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JavaScript AJAX Basics (retiring) AJAX and APIs Stage 4 Challenge Answer

Posted by Sunny Singh
Sunny Singh
1,148 Points

Not understanding .Alternate way to display search term photo not working

why does Instead of defining 

var $searchField = $('#search');

and then

 var animal = $searchField.val();

you can't simply do 

var animal = $('#search').val()

It does not work?

https://w.trhou.se/or41dhimed

2 Answers

Mikkel Rasmussen
Mikkel Rasmussen
31,772 Points
Mikkel Rasmussen
Mikkel Rasmussen
31,772 Points

Its should work maybe its because you are missing a semi colon?

var animal = $('#search').val();
Iain Simmons
MOD
Iain Simmons
Treehouse Moderator 32,305 Points
Iain Simmons
Iain Simmons
Treehouse Moderator 32,305 Points

You would use the two separate variables if you needed to access the actual element as a jQuery object for something else (like changing the value later, adding/changing classes, etc).

But otherwise, yes, your code is a little shorter.