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如何解析Android中的XML DOM解析相同的名称标签

本文介绍了如何解析Android中的XML DOM解析相同的名称标签的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我不能够解析我的XML here.It返回项目而已。
我AndroidActivity不能正常显示，因为它是非常大的，这就是为什么我只表示这是负责解析的部分。

我的XML是这样的：

 ＆LT; MyResource＆GT;
＆LT;项目＆gt;首先＆LT; /项目＆GT;
＆LT;项目＆gt;第二个＆LT; /项目＆GT;
＆LT; / MyResource＆GT;

我的ActivityClass方式：

 公共类ShowItems延伸活动{
    ListView的LV;
    ListAdapter适配器;
    静态最后弦乐KEY_RESOURCE =MyResource; //父节点
    静态最后弦乐KEY_ITEM =项目;
    ArrayList的＆LT; HashMap的＆LT;字符串，字符串＆GT;＆GT; mylist中=新的ArrayList＆LT; HashMap的＆LT;字符串，字符串＆GT;＆GT;（）;
    的String []从= {KEY_ITEM};
    INT []到= {R.id.mylist_item};
    @覆盖
    保护无效的onCreate（包savedInstanceState）{
        // TODO自动生成方法存根
        super.onCreate（savedInstanceState）;
        的setContentView（R.layout.showitems）;

        LV =（ListView控件）findViewById（R.id.lv_items）;
        parseXML（）;
        适配器=新SimpleAdapter（这一点，mylist中，R.layout.list_item，从，到）;
        lv.setAdapter（适配器）;
    }

私人无效parseXML（）{
    // TODO自动生成方法存根
    XMLParser的分析器=新XMLParser的（）;

    最终的字符串URL =htt​​p://10.0.2.2:8080/MySite/xml;
    XML字符串= parser.getXmlFromUrl（URL）;
    文档DOC = parser.getDomElement（XML）;
    NodeList的NL = doc.getElementsByTagName（KEY_RESOURCE）;
    的for（int i = 0; I＆LT; nl.getLength（）;我++）{

        HashMap的＆LT;字符串，字符串＆GT;图=新的HashMap＆LT;字符串，字符串＆GT;（）;
        元素e =（元）nl.item（我）;
        map.put（KEY_ITEM，parser.getValue（即KEY_ITEM））;
        mylist.add（图） }
}

我的XML解析器类：

 进口java.io.IOException异常;
进口java.io.StringReader中;
进口java.io.UnsupportedEncodingException;
进口javax.xml.parsers.DocumentBuilder中;
进口javax.xml.parsers.DocumentBuilderFactory中;
进口javax.xml.parsers.ParserConfigurationException;
进口org.apache.http.HttpEntity;
进口org.apache.http.Htt presponse;
进口org.apache.http.client.ClientProtocolException;
进口org.apache.http.client.HttpClient;
进口org.apache.http.client.methods.HttpGet;
进口org.apache.http.impl.client.DefaultHttpClient;
进口org.apache.http.util.EntityUtils;
进口org.w3c.dom.Document中;
进口org.w3c.dom.Element中;
进口org.w3c.dom.Node中;
进口org.w3c.dom.NodeList;
进口org.xml.sax.InputSource中;
进口org.xml.sax.SAXException;
进口android.util.Log;
公共XMLParser类{
    公共字符串getXmlFromUrl（字符串URL）{
        XML字符串= NULL;
        尝试 {
            HttpClient的HttpClient的=新DefaultHttpClient（）;
            HTTPGET HTTPGET =新HTTPGET（URL）;
            HTT presponse HTT presponse = httpClient.execute（HTTPGET）;
            HttpEntity httpEntity = HTT presponse.getEntity（）;
            XML = EntityUtils.toString（httpEntity）;
        }赶上（UnsupportedEncodingException E）{
            e.printStackTrace（）;
        }赶上（ClientProtocolException E）{
            e.printStackTrace（）;
        }赶上（IOException异常E）{
            e.printStackTrace（）;
        }
        返回XML;
    }

    公开文件getDomElement（XML字符串）{
        文档DOC = NULL;
        DocumentBuilderFactory的DBF = DocumentBuilderFactory.newInstance（）;
        尝试 {
            DocumentBuilder的DB = dbf.newDocumentBuilder（）;
            InputSource的是=新的InputSource（）;
                is.setCharacterStream（新StringReader（XML））;
                DOC = db.parse（是）;
            }赶上（的ParserConfigurationException E）{
                Log.e（错误，e.getMessage（））;
                返回null;
            }赶上（的SAXException E）{
                Log.e（错误，e.getMessage（））;
                返回null;
            }赶上（IOException异常E）{
                Log.e（错误，e.getMessage（））;
                返回null;
            }
                //返回DOM
            返回文档;
    }

    公共字符串的getValue（元素项，字符串str）{
        节点列表N = item.getElementsByTagName（STR）;
        返回this.getElementValue（n.item（0））;
    }
    公共最后弦乐getElementValue（节点ELEM）{
             子节点;
             如果（ELEM！= NULL）{
                 如果（elem.hasChildNodes（））{
                     对于（子= elem.getFirstChild（）;！孩子= NULL;孩子= child.getNextSibling（））{
                         如果（child.getNodeType（）== Node.TEXT_NODE）{
                             返回child.getNodeValue（）;
                         }
                     }
                 }
             }
             返回 ;
      }
}

