Question

  public void myClick1(View v){
    EditText txtUser = (EditText) findViewById(R.id.strUser);
    EditText txtPwd = (EditText)findViewById(R.id.strPwd);

  if (txtUser.equals("admin")){
        Toast.makeText(getApplicationContext(),"Authentication Pass", Toast.LENGTH_LONG).show();
    }else{
        Toast.makeText(getApplicationContext(),"Not valid User ID/Password: " + txtUser.getText(), Toast.LENGTH_LONG).show();
}

无论我键入文本框，它总是激活else语句。为什么呢？

Answer 1

比较时应该使用txtUser.getEditableText（）。toString（）：

if ("admin".equals(txtUser.getEditableText().toString())){
    Toast.makeText(getApplicationContext(),"Authentication Pass", Toast.LENGTH_LONG).show();
}else{
    Toast.makeText(getApplicationContext(),"Not valid User ID/Password: " + txtUser.getText(), Toast.LENGTH_LONG).show();

Answer 2

您必须获取TextView的文本，如下所示：

    EditText txtUser = (EditText) findViewById(R.id.strUser);
    EditText txtPwd = (EditText) findViewById(R.id.strPwd);

    if ("admin".equals(txtUser.getText().toString())) {
        Toast.makeText(getApplicationContext(), "Authentication Pass", Toast.LENGTH_LONG).show();
    } else {
        Toast.makeText(getApplicationContext(),"Not valid User ID/Password: " + txtUser.getText(), Toast.LENGTH_LONG).show();
    }

Answer 3

我对android atm并不熟悉，但你试过了吗？

txtUser.getText（）==“admin”

感谢您的评论（我使用的是C＃方法），修复了代码：

txtUser.getText().equals("admin")

Android - 始终有效的else语句

3 个答案: