从网站解析XML在Android的String数组,请帮助我 [英] Parsing XML from a website to a String array in Android please help me
问题描述
您好,我在做一个Android应用程序,拉从一个Wiki的一些数据,起初我打算寻找一种方法来解析HTML的过程,但是从事情,有人向我指出的是,XML会更容易的工作。现在,我坚持试图找到一种方法来正确地解析XML。我想,现在从网络地址解析来自:
Hello I am in the process of making an Android app that pulls some data from a Wiki, at first I was planning on finding a way to parse the HTML, but from something that someone pointed out to me is that XML would be much easier to work with. Now I am stuck trying to find a way to parse the XML correctly. I am trying to parse from a web address right now from:
<一个href=\"http://zelda.wikia.com/api.php?action=query&list=categorymembers&cmtitle=Category:Games&cmlimit=500&format=xml\" rel=\"nofollow\">http://zelda.wikia.com/api.php?action=query&list=categorymembers&cmtitle=Category:Games&cmlimit=500&format=xml
我试图让每一个游戏的标题为一个字符串数组,我遇到了一些麻烦。我没有code,我尝试的一个例子,它是通过使用xmlpullparser。我的应用程序崩溃每次我试图用它做任何事情。它会更好保存到本地XML,并从那里解析?或者我会好起来的,从网址去?我将如何去正确地分析此为一个字符串数组?请帮我,感谢您抽出宝贵的时间来阅读。
I am trying to get the titles of each of the games into a string array and I am having some trouble. I don't have an example of the code I was trying out, it was by using xmlpullparser. My app crashes everytime that I try to do anything with it. Would it be better to save the XML locally and parse from there? or would I be okay going from the web address? and how would I go about parsing this correctly into a string array? Please help me, and thank you for taking the time to read this.
如果您需要查看code或任何东西,我可以得到它今晚稍后,我只是不靠近我的电脑在这个时候。谢谢你。
If you need to see code or anything I can get it later tonight, I am just not near my PC at this time. Thank you.
推荐答案
无论何时你发现自己写解析器code简单的格式,比如一个在你的例子你几乎的总是的做一些事情错误的,不使用合适的框架。
Whenever you find yourself writing parser code for simple formats like the one in your example you're almost always doing something wrong and not using a suitable framework.
例如 - 有解析包含在SDK中的 android.sax
封装XML一组简单的帮手,它只是碰巧你张贴的例子可以很容易分析是这样的:
For instance - there's a set of simple helpers for parsing XML in the android.sax
package included in the SDK and it just happens that the example you posted could be easily parsed like this:
public class WikiParser {
public static class Cm {
public String mPageId;
public String mNs;
public String mTitle;
}
private static class CmListener implements StartElementListener {
final List<Cm> mCms;
CmListener(List<Cm> cms) {
mCms = cms;
}
@Override
public void start(Attributes attributes) {
Cm cm = new Cm();
cm.mPageId = attributes.getValue("", "pageid");
cm.mNs = attributes.getValue("", "ns");
cm.mTitle = attributes.getValue("", "title");
mCms.add(cm);
}
}
public void parseInto(URL url, List<Cm> cms) throws IOException, SAXException {
HttpURLConnection con = (HttpURLConnection) url.openConnection();
try {
parseInto(new BufferedInputStream(con.getInputStream()), cms);
} finally {
con.disconnect();
}
}
public void parseInto(InputStream docStream, List<Cm> cms) throws IOException, SAXException {
RootElement api = new RootElement("api");
Element query = api.requireChild("query");
Element categoryMembers = query.requireChild("categorymembers");
Element cm = categoryMembers.requireChild("cm");
cm.setStartElementListener(new CmListener(cms));
Xml.parse(docStream, Encoding.UTF_8, api.getContentHandler());
}
}
基本上,像这样调用:
Basically, called like this:
WikiParser p = new WikiParser();
ArrayList<WikiParser.Cm> res = new ArrayList<WikiParser.Cm>();
try {
p.parseInto(new URL("http://zelda.wikia.com/api.php?action=query&list=categorymembers&cmtitle=Category:Games&cmlimit=500&format=xml"), res);
} catch (MalformedURLException e) {
} catch (IOException e) {
} catch (SAXException e) {}
编辑:这是你如何创建一个列表与LT;弦乐&GT;
而不是:
This is how you'd create a List<String>
instead:
public class WikiParser {
private static class CmListener implements StartElementListener {
final List<String> mTitles;
CmListener(List<String> titles) {
mTitles = titles;
}
@Override
public void start(Attributes attributes) {
String title = attributes.getValue("", "title");
if (!TextUtils.isEmpty(title)) {
mTitles.add(title);
}
}
}
public void parseInto(URL url, List<String> titles) throws IOException, SAXException {
HttpURLConnection con = (HttpURLConnection) url.openConnection();
try {
parseInto(new BufferedInputStream(con.getInputStream()), titles);
} finally {
con.disconnect();
}
}
public void parseInto(InputStream docStream, List<String> titles) throws IOException, SAXException {
RootElement api = new RootElement("api");
Element query = api.requireChild("query");
Element categoryMembers = query.requireChild("categorymembers");
Element cm = categoryMembers.requireChild("cm");
cm.setStartElementListener(new CmListener(titles));
Xml.parse(docStream, Encoding.UTF_8, api.getContentHandler());
}
}
和则:
WikiParser p = new WikiParser();
ArrayList<String> titles = new ArrayList<String>();
try {
p.parseInto(new URL("http://zelda.wikia.com/api.php?action=query&list=categorymembers&cmtitle=Category:Games&cmlimit=500&format=xml"), titles);
} catch (MalformedURLException e) {
} catch (IOException e) {
} catch (SAXException e) {}
这篇关于从网站解析XML在Android的String数组,请帮助我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IT屋!