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我想验证窗体使用JavaScript。一旦下面的代码执行,我想从外部脚本获取PHP运行没有页面加载与成功消息谁能帮助后?javascript ajax,表单提交然后页面加载
下面的脚本工作正常,并从一个单独的函数拉notEmpty。我需要的脚本然后: 1.如果没有产生错误,然后拉动php并从列出的每个值传输数据。 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;
这听起来像你只需要一个入门教程阿贾克斯。 – Quentin 2012-07-10 11:33:32
即时通讯不是很好,所以任何帮助或指引我在正确的方向将是惊人的 – phpdev1982 2012-07-10 11:38:44
https://developer.mozilla.org/en/AJAX – Quentin 2012-07-10 11:40:18