我已经看到了更多这样的问题，但是没有对它们有任何意义，或者它们不适用于我的问题，所以就这样了。
我有一个允许您登录的活动，以及使用您的登录凭据发送POST和GET请求等的其他地方。
mainActivity：
public class MainActivity extends Activity
{
private String username;
private String password;
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
final EditText usernameField = (EditText) findViewById(R.id.enterUsername);
final EditText passwordField = (EditText) findViewById(R.id.enterPassword);
Button startButton = (Button) findViewById(R.id.startButton);
startButton.setOnClickListener(new View.OnClickListener()
{
public void onClick(View view)
{
username = usernameField.getText().toString();
password = passwordField.getText().toString();
Intent myIntent = new Intent(view.getContext(), HttpGetPost.class);
startActivityForResult(myIntent, 0);
}
});
}
public String getUser() { return this.username; }
public String getPassword() { return this.password; }
}
HttpGetPost：
public class HttpGetPost extends Activity
{
private MainActivity mainProxy = new MainActivity();
private Button postButton;
private Button getButton;
private Button getMeasureButton;
private Button getDevicesButton;
private String access_token;
private String refresh_token;
private String device_list;
private String expires_in;
private String getRequest;
private static final String TAG = "MyActivity";
private static final String USER_AGENT = "Mozilla/5.0";
private String clientID = some_id;
private String clientSecret = some_secret;
private String user = mainProxy.getUser();
private String pass = mainProxy.getPassword();
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_http);
Log.v(TAG, "mainProxy.username: "+user);
Log.v(TAG, "mainProxy.password: "+pass);
postButton = (Button) findViewById(R.id.postButton);
postButton.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View view)
{
new sendPost().execute("");
}
});
}
private class sendPost extends AsyncTask<String, Void, String>
{
@Override
protected String doInBackground(String... params)
{
try
{
String url = some_url;
URL obj = new URL(url);
HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();
//add request header
con.setRequestMethod("POST");
con.setRequestProperty("User-Agent", USER_AGENT);
con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
String urlParameters = "grant_type=password&client_id=" +clientID +"&client_secret="
+clientSecret +"&username=" +user +"&password=" +pass;
// Send post request
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(urlParameters);
wr.flush();
wr.close();
int responseCode = con.getResponseCode();
Log.v(TAG, "\nSending 'POST' request to URL : " + url);
Log.v(TAG, "Post parameters : " + urlParameters);
Log.v(TAG, "Response Code : " + responseCode);
BufferedReader in = new BufferedReader(
new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
//print result
Log.v(TAG, response.toString());
if (responseCode == 200)
{
access_token = response.substring(17, 74);
refresh_token = response.substring(93,150);
expires_in = response.substring(165, 170);
getRequest = "http://api.netatmo.net/api/getuser?access_token=" +access_token + " HTTP/1.1";
Log.v(TAG, "access token: " +access_token);
Log.v(TAG, "refresh token: " +refresh_token);
Log.v(TAG, "expires in: " +expires_in);
}
}
catch(Exception e){
e.printStackTrace();
}
return "";
}
protected void onPostExecute(String result) {
Toast.makeText(getApplicationContext(), "ENDED", Toast.LENGTH_LONG).show();
}
}
}
当我在第二个类中打印出用户名和密码时，它们都返回null，POST请求失败。
答案 0 :(得分：4)
您应该使用意图的putExtra方法将用户名和密码传递给您的活动：
Intent myIntent = new Intent(view.getContext(), HttpGetPost.class);
myIntent.putExtra("username", username);
myIntent.putExtra("password", pasword);
startActivityForResult(myIntent, 0);
在第二个活动中，在onCreate（）中（例如在setContentView之后），您可以使用getXXExtras检索它们：
Intent intent = getIntent();
String username = intent.getStringExtra("username");
String password = intent.getStringExtra("password");
答案 1 :(得分：4)
在jbihan的回答中澄清我的意思：
我正在更新你的代码：
第一次修订：
Button startButton = (Button) findViewById(R.id.startButton);
startButton.setOnClickListener(new View.OnClickListener()
{
public void onClick(View view)
{
username = usernameField.getText().toString();
password = passwordField.getText().toString();
Intent myIntent = new Intent(view.getContext(), HttpGetPost.class);
// ADDITION
myIntent.putExtra("username", username);
myIntent.putExtra("password", password);
// END ADDITION
startActivityForResult(myIntent, 0);
}
});
第二次修订：
postButton = (Button) findViewById(R.id.postButton);
// ADDITION
final String user = getIntent().getStringExtra("username");
final String password = getIntent().getStringExtra("password");
// END ADDITION
postButton.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View view)
{
// EDITED
new sendPost().execute(user, password);
}
});
第三次修订：
private class sendPost extends AsyncTask<String, Void, String>
{
@Override
protected String doInBackground(String... params)
{
// ADDITION
String user = params[0];
String password = params[1];
// END ADDITION
// use them in the request
// rest of code...
}
}
请考虑使用“username”和“password”键的常量。
答案 2 :(得分：1)
尝试使用额外内容：
Intent myIntent = new Intent(view.getContext(), HttpGetPost.class);
myIntent.putExtra("username", username);
myIntent.putExtra("password", password);
startActivityForResult(myIntent, 0);
在另一个行为中（你的HttpGetPost）
String user = getIntent().getStringExtra("username");
String password = getIntent().getStringExtra("password");
答案 3 :(得分：1)
这是关于正确使用Intents的一个很好的tutorial。
尝试调用HttpGetPost活动：
Intent myIntent = new Intent(this, HttpGetPost.class);
myIntent.putExtra("username", username);
myIntent.putExtra("password", password);
startActivity(myIntent);
使用this，您可以在Intent构造函数中传递正确的Context。将数据放入要发送到活动的Intent中。接下来的一点是不要调用startActivityForResult（），它用于调用Activity，进行一些计算并将结果发送回请求的Activity。
现在从onCreate中的HttpGetPost活动中获取数据，并将其保存到字段中：
getIntent().getExtras().getString("username");
getIntent().getExtras().getString("password");
答案 4 :(得分：1)
您不需要
private MainActivity mainProxy = new MainActivity();
在HttpGetPost中。它将创建一个新的MainActivity，它不是启动HttpGetPost的原始MainActivity。
您可以使用附加功能跨意图发送数据。这是我的解决方案：
将它放在MainActivity中
Intent myIntent = new Intent(view.getContext(), HttpGetPost.class);
myIntent.putExtra(HttpGetPost.KEY_USERNAME, username);
myIntent.putExtra(HttpGetPost.KEY_PASSWORD, password);
startActivityForResult(myIntent, 0);
这是针对HttpGetPost，KEY_USERNAME和KEY_PASSWORD可以用来存储额外的密钥，这样就可以避免拼写错误。
public static final String KEY_USERNAME = "username"; // or whatever you like for key
public static final String KEY_PASSWORD = "password"; // or whatever you like for key
private String user; // instead of private String user = mainProxy.getUser();
private String pass; // instead of private String pass = mainProxy.getPassword();
将其放入onHrepGetPost的onCreate以获取意图
的数据Intent intent = getIntent();
user = intent.getStringExtra(KEY_USERNAME);
pass = intent.getStringExtra(KEY_PASSWORD);
Here是意图的官方文件。