Android - always active else statement

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();
}

No matter what i key in into text box, it always active the else statement. Why?

Because txtUser is always an EditText not a String. You need to get the value out of the EditText (`txtUser.getText()`) before you check if it's equal. — Philip Couling, Jun 12 '15 at 15:34
An EditText cannot be ever equal to a String. Its content might be. — m0skit0, Jun 12 '15 at 15:37

score 1 · Answer 1 · answered Jun 12 '15 at 15:35

You should use txtUser.getEditableText().toString() when comparing like this:

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();

score 0 · Accepted Answer · answered Jun 12 '15 at 15:34

You have to get the text of the TextView, like this:

    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();
    }

farfanium · Answer 3 · 2015-06-15T17:30:05.697

-2

I'm not familiar with android atm, but have you tried:

txtUser.getText() == "admin"

Thanks for the comments (I was using C# approach), fixed code:

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

edited Jun 15 '15 at 17:30

answered Jun 12 '15 at 15:37

farfanium

1
1

1

This is wrong for a couple reasons. One being [how to compare strings in Java](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) – codeMagic Jun 12 '15 at 15:38

Android - always active else statement

3 Answers3