我是StackOverflow web开发的新手,我没有时间完成这项任务,所以如果我在理解web开发词汇和任何答案或提示方面有点慢,我很抱歉。我在通过ajax调用另一个php文件时遇到了问题。index.php:
<!--------------------------- HTML STARTUP ----------------------------------->
<!DOCTYPE html>
<html>
<head>
<link type="text/css" rel="stylesheet" href="finalproject.css" />
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<script type="text/javascript" src="finalproject.js"></script>
<title>Final Project</title>
</head>
<body>
<!--------------------------- LOGIN ------------------------------------------>
<?php
require_once 'dbconnection.php';
//----------------------- CREATE USERS TABLE -----------------------------------
$userspass = "CREATE TABLE usersPass (
userID VARCHAR(60),
password VARCHAR(60)
);";
$connected->query($userspass);
$selectAllInfo = "SELECT * FROM usersPass;";
$connected->query($selectAllInfo);
//-------------------------- LOG IN OR REGISTER --------------------------------
?>
<form id="trial"><button onclick="OpenRegister()">Sign Up!</button>
<script>
$(document).on("click" , "#Register", function(e)
{
var datastring = $("#trial").serialize();
$.ajax({
type: 'POST',
url: 'userlogin.php',
data: datastring,
success: function(data){
alert('Sign Up function is a success!');
}
});
e.preventDefault();
});
</script></form>
<br/><br/>
<button onclick="InputInfo()">Login</button>
<?php
$connected->close();
?>
</body>
</html>
这是调用函数时的JavaScript。finalproject.js:
/*jslint browser: true*/
/*global $, jQuery, alert*/
function OpenRegister() {
'use strict';
$('form').append("<div id=SignInContainer>
<span style='font-size: 20px'>UserID</span>
<input type='text' name='UserID' value='Type userID here...'>
<span style='font-size: 20px'>Password</span>
<input type='text' name='UserID' value='Type password here...'>
<button id='Register'>Submit</button>
</div>");
}
$(document).ready(function () {
'use strict';
$(document).on('focus', 'input', function () {
$('input').focus(function () {
$(this).val('');
});
});
});
这是我试图加载的php文件。userlogin.php:
<?php echo "If this displays, you win." ?>
dbconnection.php
<?php
$connected = new mysqli('localhost', 'Username', 'Password', 'Username');
mysqli_select_db($connected, 'cferna50');
// Check connection
if ($connected->connect_error) {
die("Connection failed: " . $connected->connect_error);
}
我只是想让"userlogin.php"文件通过AJAX运行。我在Chrome上运行这个程序。我听说在本地文件上使用AJAX有点问题,但我尝试过——从文件中访问文件,但没有帮助。我试着在FireFox和Internet Explorer上运行它,但仍然没有帮助。我确信我的所有东西都在同一个目录中。我正在使用一个在线学校服务器来完成这一切。我已经为此伤了我的头好几个小时了,任何帮助都将不胜感激。
仅仅因为你没有通过方法传递(数据),你要么序列化数据,要么单独传递数据,我想你会使用第一个方法来节省时间
<script>
$(document).on("submit","#loginForm", function(e)
{
var datastring = $("#loginForm").serialize();
$.ajax({
type: 'POST',
url: 'userlogin.php',
data: datastring,
success: function(data){
alert(data);
//or
//alert('Sign Up function is a success!');
},
error: function(){
alert("error handling");
}
});
e.preventDefault();
});
</script>
HTML
<form id="loginForm" method="POST">
<input type="text" name="userid" id="userid" placeholder="user ID" />
<input type="text" name="password" id="password" placeholder="Password" />
<button type="submit" >Submit</button>
</form>
PHP
<?php
if(isset($_POST['userid'])){
$username = $_POST['userid'];
$pass = $_POST['password'];
//do your magic now
}
?>
在.append()
处正确连接html
字符串,删除if($("#Register").click())
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
function SignUp() {
if ($("[name=username]").val().length
&& $("[name=password]").val().length) {
$.ajax({
type: 'POST',
url: 'userlogin.php',
data: {
username: $("[name=username]").val(),
password: $("[name=password]").val()
},
success: function(data) {
alert('Sign Up function is a success!');
},
error: function() {
alert("stacsnippets success")
}
});
}
}
</script>
<script>
function OpenRegister() {
'use strict';
$('body').append("<div id=SignInContainer>"
+ "<span style='font-size: 20px'>UserID</span>"
+ "<input type='text' name='username' placeholder='Type userID here...'>"
+ "<span style='font-size: 20px'>Password</span>"
+ "<input type='password' name='password' placeholder='Type password here...'>"
+ "<button onclick='SignUp()' id='Register'>Submit</button>"
+ "</div>");
}
</script>
<button onclick="OpenRegister()">Sign Up!</button>
php
if (isset($_POST["username"]) && isset($_POST["password"])) {
// do username, password validation stuff here
}
jsfiddlehttps://jsfiddle.net/qnrnn5b5/