从JS代码提交表单
从JS代码提交表单
提问于 2015-04-25 22:52:23
我有一张表格：
@model project.Models.ChannelAndLocationViewModel
@{
const string formId = "parentForm";
}
@using (Html.BeginForm("EditChannel", "Channel", FormMethod.Post, new { id = formId }))
{
Html.RenderPartial("EditChannelForm", Model);
<br /><br />
<!--<input type="submit" value="Save"/>-->
<input type="button" value="Save" onclick="editChannel('@Model.ChannelViewModel.ID', @formId)" />
}我试图在JS代码中提交这个表单，如下所示：
function editChannel(channelId, parentForm) {
$(parentForm).submit();
$.ajax({
url: '/Channel/EditChannel/',
type: "POST",
cache: false,
data: {
id: channelId
},
success: function (msg) {
alert("Msg: " + msg);
if (msg === "ChangeOfSensitiveData") {
showAlertOnChangeOfSensitiveData('sensitivDataMsgDiv');
} else {
alert("Else");
}
},
error: function (msg) {
showDialog('errorDiv');
}
});
}但似乎表单从未提交过，因为模型没有被解析给控制器。调用Action方法后，模型为null。行动看起来是这样的：
[HttpPost]
public ActionResult EditChannel(int id, ChannelAndLocationViewModel updatedModel)
{ ...}如何从JS文件中提交表单？
回答 2
Stack Overflow用户
回答已采纳
发布于 2015-04-25 22:54:25
这样做：
function editChannel(channelId, parentForm) {
$("#formId").submit(function(event) {
event.preventDefault();
$.ajax({
url: '/Channel/EditChannel/',
type: "POST",
cache: false,
data: {
id: channelId,
updatedModel: parentForm
},
success: function (msg) {
alert("Msg: " + msg);
if (msg === "ChangeOfSensitiveData") {
showAlertOnChangeOfSensitiveData('sensitivDataMsgDiv');
} else {
alert("Else");
}
},
error: function (msg) {
showDialog('errorDiv');
}
});
});
}Stack Overflow用户
发布于 2015-04-25 22:56:45
模型是空的，因为它还没有被添加到数据中。尝试：
data: {
id: channelId,
updatedModel : { ...... }
},页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/29871615
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