在应用程序打开之前 XMLParser 就崩溃了。错误是:org.apache.http.impl.conn.DefaultClientConnectionOperator.openConnection(DefaultClientConnectionOperator.java:137)XMLParser 代码:public class XMLParser {
public XMLParser(){
}
public String getXmlFromUrl(String url){
String xml = null;
try{
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
HttpResponse httpResponse = httpClient.execute(httpPost);
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 doc;
}在 main activity 中是这样调用的:XMLParser parser = new XMLParser();
String xml = parser.getXmlFromUrl(URL);
Document doc = parser.getDomElement(xml);所以问题出在哪里呢?
3 回答
幕布斯6054654
TA贡献1876条经验 获得超7个赞
LZ的网络请求返回都不用判断状态?直接就调用 HttpEntity httpEntity = httpResponse.getEntity();xml = EntityUtils.toString(httpEntity);
???
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