我不能够解析我的XML here.What是这里的问题？
它返回项目而已。
什么我需要做的，我ActivityClass特别是在这部分的code：

  XML字符串= parser.getXmlFromUrl（URL）;
    文档DOC = parser.getDomElement（XML）;
    NodeList的NL = doc.getElementsByTagName（KEY_RESOURCE）;
    的for（int i = 0; I＆LT; nl.getLength（）;我++）{


        HashMap的＆LT;字符串，字符串＆GT;图=新的HashMap＆LT;字符串，字符串＆GT;（）;
        元素e =（元）nl.item（我）;
        map.put（KEY_ITEM，parser.getValue（即KEY_ITEM））;
        mylist.add（图） }

解决方案

您的getValue（）方法获取 MyResource 元，从那里，你需要获得在 MyResource 所有项目，并做 getElementValue（）。例如：code是：

 公开地图的getValue（元素项，字符串str）{
        节点列表N = item.getElementsByTagName（STR）;
        的for（int i = 0; I＆LT; n.getLength（）;我++）{
            的System.out.println（getElementValue（n.item（我）））;
        }
        //这里将其存储在列表/图并返回字符串列表/图代替
        返回目录/ MapHere;
    }

I am not able to parse my XML here.It returns "Item" only.
My AndroidActivity cannot be shown as it is very big that's why i have only shown the part which is responsible for parsing.

My XML Looks like this :

<MyResource>
<Item>First</Item>
<Item>Second</Item>
</MyResource>

My ActivityClass method:

public class ShowItems extends Activity{
    ListView lv;
    ListAdapter adapter;
    static final String KEY_RESOURCE = "MyResource"; // parent node
    static final String KEY_ITEM = "Item";
    ArrayList<HashMap<String, String>> mylist = new ArrayList<HashMap<String, String>>();
    String[] from={KEY_ITEM };
    int[] to={R.id.mylist_item};
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        // TODO Auto-generated method stub
        super.onCreate(savedInstanceState);
        setContentView(R.layout.showitems);

        lv=(ListView) findViewById(R.id.lv_items);
        parseXML();
        adapter = new SimpleAdapter(this, mylist,R.layout.list_item,from , to);
        lv.setAdapter(adapter);
    }

private void parseXML() {
    // TODO Auto-generated method stub
    XMLParser parser = new XMLParser();

    final String URL="http://10.0.2.2:8080/MySite/xml";
    String xml = parser.getXmlFromUrl(URL);
    Document doc = parser.getDomElement(xml);
    NodeList nl = doc.getElementsByTagName(KEY_RESOURCE);
    for (int i = 0; i < nl.getLength(); i++) {

        HashMap<String, String> map = new HashMap<String, String>();
        Element e = (Element) nl.item(i);
        map.put(KEY_ITEM, parser.getValue(e, KEY_ITEM));
        mylist.add(map); }
}

My XML Parser Class:

import java.io.IOException;
import java.io.StringReader;
import java.io.UnsupportedEncodingException;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;
import android.util.Log;
public class XMLParser {
    public String getXmlFromUrl(String url) {
        String xml = null;
        try {
            HttpClient httpClient = new DefaultHttpClient();
            HttpGet httpGet = new HttpGet(url);
            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            xml = EntityUtils.toString(httpEntity);
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return xml;
    }

    public Document getDomElement(String xml){
        Document doc = null;
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        try {
            DocumentBuilder db = dbf.newDocumentBuilder();
            InputSource is = new InputSource();
                is.setCharacterStream(new StringReader(xml));
                doc = db.parse(is);
            } catch (ParserConfigurationException e) {
                Log.e("Error: ", e.getMessage());
                return null;
            } catch (SAXException e) {
                Log.e("Error: ", e.getMessage());
                return null;
            } catch (IOException e) {
                Log.e("Error: ", e.getMessage());
                return null;
            }
                // return DOM
            return doc;
    }

    public String getValue(Element item, String str) {
        NodeList n = item.getElementsByTagName(str);
        return this.getElementValue(n.item(0));
    }
    public final String getElementValue( Node elem ) {
             Node child;
             if( elem != null){
                 if (elem.hasChildNodes()){
                     for( child = elem.getFirstChild(); child != null; child = child.getNextSibling() ){
                         if( child.getNodeType() == Node.TEXT_NODE  ){
                             return child.getNodeValue();
                         }
                     }
                 }
             }
             return "";
      }
}

I am not able to parse my XML here.What is the problem here ?
It returns "Item" only.
What do i need to do in my ActivityClass specially in this part of the code :

String xml = parser.getXmlFromUrl(URL);
    Document doc = parser.getDomElement(xml);
    NodeList nl = doc.getElementsByTagName(KEY_RESOURCE);
    for (int i = 0; i < nl.getLength(); i++) {


        HashMap<String, String> map = new HashMap<String, String>();
        Element e = (Element) nl.item(i);
        map.put(KEY_ITEM, parser.getValue(e, KEY_ITEM));
        mylist.add(map); }
解决方案

Your getValue() method gets MyResource element, from there, you need to get all Items under MyResource and do getElementValue(). Example code is:

   public Map getValue(Element item, String str) {
        NodeList n = item.getElementsByTagName(str);
        for (int i = 0; i < n.getLength(); i++) {
            System.out.println(getElementValue(n.item(i)));
        }
        //Here store it in list/map and return list/map instead of String
        return list/MapHere;
    }

这篇关于如何解析Android中的XML DOM解析相同的名称标签的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

07-23 19:32
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