Reputation: 10228
Here is my code:
$('.mycls1 table tr td, .mycls2 table tr td').css('opacity', '.2');
Now I want to use two variables instead of .mycls1
and .mycls2
, Like this:
var mycls1 = $(`.mycls1`),
mycls2 = $(`.mycls2`);
So I have to use .find()
to select that, like this:
mycls1.find('table tr td').css('opacity', '.2');
mycls2.find('table tr td').css('opacity', '.2');
See? In this case (using .find()
) forces me to use to two separate lines of code for selecting those two elements.
Anyway, I want to know is it possible to use both .find()
and OR (,
) in the same line?
Noted that in my example, declaring two variables and initializing those two elements in them is not effective, but in reality it is. So I just need to learn the concept of how can I do that.
Upvotes: 0
Views: 43
Reputation: 87203
You can use add()
to add elements to collection.
mycls1.add(mycls2)....
This will add the elements from mycls2
.
var mycls1 = $(`.mycls1`),
mycls2 = $(`.mycls2`);
mycls1.add(mycls2).css('opacity', '.2');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="mycls1">
<table>
<tr>
<td>1</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
</table>
</div>
<div class="mycls2">
<table>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
</table>
</div>
Upvotes: 2
Reputation: 812
i think you should try below code
var mycls = $(`.mycls1, .mycls2`);
mycls.find('table tr td').css('opacity', '.2');
Upvotes: 0