Always shows as “Failed”
0
I'm new in Android Studio and I'm making a simple Login Form in which the user must input "admin" (Username) and "admin123" (password). If both are true, a toast message will pop up as "Success" and "Failed" otherwise.
However, when I run the code it will always say as "Failed" when all the conditions are met.
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Button btn = (Button) findViewById(R.id.btn);
btn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
EditText txtuname = (EditText) findViewById(R.id.txtuname);
EditText txtpassw = (EditText) findViewById(R.id.txtpassw)
Toast toast = new Toast(getApplicationContext());
toast.setGravity(Gravity.TOP | Gravity.LEFT, 0, 0);
if((txtuname.getText().equals("admin")) && (txtpassw.getText().equals("admin123"))){
Toast.makeText(MainActivity.this, "Success",
Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(MainActivity.this, "Failed",
Toast.LENGTH_SHORT).show();
}
}
});
}
conditional android-toast
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
add a comment |
0
I'm new in Android Studio and I'm making a simple Login Form in which the user must input "admin" (Username) and "admin123" (password). If both are true, a toast message will pop up as "Success" and "Failed" otherwise.
However, when I run the code it will always say as "Failed" when all the conditions are met.
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Button btn = (Button) findViewById(R.id.btn);
btn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
EditText txtuname = (EditText) findViewById(R.id.txtuname);
EditText txtpassw = (EditText) findViewById(R.id.txtpassw)
Toast toast = new Toast(getApplicationContext());
toast.setGravity(Gravity.TOP | Gravity.LEFT, 0, 0);
if((txtuname.getText().equals("admin")) && (txtpassw.getText().equals("admin123"))){
Toast.makeText(MainActivity.this, "Success",
Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(MainActivity.this, "Failed",
Toast.LENGTH_SHORT).show();
}
}
});
}
conditional android-toast
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
1
Code will not compile as-is, there's (at least) a missing semi-colon. You should probably cut'n'paste the code to ensure it's correct. Having said that, you can add debug statements to your code printing out (or msgboxing) the two fields. This may help you establish what the problem is. In any case, doesn'tgetText()return anEditable? Don't you have to get thetextproperty of that to get the actual string? Not sure (which is why this is a comment rather than an answer) but you may want to look into that.
– paxdiablo
Nov 24 '18 at 2:05
I have already resolved the problem. Thanks!
– Beverly Castillo
Nov 24 '18 at 3:20
add a comment |
0
0
0
I'm new in Android Studio and I'm making a simple Login Form in which the user must input "admin" (Username) and "admin123" (password). If both are true, a toast message will pop up as "Success" and "Failed" otherwise.
However, when I run the code it will always say as "Failed" when all the conditions are met.
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Button btn = (Button) findViewById(R.id.btn);
btn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
EditText txtuname = (EditText) findViewById(R.id.txtuname);
EditText txtpassw = (EditText) findViewById(R.id.txtpassw)
Toast toast = new Toast(getApplicationContext());
toast.setGravity(Gravity.TOP | Gravity.LEFT, 0, 0);
if((txtuname.getText().equals("admin")) && (txtpassw.getText().equals("admin123"))){
Toast.makeText(MainActivity.this, "Success",
Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(MainActivity.this, "Failed",
Toast.LENGTH_SHORT).show();
}
}
});
}
conditional android-toast
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
I'm new in Android Studio and I'm making a simple Login Form in which the user must input "admin" (Username) and "admin123" (password). If both are true, a toast message will pop up as "Success" and "Failed" otherwise.
However, when I run the code it will always say as "Failed" when all the conditions are met.
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Button btn = (Button) findViewById(R.id.btn);
btn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
EditText txtuname = (EditText) findViewById(R.id.txtuname);
EditText txtpassw = (EditText) findViewById(R.id.txtpassw)
Toast toast = new Toast(getApplicationContext());
toast.setGravity(Gravity.TOP | Gravity.LEFT, 0, 0);
if((txtuname.getText().equals("admin")) && (txtpassw.getText().equals("admin123"))){
Toast.makeText(MainActivity.this, "Success",
Toast.LENGTH_SHORT).show();
}
else{
Toast.makeText(MainActivity.this, "Failed",
Toast.LENGTH_SHORT).show();
}
}
});
}
conditional android-toast
conditional android-toast
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
asked Nov 24 '18 at 1:38
Beverly CastilloBeverly Castillo
114
114
1
Code will not compile as-is, there's (at least) a missing semi-colon. You should probably cut'n'paste the code to ensure it's correct. Having said that, you can add debug statements to your code printing out (or msgboxing) the two fields. This may help you establish what the problem is. In any case, doesn'tgetText()return anEditable? Don't you have to get thetextproperty of that to get the actual string? Not sure (which is why this is a comment rather than an answer) but you may want to look into that.
– paxdiablo
Nov 24 '18 at 2:05
I have already resolved the problem. Thanks!
– Beverly Castillo
Nov 24 '18 at 3:20
add a comment |
1
Code will not compile as-is, there's (at least) a missing semi-colon. You should probably cut'n'paste the code to ensure it's correct. Having said that, you can add debug statements to your code printing out (or msgboxing) the two fields. This may help you establish what the problem is. In any case, doesn'tgetText()return anEditable? Don't you have to get thetextproperty of that to get the actual string? Not sure (which is why this is a comment rather than an answer) but you may want to look into that.
– paxdiablo
Nov 24 '18 at 2:05
I have already resolved the problem. Thanks!
– Beverly Castillo
Nov 24 '18 at 3:20
1
1
Code will not compile as-is, there's (at least) a missing semi-colon. You should probably cut'n'paste the code to ensure it's correct. Having said that, you can add debug statements to your code printing out (or msgboxing) the two fields. This may help you establish what the problem is. In any case, doesn't
getText() return an Editable? Don't you have to get the text property of that to get the actual string? Not sure (which is why this is a comment rather than an answer) but you may want to look into that.– paxdiablo
Nov 24 '18 at 2:05
Code will not compile as-is, there's (at least) a missing semi-colon. You should probably cut'n'paste the code to ensure it's correct. Having said that, you can add debug statements to your code printing out (or msgboxing) the two fields. This may help you establish what the problem is. In any case, doesn't
getText() return an Editable? Don't you have to get the text property of that to get the actual string? Not sure (which is why this is a comment rather than an answer) but you may want to look into that.– paxdiablo
Nov 24 '18 at 2:05
I have already resolved the problem. Thanks!
– Beverly Castillo
Nov 24 '18 at 3:20
I have already resolved the problem. Thanks!
– Beverly Castillo
Nov 24 '18 at 3:20
add a comment |
0
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1
Code will not compile as-is, there's (at least) a missing semi-colon. You should probably cut'n'paste the code to ensure it's correct. Having said that, you can add debug statements to your code printing out (or msgboxing) the two fields. This may help you establish what the problem is. In any case, doesn't
getText()return anEditable? Don't you have to get thetextproperty of that to get the actual string? Not sure (which is why this is a comment rather than an answer) but you may want to look into that.– paxdiablo
Nov 24 '18 at 2:05
I have already resolved the problem. Thanks!
– Beverly Castillo
Nov 24 '18 at 3:20