CODEWITHSUNDEEP
javascriptjquery
Steve C.
Steve C.

Reputation: 577

Why Isn't the jQuery .index() function behaving like I expect here?

I am having issues understanding why my jQuery index() function is not performing the way I expect it to. Perhaps I'm not understanding this. Let me show you.

I set up a new array. We'll use this later:

var mySources = new Array();

I have 5 images on my page:

<div id="homeSlides">
       <img src="../images/homeImages/00.jpg" width="749" height="240" alt="myAltTag" />
       <img src="../images/homeImages/01.jpg" width="749" height="240" alt="myAltTag"  />
       <img src="../images/homeImages/02.jpg" width="749" height="240" alt="myAltTag" />
       <img src="../images/homeImages/03.jpg" width="749" height="240" alt="myAltTag" />
       <img src="../images/homeImages/04.jpg" width="749" height="240" alt="myAltTag" />

I put them all in an jQuery object, like this:

var myImages = $("#homeSlides img");

Now, for each one, I extract the src attribute and push it into the mySources array like this:

 $(myImages).each( function(index) {
        mySources[index] = $(this).attr('src');
});

Now, I will run the array to the FireBug console to make sure it worked.

This...

console.log(mySources);

...returns this...

["../images/homeImages/00.jpg", "../images/homeImages/01.jpg", "../images/homeImages/02.jpg", "../images/homeImages/03.jpg", "../images/homeImages/04.jpg"]

...which is what I expect.

Now I do this:

var myVar = $(myImages).eq(2).attr('src');

Tracing that variable now...

console.log(myVar);

...returns this...

../images/homeImages/02.jpg

But then when I do this...

console.log('Index of ' + myVar + " is: " + ($(myVar).index( mySources )) );

...it returns this:

 Index of ../images/homeImages/02.jpg is: -1

This is throwing me. Why would it return -1 for "not found" when it should be retunring 2. It does mat the second slot in the mySources array, no?

Upvotes: 1

Views: 145

Answers (2)

charlietfl
charlietfl

Reputation: 171690

You are trying to use index() on an array. It is not intended to be used for anything other than jQuery element objects.

Using $.inArray() would be the better method

Upvotes: 3

John Fable
John Fable

Reputation: 1091

You may have the function backwards. Can you try:

$(mySources).index( myVar )

Upvotes: 2

Related Questions