I am misunderstanding a jQuery array

Question

This code doesn't work

var next = $("#orders").find(".next");
            if (next.length == 1) {
                var address = $(next[0]).find(".directionsAddress");
                var destination = $(address[0]).text();
            }

It is suppose to find one div with a class of "next" that I know exists on the page, then within that one item of the result set array, there will be one div with a class name of directionsAddress. The "next" array is coming back with a length of 1, so it looks like the problem is with my $(next[0]).find because the address array is coming back as 0 length and I am making a syntax error of some sort that I don't understand.

1. You can simplify your code to this: var destination = $('#orders .next:first .directionsAddress:first').text(). 2. If your element exists it will return the text. 3. Don't forget to wrap all with $(document).ready(). — matewka
– matewka, Commented Aug 20, 2015 at 19:11

indubitablee · Accepted Answer · 2015-08-20 19:18:47Z

0

are you looking to do something like this?

$(document).ready(function() {
	alert($('#orders div.next:first div.directionsAddress:first').text());
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id = "orders">
    <div class="next">
        <div class="directionsAddress">THIS IS WHAT I WANT</div>
        <div class="directionsAddress">This is not the one I want</div>
    </div>
    <div class="next">
        <div class="directionsAddress">This is not the one I want</div>
    </div>
    <div class="next">
        <div class="directionsAddress">This is not the one I want</div>
    </div>
</div>

answered Aug 20, 2015 at 19:18

indubitablee

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