我曾经做过类似的事情;

EditText usernameEditText = (EditText) findViewById(R.id.editUsername);
sUsername = usernameEditText.getText().toString();
if (sUsername.matches("")) {
    Toast.makeText(this, "You did not enter a username", Toast.LENGTH_SHORT).show();
    return;
}

对于验证EditText使用edittext# setError方法来显示错误，对于检查空值或空值使用内置的android类TextUtils.isEmpty(strVar)，如果strVar为空或零长度返回true

EditText etUserName = (EditText) findViewById(R.id.txtUsername);
String strUserName = etUserName.getText().toString();

 if(TextUtils.isEmpty(strUserName)) {
    etUserName.setError("Your message");
    return;
 }

如果EditText为空，为什么不禁用按钮呢?恕我直言，这看起来更专业:

        final EditText txtFrecuencia = (EditText) findViewById(R.id.txtFrecuencia);  
        final ToggleButton toggle = (ToggleButton) findViewById(R.id.toggleStartStop);
        txtFrecuencia.addTextChangedListener(new TextWatcher() {
        @Override
        public void afterTextChanged(Editable s) {
            toggle.setEnabled(txtFrecuencia.length() > 0);
        }

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count,
                int after) {
        }

        @Override
        public void onTextChanged(CharSequence s, int start, int before,
                int count) {

        }
       });

我使用TextUtils来做这个:

if (TextUtils.isEmpty(UsernameInfo.getText())) {
    validationError = true;
    validationErrorMessage.append(getResources().getString(R.string.error_blank_username));
}

尝试使用If ELSE If条件。您可以轻松地验证editText字段。

 if(TextUtils.isEmpty(username)) {
                userNameView.setError("User Name Is Essential");
                return;
         } else  if(TextUtils.isEmpty(phone)) {
                phoneView.setError("Please Enter Your Phone Number");
                return;
            }

EditText txtUserID = (EditText) findViewById(R.id.txtUserID);

String UserID = txtUserID.getText().toString();

if (UserID.equals("")) 

{
    Log.d("value","null");
}

您将在LogCat....中看到该消息