我有一个观点
@using staffInfoDetails.Models
@model staffInfo
<link href="../../Content/myOwn.css" rel="stylesheet" type="text/css" />
@{staffInfo stf = Model;
}
<div id="education1">
@using (Html.BeginForm("addNewEdu","Home",FormMethod.Post))
{
@Html.HiddenFor(x=>x.StaffId)
<table>
<tr>
<th>Country</th>
<th>Board</th>
<th>Level</th>
<th>PassedYear</th>
<th>Division</th>
</tr>
<tr>
@Html.EditorFor(x => x.eduList)
</tr>
<tr>
@*<td><input type="submit" value="create Another" id="addedu"/> </td>*@
@*<td>@Html.ActionLink("Add New", "addNewEdu", new { Model })</td>*@
</tr>
</table>
}
<button id="addedu">Add Another</button>
</div>
我想使用jquery将模型staffInfo传递给控制器,如下所示
<script type="text/javascript">
$(document).ready(function () {
$("#addedu").live('click', function (e) {
// e.preventDefault();
$.ajax({
url: "Home/addNewEdu",
type: "Post",
data: { model: stf },//pass model
success: function (fk) {
// alert("value passed");
$("#education").html(fk);
}
});
});
});
</script>
jquery似乎只传递元素而不是整个模型,所以如何将模型从视图传递给控制器,这样我就不必在jquery中编写整个参数列表
答案 0 :(得分:5)
您可以使用此
为表单提供ID@using (Html.BeginForm("addNewEdu", "Home", FormMethod.Post, new { id = "addNewEduForm" }))
{
}
然后在剧本中
<script type="text/javascript">
$('#addedu').click(function(e) {
e.preventDefault();
if (form.valid()) { //if you use validation
$.ajax({
url: form.attr('action'),
type: form.attr('method'),
data: $("#addNewEduForm").serialize(),
success: function(data) {
}
});
}
});
</script>
答案 1 :(得分:2)
我可以看到您尝试使用AJAX提交表单?查看serialize函数。
$('#addedu').click(function(e) {
e.preventDefault();
var form = $('form');
if (form.valid()) { //if you use validation
$.ajax({
url: form.attr('action'),
type: form.attr('method'),
data: form.serialize(),
success: function(r) {
}
});
}
});
答案 2 :(得分:0)
您可以使用getElementById
获取您的值,然后发送您的操作:
$(document).ready(function () {
var select = document.getElementById('...');
$.ajax({
type: "get",
url: "get_street_name",
data: { a: select },
success: function (data) {
$('#result').html(data);
}
});
}