我试图解析我从HttpPost到一个服务器(last.fm)的一个last.fm
android应用程序的
XML HttpResponse.如果我只是解析它到字符串我可以看到它是一个正常的xml字符串,所有所需的信息.但我只是不能解析单个NameValuePairs.这是我的HttpResponse对象：
HttpResponse response = client.execute(post); HttpEntity r_entity = response.getEntity();@H_502_3@我尝试了两个不同的东西,而不是他们的工作.首先我尝试检索NameValuePairs：
List<NameValuePair> answer = URLEncodedUtils.parse(r_entity); String name = "empty"; String playcount = "empty"; for (int i = 0; i < answer.size(); i++){ if (answer.get(i).getName().equals("name")){ name = answer.get(i).getValue(); } else if (answer.get(i).getName().equals("playcount")){ playcount = answer.get(i).getValue(); } }@H_502_3@在此代码之后,名称和播放计数仍保持“空”.所以我试图使用XML解析器：
DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder(); Document answer = db.parse(new DataInputStream(r_entity.getContent())); NodeList nl = answer.getElementsByTagName("playcount"); String playcount = "empty"; for (int i = 0; i < nl.getLength(); i++) { Node n = nl.item(i); Node fc = n.getFirstChild(); playcount Url = fc.getNodeValue(); }@H_502_3@这似乎失败得太早了,因为它甚至没有设置playcount变量.但是就像我说的,如果我这样做：
EntityUtils.toString(r_entity);@H_502_3@我会得到一个完美的xml字符串.因此,HttpResponse包含正确的信息应该没有问题.我究竟做错了什么？
解决方法
我解决了DOM XML解析器需要更多的调整：
HttpResponse response = client.execute(post);
HttpEntity r_entity = response.getEntity();
String xmlString = EntityUtils.toString(r_entity);
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder db = factory.newDocumentBuilder();
InputSource inStream = new InputSource();
inStream.setCharacterStream(new StringReader(xmlString));
Document doc = db.parse(inStream);
String playcount = "empty";
NodeList nl = doc.getElementsByTagName("playcount");
for(int i = 0; i < nl.getLength(); i++) {
if (nl.item(i).getNodeType() == org.w3c.dom.Node.ELEMENT_NODE) {
org.w3c.dom.Element nameElement = (org.w3c.dom.Element) nl.item(i);
playcount = nameElement.getFirstChild().getNodeValue().trim();
}
}@H_502_3@
