我正在使用javascript,jquery,ajax
如何创建此加载微调器?在以下URL
上看到了这一点我在下面编写了以下代码,但我不确定如何在成功功能中添加代码,它们有超时功能,但它看起来与我的代码不同。
$('#content').html('<img id="loader-img" alt="" src="http://adrian-design.com/images/loading.gif" width="100" height="100" align="center" />');
$.ajax({
type: "POST",
url: "Checkout_Payment.aspx/ChargeCreditCard",
data: '{NameVB: "' + NameJS + '"}',
contentType: "application/json; charset=utf-8",
dataType: "json",
success: OnSuccess,
failure: OnFailure,
error: OnError
});
function OnSuccess(response) {
var getResponse = response.d;
if (getResponse.indexOf('success') >= 0) {
var order_ID_Array = getResponse.split("_");
var order_ID = order_ID_Array[1];
window.location.href = "Checkout_Receipt.aspx?OID=" + order_ID ;
}
}
答案 0 :(得分:1)
你几乎就在那里,你需要hide
ajax success
之后的图像
$('#content').html('<img id="loader-img" alt="" src="http://adrian-design.com/images/loading.gif" width="100" height="100" align="center" />');
$.ajax({
type: "POST",
url: "Checkout_Payment.aspx/ChargeCreditCard",
data: '{NameVB: "' + NameJS + '"}',
contentType: "application/json; charset=utf-8",
dataType: "json",
success: OnSuccess,
failure: OnFailure,
error: OnError
});
function OnSuccess(response) {
$("#loader-img").hide(); //REMOVES THE LOADING UPON RECEIVING RESPONSE
var getResponse = response.d;
if (getResponse.indexOf('success') >= 0) {
var order_ID_Array = getResponse.split("_");
var order_ID = order_ID_Array[1];
window.location.href = "Checkout_Receipt.aspx?OID=" + order_ID ;
}
}