需要帮助的读取XML文件 [英] Need help in reading an xml file

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本文介绍了需要帮助的读取XML文件的处理方法,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

我创建资源下的XML文件,并放置在myxml.xml XML文件夹中。
RES / XML / myxml.xml

我想根据ID号读取的描述。我怎样才能做到这一点?
请帮助。

myxml.xml是这样的;

 <?XML版本=1.0编码=UTF-8&GT?;
<图书>
  &所述;数ID =1>
    <说明>这是儿童的书。
    < /说明>
  < /数字>
  &所述;数ID =2>
    <说明>这是关于烹饪。
   < /说明>
< /图书>

下面是code,我有工作(但不似乎做什么)

  {尝试
        XmlPullParser XPP = getResources()的getXML(R.xml.myxml)。
                而(xpp.getEventType()!= XmlPullParser.END_DOCUMENT){
            如果(xpp.getEventType()== XmlPullParser.START_TAG){
              如果(xpp.getName()。等于(说明)){                  show1.setText(+ xpp.getAttributeValue(1));
              }
            }
        xpp.next();
         }
        }
        捕获(的Throwable t)的{        }


解决方案

如何将它甚至工作时,活动中有空..üř干什么null.getResources();

I have created an xml folder under res and placed myxml.xml in xml folder. res/xml/myxml.xml

I want to read the description according to the id number. How can I achieve that? Please help.

myxml.xml is like this;

<?xml version="1.0" encoding="utf-8" ?>
<Books>
  <Number id ="1">
    <Description>This is childrens book.
    </Description>
  </Number> 
  <Number id = "2">
    <Description>This is about cooking.
   </Description>
</Books>

Here is the code I am working with (but doesn't seems to be doing anything)

try{
        XmlPullParser xpp = getResources().getXml(R.xml.myxml);
                while(xpp.getEventType() != XmlPullParser.END_DOCUMENT){
            if(xpp.getEventType() == XmlPullParser.START_TAG){ 
              if(xpp.getName().equals("Description")){

                  show1.setText(" " + xpp.getAttributeValue(1));
              }
            }
        xpp.next();
         }
        }
        catch(Throwable t){

        }

解决方案

How will it even work when activity has null in it.. u r doin null.getResources();

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