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java - android http 请求的未知主机异常应返回 XML

转载 作者:行者123 更新时间:2023-12-01 15:11:25 27 4
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我使用以下类发送 http 请求并在我的 android 项目中获取返回的 XML 响应。但是当它尝试发送请求时会抛出 UnknownHostException。请帮我解决这个问题。

import java.io.ByteArrayOutputStream;
import java.io.File;
import java.io.IOException;

import org.apache.http.HttpResponse;
import org.apache.http.HttpStatus;
import org.apache.http.StatusLine;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.xml.sax.Attributes;
import org.xml.sax.ContentHandler;
import org.xml.sax.Locator;
import org.xml.sax.SAXException;

import android.os.AsyncTask;
import android.util.Xml;

class RequestTask extends AsyncTask<String, String, String>{

@Override
protected String doInBackground(String... uri) {
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response;
String responseString = null;
try {
response = httpclient.execute(new HttpGet(uri[0]));
StatusLine statusLine = response.getStatusLine();
if(statusLine.getStatusCode() == HttpStatus.SC_OK){
ByteArrayOutputStream out = new ByteArrayOutputStream();
response.getEntity().writeTo(out);
out.close();
responseString = out.toString();
} else{
//Closes the connection.
response.getEntity().getContent().close();
throw new IOException(statusLine.getReasonPhrase());
}
} catch (ClientProtocolException e) {
//TODO Handle problems..
} catch (IOException e) {
String s=e.toString();
System.out.println(s);
//TODO Handle problems..
}
return responseString;
}

@Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
//Do anything with response..
System.out.println(result);





}


}

最佳答案

确保您已在 AnroidManifest 文件中声明 INTERNET 权限,并且具有互联网连接。

您应该在 AndroidManifest 中声明此权限:

<uses-permission android:name="android.permission.INTERNET" />

关于java - android http 请求的未知主机异常应返回 XML,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12294510/

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