java - 解析 xml 文档的更优雅的方式

Question

目前我递归地解析我的 xml 文档。我询问是否有任何子节点，如果有，它将迭代子节点列表，并检查这些子节点是否有子节点。我的基本情况是它不再有 child 了。现在解析是正确的，我得到了一个按正确顺序排列的大节点列表。

我想做的是让每个节点成为它自己的对象。因此实体将是一个包含索引对象、列对象等的对象。我的第一种思维方式是迭代一个循环，并使用一个 switch 语句检查它命中的每个节点并创建将添加到实体的对象。我认为大的 switch 语句是不正确的。

这里是 xml 的示例:

<ENTITY_DEFS>
    <ENTITY name="WORKORDER">
        <INDEXES>           
            <INDEX name="IDX_WO_1">WORKORDERID</INDEX>
            <INDEX name="IDX_WO_2">WONUM, SITEID</INDEX>
            <INDEX name="IDX_WO_3">ASSETNUM, SITEID</INDEX>
            <INDEX name="IDX_WO_4">LOCATION, SITEID</INDEX>
        </INDEXES>      
        <COLUMNS>
            <COLUMN autokey="Y" autokeyprefix="LOCAL" parentkeycol="PARENTID" uniquekey="Y">WORKORDERID</COLUMN>
            <COLUMN displaykey="Y" autokey="Y" autokeyprefix="OFFLN">WONUM</COLUMN>
            <COLUMN>PARENTID</COLUMN>
            <COLUMN>PARENT</COLUMN>
            <COLUMN>DESCRIPTION</COLUMN>
            <COLUMN>LONGDESCRIPTION</COLUMN>
            <COLUMN>LOCATION</COLUMN>
            <COLUMN>ASSETNUM</COLUMN>
            <COLUMN>PERSONGROUP</COLUMN>
            <COLUMN>CREWID</COLUMN>
            <COLUMN>SUPERVISOR</COLUMN>
            <COLUMN>LEAD</COLUMN>
            <COLUMN>PHONE</COLUMN>
            <COLUMN>REPORTEDBY</COLUMN>
            <COLUMN>REPORTDATE</COLUMN>
            <COLUMN>STATUS</COLUMN>
            <COLUMN>STATUSDATE</COLUMN>
            <COLUMN>WOPRIORITY</COLUMN>
            <COLUMN>WORKTYPE</COLUMN>
            <COLUMN>FAILURECODE</COLUMN>
            <COLUMN>PROBLEMCODE</COLUMN>
            <COLUMN>ISTASK</COLUMN>
            <COLUMN>ORIGRECORDID</COLUMN>
            <COLUMN>ESTDUR</COLUMN>
            <COLUMN>OBSERVATION</COLUMN>
            <COLUMN>MEASUREMENTVALUE</COLUMN>
            <COLUMN>TASKID</COLUMN>
            <COLUMN>SITEID</COLUMN>
            <COLUMN>ORGID</COLUMN>
        </COLUMNS>
        <INIT_ACTION>
            <ACTION>offline/getMyWorkOrderListJson.action</ACTION>
            <ACTION>offline/getMyTaskWorkOrderListJson.action</ACTION>
        </INIT_ACTION>
        <DELTA_ACTION>
            <ACTION>offline/getDeltaMyWorkOrderListJson.action</ACTION>
            <ACTION>offline/getDeltaMyTaskWorkOrderListJson.action</ACTION>
        </DELTA_ACTION>
        <SYNC_ACTIONS transentityname="WORKORDER" transentitykey="WORKORDERID">
            <ACTION name="INSERT" beanname="OfflineDataSyncAction" methodname="addWorkOrder"/>
            <ACTION name="EDIT" beanname="OfflineDataSyncAction" methodname="editWorkOrder"/>
            <ACTION name="INSERT_TASK" beanname="OfflineDataSyncAction" methodname="addTaskWorkOrder"/>
            <ACTION name="UPDATE_STATUS" beanname="OfflineDataSyncAction" methodname="updateWorkOrderStatus"/>
            <ACTION name="UPDATE_TASK" beanname="OfflineDataSyncAction" methodname="updateWorkOrderTask"/>
            <ACTION name="UPDATE_TASK_STATUS" beanname="OfflineDataSyncAction" methodname="updateWorkOrderStatus"/>
            <ACTION name="EMM_FILE_UPLOAD"/>
        </SYNC_ACTIONS>
    </ENTITY>

我的递归语句:

public void parse(final Document doc,final List<Node> nodelist, final Element el)
{
    final NodeList children = el.getChildNodes();
    for(int i = 0; i < children.getLength(); i++)
    {
        final Node node = children.item(i);

        if(node.getNodeType() == Node.ELEMENT_NODE)
        {
            nodelist.add(node);
            parse(doc,nodelist,(Element)node);
        }
    }       
}

Answer 1

更优雅的方法是仅传递节点，因为您的 nodelist 是通过单个对象遍历的。这样，您就可以避免冗余参数。

此外，您可以将 el 设置为要捕获的 nodename 的静态字符串:

public static void parse(Node node) {
        NodeList children = node.getChildNodes();
        for (int i = 0; i < children.getLength(); i++) {
            Node currentNode = children.item(i);
            if (node.getNodeType() == Node.ELEMENT_NODE)
                parse(currentNode);

        }

        if (node.getNodeName().equals(EL_NAME)) {
            System.out.println("This -> " + node.getTextContent());
        }

    }