Question

所以我用Java 7开始了一些Android编程。我有Eclipse Juno（我认为是4.2）。

问题在于它给我一个错误“此行的多个标记 - 令牌上的语法错误“）”,; 预期 - 令牌上的语法错误“）”,; 预期

使用sendMessage方法。这是代码：

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.");
            }   
        }   
    });



}

我没有在任何地方找到解决方案，而且我完全按照教程告诉我的那样做了。我尝试清理项目三次并重新启动eclipse / computer没有任何效果。

Answer 1

您在onclick中的调用方法是错误的，请尝试以下方法： -

@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

您的代码中有2个问题：

你不是在onClick方法中调用函数，只是定义它，你不应该这样做。这个问题可以通过调用onClick中的方法并定义它来解决。在onClickListener。
你不应该在结束大括号后用半冒号终止你的sendMessage方法声明。这个问题可以通过从那里删除分号来解决。

所以你的代码最终如下所示：

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
}

Java新手错误Android编程sendMessage方法

2 个答案: