基于键将XML节点合并为一个节点

我有一个XML文件,其中包含多个具有子节点的节点。如果节点具有相同的键,则必须将其子节点合并到一个节点中。

<availexport>
  <date>090423121513</date>
  <employee emp_cd="9120004">
    <day date="20050406" start="10" end="20"/>
    <day date="20050406" start="21" end="23"/>
    <day date="20050511" start="12" end="23"/>
  </employee>
  <records>3</records>
</availexport>

这里的两个记录有相同的日期。如何将它们合并为一个?我有一个作为键/值对的列表。

for (int k = this.employeesList.size(); k > 0; k--) {
    Map empInfo1 = new HashMap();
    Map empInfo = (Map) this.employeesList.remove(0);
    this.empID = (Long) empInfo.get(EMP_ID);
    this.hrID = (String) empInfo.get(HR_EMP_ID);
    this.avail_iDate = (Long) empInfo.get(AVAIL_IDATE);
    this.start_ITime = (Long) empInfo.get(START_ITIME);
    this.end_ITime = (Long) empInfo.get(END_ITIME);
    List availList = new ArrayList();
    Map availList1 = new HashMap();
    if (empAvailRecord.containsKey(empID)) {
        empInfo1 = (Map) empAvailRecord.get(empID);
        availList = (List) empInfo1.get("DAY");
        availList1.put(AVAIL_IDATE, avail_iDate);
        availList1.put(START_ITIME, start_ITime);
        availList1.put(END_ITIME, end_ITime);
        availList.add(availList1);
    } else {

        availList1.put(AVAIL_IDATE, avail_iDate);
        availList1.put(START_ITIME, start_ITime);
        availList1.put(END_ITIME, end_ITime);
        availList.add(availList1);

        empInfo1.put("HR_ID", hrID);
        empInfo1.put("DAY", availList);
        empAvailRecord.put(empID, empInfo1);
    }

}