在Ajax發佈請求之後調用Submit Action時。它返回值爲「Bad Request」的錯誤方法。爲什麼它不成功?如何解決這個問題?Http發佈請求失敗後返回部分視圖
Ajax請求:
<p>
@using (Html.BeginForm("Submit", "Test"))
{
<span>
Price: @Html.TextBox("Price", string.Empty)
</span>
<input id="submit" type="submit" value="test" />
}
</p>
<div id="testPost">
</div>
@section Scripts {
<script type="text/javascript">
$(function() {
$('#submit').on("click", function() {
var form = $(this).parent("form");
$.ajax({
type: "POST",
url: form.attr('action'),
data: form.serialize()
})
.success(function (html) {
$('#testPost').replaceWith(html);
alert("Success");
})
.error(function() {
alert("Bad Request");
});
return false;
});
});
</script>
}
提交行動:
[HttpPost]
public ActionResult Submit(Auction auction)
{
return PartialView("_CurrentPrice",auction);
}
管窺:
<p class="current-price">
<h2>ACTIVE HERE</h2>
</p>
檢查你的預期和發送值,記住'form.serialize()'返回一個字符串 – Sarath