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Mark Bradshaw
Courses Plus Student 7,658 PointsStuck on a jquery quiz question.
Here's the task: On line 11 in app.js, implement the isUsernamePresent function. Return true if the username length is greater than 0.
Here's my code: My answer:
function isUsernamePresent() {
if(isUsernamePresent().val().length > 0;)
}
Return true;
4 Answers
Jason Anello
Courses Plus Student 94,610 PointsHi Mark,
You have a few errors.
You used a capital 'R' for `return. You have also placed your return statement outside your function. You will need to get it back inside the closing curly brace.
You also have a misplaced semicolon at the end of your if condition and you don't want to call the function inside of itself.
You should use the $username variable which is storing the username input as pointed out by Michael.
function isUsernamePresent() {
if ($username.val().length > 0) {
return true;
}
}
Also, whenever you have this pattern of, "if something is true, return true", you can just return that something directly. Making your code a little simpler.
function isUsernamePresent() {
return $username.val().length > 0;
}
Jonathan Grieve
Treehouse Moderator 91,252 PointsYou have a capital R on your return keyword.
return is a specific keyword which isn't recognised if you capitalise it. :)
Michael Escoto
30,028 PointsI think the problem is you're calling the function name in the if statement.
Instead you'll need just use:
if($username.val().length > 0)
Vittorio Somaschini
33,371 PointsYes Michael and Jonathan are both right Mark.
return needs to be all lowercase.
Also, the if condition should check the length of the value inside the $username variable (declared by the code at the beginning on line 5.
So $username.val().length is the right thing to do here
Mark Bradshaw
Courses Plus Student 7,658 PointsMark Bradshaw
Courses Plus Student 7,658 PointsThis clarification was exactly what I needed. Thanks.