我正在尝试使用Ajax提交表单。但是,当我提交页面时,页面会重新加载,并且网址也会更改。我认为网址是由于 @ Html.AntiForgeryToken();
请参阅下面的代码:
这是我的表单的样子:
@model PersonModel
....
<form action="SubmitLead" class="new-lead">
@Html.AntiForgeryToken();
<div class="col-md-12">
<p>
<input type="hidden" value="@Model.TrackingCode" id="hdnTrackingCode" />
My name is @Html.TextBoxFor(model => model.FirstName,
new { @placeholder =
Html.DisplayNameFor(model => model.FirstName) })
@Html.TextBoxFor(model => model.Surname, new { @placeholder = Html.DisplayNameFor(model => model.Surname) })
</p>
</div>
<div class="clearfix"></div>
<div class="col-md-12 text-center">
<button type="submit" id="btnSubmit" class="orange-button">Get Quotes Now</button>
</div>
</form>
@if (Model.Results != null &&
Model.Results.IsSuccessful)
{
<div class="col-md-12 text-center">
<img src="~/Content/Images/Products/new-success.png" height="24px" />
<p id="result"></p>
</div>
}
请在这里查看我的脚本:
@section Scripts{
<script type="text/javascript">
$(document).ready(function () {
$('.new-lead').submit(function (event) {
$.ajax({
url: '@Url.Action("Lead/SubmitLead")',
type: 'POST',
data: $(this).serialize(),
dataType: 'json',
success: function (result) {
var resultMessage = "success";
$('result').html(resultMessage);
}
})
})
})
</script>
答案 0 :(得分:2)
这样做
<form onsubmit="return submit(thi)" class="new-lead">
....
</form>
<script>
function submit(e){
$.ajax({
url: '@Url.Action("Lead/SubmitLead")',
type: 'POST',
data: $(e).serialize(),
dataType: 'json',
success: function (result) {
var resultMessage = "success";
$('result').html(resultMessage);
}
})
return false;
}
</script>