业务需求是这样的：根据前面的列合并后面的列
代码：
<body onl oad="fixRowspan()">
<display:table name="dataList" id="tableList" cellspacing="0" cellpadding="0" requestURI="/x/xx_xxx.action" >
<display:column media="html" title="<input type='checkbox' class='chk' name='selChkbox' >" style="width:40px"/>
<display:column property="invoiceNo" title="比较的列" style="color:red"/>
<display:column property="weight" title="不重要的列1"/>
<display:column property="makeHours" title="不重要的列2"/>
<display:column property="factMoney" title="合并列1"/>
<display:column media="html" title="合并列2" style="width:6%">
<s:if test='%{#attr.bid != ""}'>
<s:a ><s:property value="#i18n_print"/></s:a>
</s:if>
</display:column>
</display:table>
</body>
js代码：
<script>
function fixRowspan() {
var tb = document.getElementById("tableList");
var contrastIndex = 1;//对比的列序号
var mergeIndexs = [4,5];//需要合并的列序号，注意：序号必须按从小到大放置 @1
var row_span_num = 1;
var first_row_title = "";
var first_row_obj_list = [];
for ( var i = 1; i < tb.rows.length; i++) {
var first_new_row_title = tb.rows[i].cells[contrastIndex].innerHTML;
if (first_row_title != "" && first_row_title == first_new_row_title) {
for(var j = 0; j < mergeIndexs.length; j++){
tb.rows[i].deleteCell(mergeIndexs[j] - j);//删除 @2
}
row_span_num++;
} else {
if (first_row_title != "") {
for(var j = 0; j < mergeIndexs.length; j++){
first_row_obj_list[j].setAttribute("rowSpan", row_span_num);
}
row_span_num = 1;
}
for(var j = 0; j < mergeIndexs.length; j++){
first_row_obj_list[j] = tb.rows[i].cells[mergeIndexs[j]];
}
first_row_title = first_new_row_title;
}
}
}
</script>
要注意的点：
在删除多余的cell时，因为row中的cell是按照下标来的，如果删除一个cell，那么后面的cell的下标就会向前移，所以在@2时需要实时把下标 - j ,而@1处则必须按顺序放，这样才不会出现错乱的情况；
当然，我们也可以按从大到小排，这样在@2处就不需要 - j 了
标签：某列,title,mergeIndexs,jsp,var,table,row,tb,first From： https://www.cnblogs.com/wdk2020/p/17346119.html