我已经编写了这个javascript来获取表格的ID然后循环遍历tr,然后是td' s。我不是起诉写什么逻辑来获取td中下拉列表的选定值。不是所有的tds,都有一个下拉列表。
这是我的javascript
function submitPanel(value) {
$('#' + value + '> tbody > tr').each(function () {
alert($(this).html());
$(this).find('td').each(function () {
alert($(this).html());
})
});
}
哪个输出:
该表格采用MVC 4剃须刀
@model IMEModels.InterviewManagement.InterviewManagement
<hr />
@using (Html.BeginForm("SubmittedInterviews", "InterviewManagement", FormMethod.Post))
{
if (Model.InterviewSchedules.Count > 0)
{
<table>
<tr>
<td>@Html.Label("Show dates without Chair or Co-panelist") </td>
<td>@Html.RadioButton("Show dates without Chair or Co-panelist", new {Id = "rdoShow" })</td>
</tr>
</table>
for (int i = 0; i < Model.Centres.Count; i++)
{
@Html.Label(Model.Centres[i].CentreName)
for (int ii = 0; ii < Model.Centres[i].Locations.Count; ii++)
{
@Html.Label(Model.Centres[i].Locations[ii].LocationName)
for (int iii = 0; iii < Model.Centres[i].Locations[ii].InterviewDates.Count; iii++)
{
var ChairList = Model.Interviewers.Join(Model.DatePreferences, m => m.InterviewerId, d => d.InterviewersInterviewerId, (m, d) => new
{
Interviewer = m,
DatePreferences = d
})
.Where(d => d.DatePreferences.LocKey == Convert.ToString(Model.Centres[i].Locations[ii].LocationKey) && d.Interviewer.IsChair && d.DatePreferences.Date == Model.Centres[i].Locations[ii].InterviewDates[iii].Date)
.GroupBy(x => new { x.Interviewer.InterviewerId, x.Interviewer.Name })
.ToDictionary(a => a.Key.InterviewerId, b => b.Key.Name);
var NonChairList = Model.Interviewers.Join(Model.DatePreferences, m => m.InterviewerId, d => d.InterviewersInterviewerId, (m, d) => new
{
Interviewer = m,
DatePreferences = d
})
.Where(d => d.DatePreferences.LocKey == Convert.ToString(Model.Centres[i].Locations[ii].LocationKey) && d.DatePreferences.Date == Model.Centres[i].Locations[ii].InterviewDates[iii].Date)
.GroupBy(x => new { x.Interviewer.InterviewerId, x.Interviewer.Name })
.ToDictionary(a => a.Key.InterviewerId, b => b.Key.Name);
@:<div class="date-wrap @(ChairList.Count == 0 || NonChairList.Count == 0 ? "nochairspanelists" : "chairspanelists") >
if (NonChairList.Count == 0)
{
NonChairList.Add(new Guid(), "No panelists available.");
}
if (ChairList.Count == 0)
{
ChairList.Add(new Guid(), "No panelists available.");
}
@Html.Label(Model.Centres[i].Locations[ii].InterviewDates[iii].Date.ToLongDateString())
<table id="tbl@(Model.Centres[i].Code + "-" + Model.Centres[i].Locations[ii].LocationKey + "-" + Model.Centres[i].Locations[ii].InterviewDates[iii].Date.Ticks)" class="tblInterviewManager">
<tr>
<td>
Chair
</td>
<td>
Co-panelist
</td>
<td></td>
</tr>
<tr>
<td>
@Html.DropDownListFor(m => m.InterviewSchedules[iii].ChairId, new SelectList(ChairList, "Key", "Value"))
<br />
</td>
<td>
@Html.DropDownListFor(m => m.InterviewSchedules[iii].CofacilitatorId, new SelectList(NonChairList, "Key", "Value"))
</td>
@if (ChairList.ElementAt(0).Value == "No panelists available." || NonChairList.ElementAt(0).Value == "No panelists available.")
{
<td>
<input type="submit" value="Save panel" disabled="disabled" />
</td>
}
else
{
<td>
<input type="button" value="Save panel" id="btnSubmit" onclick="return submitPanel('tbl@(Model.Centres[i].Code + "-" + Model.Centres[i].Locations[ii].LocationKey + "-" + Model.Centres[i].Locations[ii].InterviewDates[iii].Date.Ticks)');"/>
</td>
}
</tr>
</table>
@:</div>
}
}
<br />
}
<div class="clear"></div>
<hr />
}
}
任何人都知道获得我想要的价值的好方法。
答案 0 :(得分:2)
使用Descendant Selector (“ancestor descendant”)直接获取selects
中的table
。
function submitPanel(value) {
$('#' + value + ' select').each(function () {
alert($(this).val());
});
}
选择作为给定祖先A后代的所有元素 元素的后代可以是孩子,孙子, 这个元素的曾孙,等等,jQuery docs。