【发布时间】：2014-05-22 10:59:49
【问题描述】：
搜索后我能够从我的 servlet 检索响应，但我无法将参数（用户名和密码参数）从 android 发送到 servlet！我的 logcat 显示此错误：
04-0java.lang.ClassCastException: org.apache.http.client.methods.HttpGet cannot be cast to org.apache.http.HttpResponse
at com.example.httpgetandroidexample.MainActivity$AsyncTaskRunner.doInBackground(MainActivity.java:76)
at com.example.httpgetandroidexample.MainActivity$AsyncTaskRunner.doInBackground(MainActivity.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:288)
at java.util.concurrent.FutureTask.run(FutureTask.java:237)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:841)
我不明白为什么！ 这是我的 android 主要活动：
package com.example.httpgetandroidexample;
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.net.URLEncoder;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
public class MainActivity extends Activity {
public TextView content;
EditText name,pass;
String URL,nameValue,passValue;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
name = (EditText)findViewById(R.id.name);
pass = (EditText)findViewById(R.id.pass);
content = (TextView)findViewById(R.id.text);
Button button=(Button)findViewById(R.id.but);
try {
nameValue =URLEncoder.encode(name.getText().toString(), "UTF-8");
passValue =URLEncoder.encode(pass.getText().toString(), "UTF-8");
URL = "http://10.0.2.2:8080/login/web";
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AsyncTaskRunner runner = new AsyncTaskRunner();
runner.execute(new String[ ] { URL });
}
});
}
private class AsyncTaskRunner extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... urls) {
String response = "";
for (String url : urls) {
DefaultHttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet(url);
try {
List<NameValuePair> postParameters =
new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("user",nameValue));
postParameters.add(new BasicNameValuePair("pass",passValue));
UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(
postParameters);
((HttpResponse) httpGet).setEntity(formEntity);
HttpResponse execute = client.execute(httpGet);
InputStream content = execute.getEntity().getContent();
BufferedReader buffer = new BufferedReader(new InputStreamReader(content));
String s = "";
while ((s = buffer.readLine()) != null) {
response += s;
}
} catch (Exception e) {
e.printStackTrace();
}
}
return response;
}
@Override
protected void onPostExecute(String result) {
content.setText(result);
}
}
}
有人知道吗？
更新：
now i have changed my android code like this:
package com.example.httpgetandroidexample;
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.net.URLEncoder;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
public class MainActivity extends Activity {
public TextView content;
EditText name,pass;
String URL;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
name = (EditText)findViewById(R.id.name);
pass = (EditText)findViewById(R.id.pass);
content = (TextView)findViewById(R.id.text);
Button button=(Button)findViewById(R.id.but);
try {
String nameValue ="user="+URLEncoder.encode(name.getText().toString(), "UTF-8");
String passValue ="&pass="+URLEncoder.encode(pass.getText().toString(), "UTF-8");
URL = "http://10.0.2.2:8080/login/web?"+nameValue+passValue;
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AsyncTaskRunner runner = new AsyncTaskRunner();
runner.execute(new String[ ] { URL });
}
});
}
private class AsyncTaskRunner extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... urls) {
String response = "";
for (String url : urls) {
DefaultHttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet(url);
try {
HttpResponse execute = client.execute(httpGet);
InputStream content = execute.getEntity().getContent();
BufferedReader buffer = new BufferedReader(new InputStreamReader(content));
String s = "";
while ((s = buffer.readLine()) != null) {
response += s;
}
} catch (Exception e) {
e.printStackTrace();
}
}
return response;
}
@Override
protected void onPostExecute(String result) {
content.setText(result);
}
}
}
这次运行没有logcat错误但是参数没有传到servlet！
【问题讨论】：
-
我认为您在 HTTP GET 和 HTTP POST 之间混用。选择其中之一，但不能同时选择两者。你想做什么？我看到您正在尝试传递
user和pass。 -
我只是试图通过 GET 方法将 2 个参数（用户，传递）发送到 servlet，并从 servlet 获得响应并显示它。我可以从 servlet 获得响应，但我无法发送参数到服务器！