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  • 这是我的xml，

    <root>
  <child>
      <Lunchmenu>
           <id>2</id>
           <lunch_date>2013-10-24</lunch_date>
           <break_name>Lunch</break_name>
           <class_id/>
           <school_id>1</school_id>
           <batch_id/>
     </Lunchmenu>
     <Eatable>
           <id>2</id>
           <eatable_name>Apples</eatable_name>
     </Eatable>
 </child>
  <child>
      <Lunchmenu>
           <id>2</id>
           <lunch_date>2013-10-24</lunch_date>
           <break_name>Lunch</break_name>
           <class_id/>
           <school_id>1</school_id>
           <batch_id/>
     </Lunchmenu>
     <Eatable>
           <id>3</id>
           <eatable_name>Orange</eatable_name>
     </Eatable>
</child>

  • 我需要知道，有没有办法解析上面的 xml 的孩子，因为还有两个单独的标签，比如午餐菜单和可食用的。

  • 我正在使用 sax 解析器来解析这个 xml。
  • 我知道如何解析具有数据的单个子标签并再次对其进行迭代，但在这里我很困惑如何做到这一点？
  • 如果有人知道如何解析它，请提出任何解决方案？

  • 谢谢你

  • 我的解析器类是：SchoolParser.java

公共类 SchoolParser 扩展 DefaultHandler {

private List<School> schoolListData ;
private boolean isSuccess;
private School school;
StringBuilder tempData;

@Override
public void startDocument() throws SAXException {

    super.startDocument();
    Log.e("StudentListParser","startDocument");
    schoolListData = new ArrayList<School>();
}

public List<School> getSchoolListData(){
    return schoolListData;
}

@Override
public void startElement(String uri, String localName, String qName,
        org.xml.sax.Attributes attributes) throws SAXException {

    super.startElement(uri, localName, qName, attributes);

    if (localName.equals("School")) {
        school = new School();
        Log.e("SchoolParser", "-----START----");
    } 
    tempData = new StringBuilder();

}

@Override
public void characters(char[] ch, int start, int length)
        throws SAXException {
    super.characters(ch, start, length);
    tempData.append(new String(ch, start, length));
}

@Override
public void endElement(String uri, String localName, String qName)
        throws SAXException {
    super.endElement(uri, localName, qName);
    if(localName.equals("School")){
        schoolListData.add(school);
        Log.d("localName",localName);

    }
    else if (localName.equals("id")) {
        Log.e("id", localName);

        school.id = Integer.parseInt(tempData.toString());

    } else if (localName.equals("school_name")) {
        school.schoolName = tempData.toString();
        Log.e("name", localName);

    } else if (localName.equals("logo")) {
        school.logo = tempData.toString().getBytes();
        Log.e("logo", localName);

    }else if (localName.equals("phone")) {
        school.phn_no = tempData.toString();
        Log.e("phn no", localName);

    } 
    else if (localName.equals("School")) {
        Log.e("SchoolParser", "----END---");
    }
    //          int size = buffer.length(); 
    //          buffer.delete(0, size);
    //          Log.i("buffer is empty", ""+buffer.toString());
}

@Override
public void endDocument() throws SAXException {
    super.endDocument();
    isSuccess = false;
    Log.e("StudentListParser", "endDocument");
}

}

4

1 回答 1

0

可能您可以使用序列化逻辑来解析 xml 以直接 java 类对象。

你可以使用下面的类..

package com.lib.android.util;

import java.io.InputStream;
import org.apache.log4j.Logger;
import org.simpleframework.xml.Serializer;
import org.simpleframework.xml.core.Persister;
import com.lib.android.util.logger.PLogger;

public class XmlParser
{
    private static Logger log = new PLogger(XmlParser.class).getLogger();

    public static Object Parse(Class<? extends Object> parseInto, InputStream is)
    {
        Serializer serializer = new Persister();
        try
        {
            Object obj = serializer.read(parseInto, is);
            return obj;
        } catch (Exception e)
        {
            log.error(e.getMessage());
        }
        return null;
    }
}

希望对你有帮助。。

于 2013-10-23T13:30:28.890 回答