如何阻止帐户当用户名和密码错误时

Question

我的要求是使用count.when我们按三次按钮我想要出现toast消息为Blocked ??请任何人都可以显示代码或条件。编程中有任何错误？

public void empbtn(View v) {
       EditText et11,et22;
        int counter = 3;
        int i;
        String unames[] = {"Anil","Ramesh","Khaja"};
        String passwds[] = {"anil","ramesh","khaja"};
        et11 = (EditText)findViewById(R.id.empname);
        et22 = (EditText)findViewById(R.id.emppass);
        String uname = et11.getText().toString();
        String pass = et22.getText().toString();
        if(uname.equals(""))
        {
            et11.setError("PLS ENTER NAME");
        }
      else if (pass.equals(passwds)) 
       {
        //et22.setError("pls enter password");
        Toast.makeText(main.this, "valid", Toast.LENGTH_SHORT).show();
       } 
     else {
        counter++;
        if (counter > 3) 
          {
            Toast.makeText(main.this, "blocked", Toast.LENGTH_SHORT).show();
        }
        else {
            for (i = 0; i < unames.length; i++) {
                if (uname.equals(unames[i]) && pass.equals(passwds[i])) {
                    setContentView(R.layout.home);
                    TextView tv = (TextView) findViewById(R.id.names);
                    tv.setText(uname);
                    break;
                }
            }
            if (i == unames.length) {
                Toast.makeText(main.this, "INVALID", Toast.LENGTH_SHORT).show();
            }
         }
    }

Answer 1

您可以将布尔值作为标志来知道何时有效：

boolean isValid = false;

for (i = 0; i < unames.length; i++) {
            if (uname.equals(unames[i]) && pass.equals(passwds[i])) {
                setContentView(R.layout.home);
                TextView tv = (TextView) findViewById(R.id.names);
                tv.setText(uname);
                isValid=true
                break;
            }
        }
        if (!isValid) {
            Toast.makeText(main.this, "INVALID", Toast.LENGTH_SHORT).show();
        }

修改

试试这个：

    if (i == unames.length-1) {
        Toast.makeText(main.this, "INVALID", Toast.LENGTH_SHORT).show();
    }