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jquery - 如何从表中获取日期 td 值

转载 作者:太空宇宙 更新时间:2023-11-04 02:42:42 25 4
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我正在尝试从表中获取日期

表格中的所有日期都相同

我正在尝试获取值“2015-12-21”

这是我的表格数据

<table id="1dtopg" class="gaintable table table-striped">
<thead>
<tr>
<th class="thheaders">Symbol</th>
<th class="thheaders">Close</th>
<th class="thheaders">Rs Chg</th>
<th class="thheaders">% Chg</th>
<th class="thheaders">As On</th>
</tr>
</thead>
<tbody>
<tr>
<td>NILKAMAL</td>
<td>1243.30</td>
<td class="greenclass">172.45</td>
<td class="greenclass">13.87</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>LAXMIMACH</td>
<td>3648.90</td>
<td class="greenclass">148.95</td>
<td class="greenclass">4.08</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>TTKHLTCARE</td>
<td>1076.30</td>
<td class="greenclass">116.65</td>
<td class="greenclass">10.84</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>MONTECARLO</td>
<td>536.45</td>
<td class="greenclass">78.55</td>
<td class="greenclass">14.64</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>TRENT</td>
<td>1647.60</td>
<td class="greenclass">74.80</td>
<td class="greenclass">4.54</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>BOSCHLTD</td>
<td>18287.40</td>
<td class="greenclass">71.85</td>
<td class="greenclass">0.39</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>WABCOINDIA</td>
<td>6025.10</td>
<td class="greenclass">58.90</td>
<td class="greenclass">0.98</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>BBL</td>
<td>852.40</td>
<td class="greenclass">58.65</td>
<td class="greenclass">6.88</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>BAJFINANCE</td>
<td>5869.80</td>
<td class="greenclass">56.00</td>
<td class="greenclass">0.95</td>
<td class="">2015-12-21</td>
</tr>
<tr>
<td>HEROMOTOCO</td>
<td>2697.95</td>
<td class="greenclass">54.80</td>
<td class="greenclass">2.03</td>
<td class="">2015-12-21</td>
</tr>
</tbody>
</table>

我试过

var date = $('#1dtopg').find('tbody tr td:eq(5)').html();

alert(date);

但它给了我象征值(value)

这是我的 jsfiddle

https://jsfiddle.net/gr1L23us/27/

最佳答案

你可以使用:last选择器

var date = $('#1dtopg').find('tbody tr td:last').html();

.last()

var date = $('#1dtopg').find('tbody tr td').last().html();

或者您可以使用 :nth-child(starts from 1) 而不是 :eq(starts from 0)

var date = $('#1dtopg').find('tbody tr td:nth-child(5)').html();

关于jquery - 如何从表中获取日期 td 值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34414448/

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