2017-03-06 62 views
0
新的一页

我有一个自举表看起来像这样:内容下载到

<table style="width:90%" align="center" class="table table-bordered" id="example_table" data-toggle="table" 
     data-url="example" data-search="true" data-show-columns="true"> 
    <thead> 
    <tr> 
     <th rowspan="3" data-field="name">Name</th> 
     <th colspan="9" id="title">Title</th> 
    </tr> 
    <tr class="table_headers"> 
     <th colspan="3">Header 1</th> 
     <th colspan="3">Header 2</th> 
     <th rowspan="2" style="vertical-align: middle" data-field="header3">Header 3</th> 
     <th rowspan="2" style="vertical-align: middle" data-field="header4">Header 4</th> 
     <th rowspan="2" style="vertical-align: middle" data-field="header5">Header 5</th> 
    </tr> 
    <tr class="table_headers"> 
     <th data-field="sub1">Sub 1</th> 
     <th data-field="sub2">Sub 2</th> 
     <th data-field="sub3">Sub 3</th> 
     <th data-field="sub4">Sub 4</th> 
     <th data-field="sub5">Sub 5</th> 
     <th data-field="sub6">Sub 6</th> 
    </tr> 
    </thead> 
</table> 

我希望用户能够下载这个CSV格式的表格。我还想在“名称”中指定另一个页面的链接。我该如何去做呢?

回答