Java新手錯誤Android編程sendMessage方法

Question

所以我用Java 7開始了一些Android編程。我有Eclipse Juno（我認為是4.2）。

問題是它給我一個錯誤“此行有多個標記-令牌“）”上的語法錯誤；;預期-令牌“）”上的語法錯誤；;預期

與sendMessage方法一起就行。 這是代碼：

public class MainActivity extends ActionBarActivity {


int counter;
Button login; 
EditText username, password;
String success;


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    counter = 0;

    username = (EditText)findViewById(R.id.getEmail);
    password = (EditText)findViewById(R.id.getPassword);        

    login.setOnClickListener(new View.OnClickListener() {       
        public void onClick (View v) {
            if(username.getText().toString().equals("admin") && password.getText().toString().equals("admin")){
                success = "Successful";
                counter = 0;
                public void sendMessage (View view){
                    Intent intent = new Intent("com.example.linked1n.SCREENAFTLOG");
                    startActivity(intent);
                };
            } else {
                    counter++;
                    login.setText("Unsuccessful. Try again. " + 3-counter + " tries left.");
            }   
        }   
    });



}

我在任何地方都找不到解決方案，而且完全按照本教程告訴我的方法進行。 我試圖清理項目三次，然后重新啟動Eclipse /計算機，但沒有任何效果。

Answer 1

您在onclick中的調用方法錯誤，因此請嘗試以下方式：-

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    counter = 0;

    username = (EditText)findViewById(R.id.getEmail);
    password = (EditText)findViewById(R.id.getPassword);        

    login.setOnClickListener(new View.OnClickListener() {       
        public void onClick (View v) {
            if(username.getText().toString().equals("admin") && password.getText().toString().equals("admin")){
                success = "Successful";
                counter = 0;
               sendMessage(v);
            } else {
                    counter++;
                    login.setText("Unsuccessful. Try again. " + 3-counter + " tries left.");
            }   
        }   
    });

 public void sendMessage (View view){
                        Intent intent = new Intent("com.example.linked1n.SCREENAFTLOG");
                        startActivity(intent);
                    }

Answer 2

您的代碼中有2個問題：

1. 你是不是調用該函數，只是定義它，太內部onClick方法，你應該不do.This問題可以通過調用里面的方法去解決onClick和定義外面onClickListener 。
2. 您不應該在花括號結束后用分號來終止sendMessage方法的聲明。可以通過從該位置刪除分號來解決此問題。

因此，您的代碼最終如下所示：

public class MainActivity extends ActionBarActivity { 


int counter;
Button login; 
EditText username, password;
String success;


@Override 
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    counter = 0;

    username = (EditText)findViewById(R.id.getEmail);
    password = (EditText)findViewById(R.id.getPassword);        

    login.setOnClickListener(new View.OnClickListener() {        
        public void onClick (View v) { 
            if(username.getText().toString().equals("admin") && password.getText().toString().equals("admin")){ 
                success = "Successful"; 
                counter = 0; 
                sendMessage(v);//Method sendMesaage is called inside onClick

            } else { 
                    counter++; 
                    login.setText("Unsuccessful. Try again. " + 3-counter + " tries left."); 
            }    
        }    
    }); 
     //Definition of method sendMessage
     public void sendMessage (View view){ 
                    Intent intent = new Intent("com.example.linked1n.SCREENAFTLOG"); 
                    startActivity(intent); 
                }//Removed semicolon from the method sendMessage
}