Question

public class LoginActivity extends AppCompatActivity implements View.OnClickListener {

public EditText studentNumber, passWord;
//defining AwesomeValidation object
public AwesomeValidation awesomeValidation;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    awesomeValidation = new AwesomeValidation(ValidationStyle.BASIC);
    studentNumber = (EditText) findViewById(R.id.studentNumber);
    passWord = (EditText) findViewById(R.id.passWord);

    awesomeValidation.addValidation(this, R.id.studentNumber, "^[0-9]{9}", R.string.studentnumberError);
    String regexPassword = ".{8,}";
    awesomeValidation.addValidation(this, R.id.passWord, regexPassword, R.string.invalid_password);

    signIn.setOnClickListener(this);
}

// Triggers when LOGIN Button clicked
@Override
public void onClick(View view) {
    if (view == signIn) {
        checkLogin();
    }
}

public void checkLogin() {

    // Get text from studentNumber and passWord field
    final String studentNumber = studentNumber.getText().toString();
    final String passWord = passWord.getText().toString();

    // Initialize  AsyncLogin() class with studentNumber and password
    new AsyncLogin().execute(studentNumber,passWord);

    }
}

Android Studio表示可能尚未初始化变量studentNumber和passWord。错误显示在这一行：

final String studentNumber = studentNumber.getText().toString();
final String passWord = passWord.getText().toString();

似乎有什么问题？它是表格的验证吗？谢谢。

Answer 1

在onClick方法中更改字符串变量名称，它与edittext变量名称冲突：

public void checkLogin() {

// Get text from studentNumber and passWord field
final String studentNumberString = studentNumber.getText().toString();
final String passWordString = passWord.getText().toString();

// Initialize  AsyncLogin() class with studentNumber and password
new AsyncLogin().execute(studentNumberString,passWordString);

}
}

Answer 2

你得到变量“studentNumber”和“passWord”可能尚未初始化？

因为您的String变量名和您的Edit-text名都相同所以它们是冲突的

只需更改您的String变量名称

使用此

final String studentNumber2 = studentNumber.getText().toString();
final String passWord2 = passWord.getText().toString();

而不是

final String studentNumber = studentNumber.getText().toString();
final String passWord = passWord.getText().toString();

Answer 3

您收到错误，因为字符串名称和编辑文字名称是示例请更改您的方法。

public void checkLogin() {

    // Get text from studentNumber and passWord field
    final String studentNum = studentNumber.getText().toString();
    final String pass= passWord.getText().toString();

    // Initialize  AsyncLogin() class with studentNumber and password
    new AsyncLogin().execute(studentNum,pass);

    }

变量“studentNumber”和“passWord”可能尚未初始化？

3 个答案: