align = "center"> Please enter information for the new lock </h2>
<table border="0" align= "center">
<form id="newlockform" action="insert_new_lock.php" method="post">
<tr><td>Lock Name:</td> <td><input type="text" name="name"></td></tr>
<tr><td>Lock IP:</td> <td><input type="text" name="IP"></td></tr>
<tr><td><input type="reset" value="Clear"> <input type="submit" name="submitted" value="Submit"></td></tr>
</form>
</table>
</html>
<script>
$("#newlockform").submit(function() {
var name = $("#name").val();
var IP = $("#IP").val();
var dataString = 'name=' + name + '&IP=' +IP;
$.post('insert_new_lock.php', dataString, function(data) {
$("#tabs-1").html(data).fadeIn('100');
$('#name,#IP').val('');
}, 'text');
return false;
});
</script>
new lock.php
<script>
function newlock() {
$("#new-lock").load("new-lock.php");
}
</script>
<h3>Lock Settings</h3>
<div>
<a href="#" onclick="newlock();">New Lock</a>
</div>
//code in between
<div id="tabs-1">
<div id="new-lock"></div>
</div>
require "connect.php";
$name = $_POST['name'];
echo $name;
$IP = $_POST['IP'];
echo $name;
$sql="INSERT INTO door_lock (lock_IP, lock_name)
VALUES
('$name','$IP')";
if (!mysql_query($sql,$conn))
{
die('Error: ' . mysql_error());
}
mysql_close($conn);
?>
<!doctype html>
<html lang="us">
<h2 align = "center"> Please enter information for the new lock </h2>
<table border="0" align= "center">
<form id="newlockform" action="insert_new_lock.php" method="post">
<tr><td>Lock Name:</td> <td><input type="text" name="name"></td></tr>
<tr><td>Lock IP:</td> <td><input type="text" name="IP"></td></tr>
<tr><td><input type="reset" value="Clear"> <input type="submit" name="submitted" value="Submit"></td></tr>
</form>
</table>
</html>
<script>
$("#newlockform").submit(function() {
var name = $("#name").val();
var IP = $("#IP").val();
var dataString = 'name=' + name + '&IP=' +IP;
$.post('insert_new_lock.php', dataString, function(data) {
$("#tabs-1").html(data).fadeIn('100');
$('#name,#IP').val('');
}, 'text');
return false;
});
</script>
请输入新锁的信息
锁名称:
锁定IP:
$(“#newlockform”).submit(函数(){
var name=$(“#name”).val();
var IP=$(“#IP”).val();
var dataString='name='+name+'&IP='+IP;
$.post('insert\u new\u lock.php',数据字符串,函数(数据){
$(“#tabs-1”).html(数据).fadeIn('100');
$('#name,#IP').val('');
}“文本”);
返回false;
});
任何帮助都将不胜感激D您的表单字段上没有设置任何id
,因此
var name = $("#name").val();
将返回null,然后将其作为空字符串提交。表单字段应该与loo类似
<input type="text" name="whatever" id="name" />
^^^^^^^^^^^
^^^^^^^^^^^
为了让JS代码正常工作。在处理其余内容之前,您确实需要仔细阅读。否则,请享受服务器pwn3d.com的许可。我有很多东西要学D