Question

ajax提交form表单，到底怎么做

Answer 1

$.ajax({
cache: true,
type: "POST",
url:yopurAjaxUrl,
data:$('#yourformid').serialize(),// 这句很重要
async: false,
error: function(request) {
alert("Connection error");
},
success: function(data) {
...................
}
});

Answer 2

$.ajax({
cache: true,
type: "POST",
url:ajaxCallUrl,
data:$('#yourformid').serialize(),// 你的formid
async: false,
error: function(request) {
alert("Connection error");
},
success: function(data) {
$("#commonLayout_appcreshi").parent().html(data);\
}
});