2017-08-27 79 views
1

所以,我相信答案可能會相當簡單。儘管如此,我一直試圖弄清楚這個問題已經一個多星期了,而且沒有運氣。我正在嘗試爲桌面/操作系統模擬器遊戲創建一個FileSystem。Java + XML |從同名節點的深層樹中訪問特定節點

目前,文件系統看起來是這樣的:(精簡略)

<system> 
    <info> <!-- Default information/data for this system --> 
     <top dir="/"/> 
    </info> 

    <files> <!-- File and Directory structure for this system --> 
     <dir name="home" owner="0" group="0" ownerp="rwx" groupp="r-x" otherp="r-x"> 
      <dir name="cyanite" owner="1000" group="1000" ownerp="rwx" groupp="rwx" otherp="r-x"> 
       <dir name="folder" owner="1000" group="1000" ownerp="rwx" groupp="rwx" otherp="r-x"> 
        <dir name="folder" owner="1000" group="1000" ownerp="rwx" groupp="rwx" otherp="r-x"> 
         <dir name="folder" owner="1000" group="1000" ownerp="rwx" groupp="rwx" otherp="r-x"> 
         </dir> 
        </dir> 
       </dir> 
      </dir> 
     </dir> 
     <dir name="bin" owner="0" group="0" ownerp="rwx" groupp="r-x" otherp="r-x"> 
     </dir> 
     <dir name="etc" owner="0" group="0" ownerp="rwx" groupp="r-x" otherp="r-x"> 

     </dir> 
     <dir name="usr" owner="0" group="0" ownerp="rwx" groupp="r-x" otherp="r-x"> 
     </dir> 
    </files> 
</system> 

這裏是我馬虎WIP代碼嘗試從文件中讀取:

System.out.println(fileNavigator.getDirectory("home")); 

public ArrayList<String> getDirectory(String path) { 
    Document doc = ExitParser.getDocFromZip(system); // This loads the save file and returns the doc 
    NodeList dataList = doc.getElementsByTagName("files"); 
    NodeList dataList2 = parsePath(dataList, path); 
    ArrayList<String> stringList = new ArrayList<String>(); 
    try { 
     for(int i=0; i < dataList2.getLength(); i++) { 
      Node dataItem = dataList2.item(i); 
      Element elementDataItem = (Element)dataItem; 
      stringList.add(elementDataItem.getAttribute("name")); 
     } 
    } catch (NullPointerException e) { 
     stringList.add("NullPointerException"); 
    } 
    return stringList; 
} 

public NodeList parsePath(NodeList nodelist, String path) { 
    Document doc = ExitParser.getDocFromZip(system); // This loads the save file and returns the doc 
    NodeList dataList = doc.getElementsByTagName("info"); 
    if (path.startsWith(ExitParser.getAttributeValue(dataList, "top", "dir"))) { 
     // This simply grabs "top" from the FileSystem <info> area and strips it. 
     path.replaceFirst(Pattern.quote(ExitParser.getAttributeValue(dataList, "top", "dir")), ""); 
    } 
    List<String> pathList = new ArrayList<String>(); 
    if (path.contains("/")) { 
     pathList = new ArrayList<String>(Arrays.asList(path.split("/"))); 
    } else if (path.contains("\\")) { 
     pathList = new ArrayList<String>(Arrays.asList(path.split("\\"))); 
    } else { 
     pathList.add(path); 
    } 


    for (String string : pathList) { 
     System.out.println(string); 
     nodelist = ExitParser.getNextSetAttr(nodelist, string); 
    } 


    return nodelist; 
} 

public static NodeList getNextSetAttr(NodeList data, String attrName) { 
    try { 
     for(int i=0; i < data.getLength(); i++){ 
      Element dataElement = (Element)data.item(i); 
      NodeList nodeList = dataElement.getElementsByTagName("dir"); 
      Element nodeElement = (Element)nodeList.item(0); 
      if(nodeElement.hasAttribute("name")){ 
       if (nodeElement.getAttribute("name").contains(attrName)) { 
        System.out.println(true); 
        return nodeList; 
       } 
      } 
     } 
    } 
    catch(Exception ex) { 
     System.out.println(ex.getMessage()); 
    } 
    return null; 
} 

使用上面的代碼,我預計輸出爲[cyanite],但我得到[home, cyanite, folder, folder, folder, bin, etc, usr]

如果我將System.out.println(fileNavigator.getDirectory("home"));更改爲System.out.println(fileNavigator.getDirectory("home/cyanite"));,那麼我會得到[cyanite]。如果我將"home/cyanite"更改爲"bin",我會得到[NullPointerException]

對於[home, cyanite, folder, folder, folder, bin, etc, usr]這個問題似乎是父節點被添加到列表(即home,bin等,usr)以及所有子節點(即:文件夾,文件夾,文件夾)的問題。在我的使用案例中,我只希望獲取當前節點的子節點,而不是節點的子節點。

雖然我不確定爲什麼我不能訪問home以外的其他文件夾。

回答

0

該問題是由於代碼一次接收所有同名節點而造成的。但是,有人可以通過做類似於this answer(StackOverflow |按名稱僅獲取XML的直接子元素)中建議的內容來獲得直接子項。在那裏,他們利用.getFirstChild().getNextSibling()以僅循環通過當前級別。