javascript ajax,表单提交然后页面加载

javascript ajax, form submission then page load

本文关键字:加载 然后 表单提交 ajax javascript      更新时间:2023-09-26

我正在尝试使用javascript验证表单。一旦执行了下面的代码,我想从外部脚本中获取php,以便在没有页面加载的情况下运行,并在之后显示成功消息。有人能帮忙吗?

下面的脚本运行良好,并从一个单独的函数中提取notEmpty。我需要脚本,然后:1.如果没有产生错误,则提取php并从列出的每个值中传输数据。2.显示成功消息

var year = document.getElementById('year');
var period = document.getElementById('period');
var live = document.getElementById('live');
var start = document.getElementById('start');
var todate = document.getElementById('todate');
var sdeadline = document.getElementById('sdeadline');
var cdeadline = document.getElementById('cdeadline');
var circ = document.getElementById('circ');
var line = document.getElementById('line');
// Check each input in the order that it appears in the form!
if(notEmpty(year, "Please fill in Year")){
if(notEmpty(period, "Please fill in Period")){
if(notEmpty(live, "Please fill in live Date")){
if(notEmpty(start, "Please fill in Start Date")){
if(notEmpty(todate, "Please fill in End Date")){
if(notEmpty(sdeadline, "Please fill in Supplier Deadline")){
if(notEmpty(cdeadline, "Please fill in Commerical Deadline")){
if(notEmpty(circ, "Please fill in Circulars Due")){
if(notEmpty(line, "Please fill in Line Listing Downloads")){
}}}}}}}}}

return false;

ideea是用要传输到php脚本的数据创建一个字符串:

var datasend = 'year='+year+'&period='+period+'&live='+live+ ...;

然后,如果没有错误,调用一个Ajax函数,该函数通过Post将字符串发送到服务器。

请参阅关于以下内容的教程:AJAX with POST and PHP