需求合併行，在上一篇的基礎上改進。

解決思路

1、獲取前排列的rowspan值，存入陣列，

2、再迴圈取出陣列，根據rowspan的值合併

3、關鍵點用到jQuery的$("#xxx tbody td:nth-child(xx)")方法獲取指定列，在迴圈遍歷處理

程式碼：

   /**
     * 依據前面的列內容合併後面的列內容
     * @param table_id tableID
     * @param table_colnum 待合併的列
     * @param table_colnum_decide 依據的列
     */
function servermergetable_rowspan
 
(table_id, table_colnum, table_colnum_decide) {
//    查詢依據列是否有相同待合併的
var table_firsttd_decide = "";
        var table_currenttd_decide = "";
        var table_SpanNum_decide = 0;
        var array = [];
        var table_Obj_decide = $(table_id + " tbody td:nth-child(" + table_colnum_decide + ")");
        table_Obj_decide
 
.each(function (i) {
            if (i == 0) {
                table_firsttd_decide = $(this);
                table_SpanNum_decide = 1;
            } else {
                table_currenttd_decide = $(this);
                if (table_firsttd_decide.text() == table_currenttd_decide.text()) {
                    table_SpanNum_decide
 
++;
                } else {
                    table_firsttd_decide = $(this);
                    table_SpanNum_decide = 1;
                }
            }
            //將依據列td的rowspan屬性加入陣列
array.push(table_SpanNum_decide);
        });
        var table_firsttd = "";
        var table_currenttd = "";
        var table_SpanNum = 0;
        var table_Obj = $(table_id + " tbody td:nth-child(" + table_colnum + ")");
        table_Obj.each(function (i) {
            if (i == 0) {
                table_firsttd = $(this);
                table_SpanNum = array[i];
            } else {
                table_currenttd = $(this);
                table_SpanNum = array[i];
                if (table_SpanNum == 1) {//rowspan值為1的不合並
table_firsttd = $(this);
                } else {//否則合併
table_firsttd.attr("rowSpan", table_SpanNum);
                    table_currenttd.remove();
                }
            }

        });
    }