问题
In android studio im making a simple username and password code that toasts something if both the values in the username and password edit text views are certain values ("cake" and "robot" for instance) however when I test it and enter the correct username and password values im being told that I have entered the wrong match?
Here is the main activity
public class MainActivity extends AppCompatActivity {
EditText usernameET;
EditText passwordET;
Button SignIn;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
usernameET = (EditText) findViewById(R.id.username);
passwordET = (EditText) findViewById(R.id.password);
SignIn = (Button) findViewById(R.id.signin);
}
public void SignIn(View view) {
String username = usernameET.getText().toString();
String password = passwordET.getText().toString();
if(username == "cake" && password == "robot"){
Toast.makeText(this, "You Signed In", Toast.LENGTH_SHORT).show();
}else{
Toast.makeText(this, "Invalid Login", Toast.LENGTH_SHORT).show();
}
}
}
main activity layout:
<EditText
android:id="@+id/username"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_above="@+id/password"
android:layout_alignStart="@+id/password"
android:layout_marginBottom="15dp"
android:hint="@string/username"
android:inputType="text" />
<EditText
android:id="@+id/password"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_centerHorizontal="true"
android:layout_centerVertical="true"
android:hint="@string/password"
android:inputType="text" />
<Button
android:id="@+id/signin"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/sign_in"
android:onClick="SignIn"
android:layout_below="@+id/password"
android:layout_centerHorizontal="true" />
Why is this happening?
回答1:
I think you have to write
username.equals("cake") && password.equals("robot")
Reason is that you have to check the content of the edittext not the reference
when you use == it checks reference in the memory whereas you have to check the data.
回答2:
String username = usernameET.getText().toString();
String password = passwordET.getText().toString();
"==" matches the reference id of the Strings. so,the String username is an object in the heap whose reference_id will be different from the String "cake" whereas the .eqauls() method of the String class compares the content of the string. So , in your case use
if(username.equals("cake") && password.equals("robot")){
Toast.makeText(this, "You Signed In",Toast.LENGTH_SHORT).show();
}else{
Toast.makeText(this, "Invalid Login", Toast.LENGTH_SHORT).show();
}
for more reference http://www.studytonight.com/java/string-handling-in-java.php
回答3:
Reeplace username == "cake" && password == "robot"
with
username.equals("cake") && password.equals("robot")
回答4:
You are comparing string.. so you need to use .equals
public void SignIn(View view) {
String username = usernameET.getText().toString();
String password = passwordET.getText().toString();
if(username.equals("cake") && password.equals("robot")){
Toast.makeText(this, "You Signed In", Toast.LENGTH_SHORT).show();
}else{
Toast.makeText(this, "Invalid Login", Toast.LENGTH_SHORT).show();
}
}
回答5:
In java, a==b is used to compare 2 references, not the objects themselves.
so if you have 2 strings that you want to compare, use the equals() method on String. For e.g , username.equals("cake") && password.equals("robot")
回答6:
Change your if condition like
if(username.equalsIgnoreCase("cake") && password.equalsIgnoreCase("robot")
来源:https://stackoverflow.com/questions/44900405/simple-username-password-code-in-android-not-working