我需要帮助提交表单。我不想页面刷新，我不知道如何做到这一点 [英] I need help submitting a form. I do not want the page to refresh, and I do not know how to accomplish this
问题描述
我使用一个按钮提交表单，而不是一个提交按钮。不过，我的问题是，我当我preSS进，形式不不会提交，所以不是我实际上是点击提交按钮，它可以是不方便。我想知道我怎么可以用一个按钮从提交我而无需刷新页面，当我preSS进入，也有形式提交。我觉得这是一个非常简单的解决方法，我只是找不到。这里是URL以在其上发生问题的页面。 http://www.pearlsquirrel.com/register.php
I am using a button to submit a form, not a submit button. However, my problem is that I when I press enter, the form does not not submit, so instead i actually have to click the submit button, which can be inconvenient. I am wondering how I can use a button the submit my from without having the page refresh and when I press enter, also have the form submit. I feel like it is a very easy fix that I just can not find. Here is the URL to the page on which the problem is occurring. http://www.pearlsquirrel.com/register.php
下面是code到register.php
Here is the code to register.php
<div id="centered3">
<h1><div align='center' style="color:white;">Join PearlSquirrel</align></h1>
<h4><div id='response'></div></h4>
<script language='javascript' type='text/javascript'>
function ajaxregister(){
var ajaxRequest;
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject('Msxml2.XMLHTTP');
} catch (e) {
try{
ajaxRequest = new ActiveXObject('Microsoft.XMLHTTP');
} catch (e){
// Something went wrong
alert('Your browser broke!');
return false;
}
}
}
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('response');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
var username = document.getElementById('username').value;
var password = document.getElementById('password').value;
var repeatpassword = document.getElementById('repeatpassword').value;
var queryString = '?username=' + username + '&password=' + password + '&repeatpassword=' + repeatpassword;
ajaxRequest.open('GET', 'ajaxregister.php' + queryString, true);
ajaxRequest.send(null);
}
</script>
<form method='GET'>
<table>
<tr>
<td>
Choose a username:
</td>
<td>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.js"></script>
<script>
$(document).ready(function(){
$('#username').keyup(username_check);
});
function username_check(){
var username = $('#username').val();
if(username == "" || username.length < 4){
$('#username').css('border', '3px #CCC solid');
$('#tick').hide();
}else{
jQuery.ajax({
type: "POST",
url: "check.php",
data: 'username='+ username,
cache: false,
success: function(response){
if(response == 1){
$('#username').css('border', '3px #C33 solid');
$('#tick').hide();
$('#cross').fadeIn();
}else{
$('#username').css('border', '3px #090 solid');
$('#cross').hide();
$('#tick').fadeIn();
}
}
});
}
}
</script>
</head>
<script>
function nospaces(evt)
{
var charCode = (evt.which) ? evt.which : event.keyCode
if (charCode == 32)
return false;
return true;
}
</script><body>
<input name="username" id="username" type="text" style="height:30px; width:212px;" value='<?php echo $username; ?>' onkeypress="return nospaces(event)"/>
<img id="tick" src="tick.png" width="16" height="16"/>
<img id="cross" src="cross.png" width="16" height="16"/>
</body>
</html>
</td>
</tr>
<tr>
<td>
Choose a password:
</td>
<td>
<input type='password' name='password' id='password' style="height: 30px; width:220px;" >
</td>
</tr>
<tr>
<td>
Repeat your password:
</td>
<td>
<input type='password' name='repeatpassword' id='repeatpassword' style="height: 30px; width:220px;" >
</td>
</tr>
<tr>
<td>
</td>
<td>
By registering, you agree to the<br><a href="termsofservice.php">terms of service</a> & <a href="privacypolicy.php">privacy policy</a>
</td>
</tr>
<tr>
<td>
</td>
<td>
<input type='button' name='submit' value='Register' class='registerbutton' onclick='ajaxregister()'>
<div id="msg"></div>
</td>
</tr>
</table>
<p>
有code为页面的一部分块。任何帮助将大大AP preciated。谢谢！
There is the block of code for that part of the page. Any help would be greatly appreciated. Thanks!
推荐答案
您得添加安其preSS处理程序输入字段，检查重点code：如果是13（回车），执行Submit按钮处理程序
You gotta add onKeyPress handlers for input fields and check the keyCode: if it is 13 (Enter), execute Submit button handler.
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