How to use muliple variables in same function

Question

What is the proper way to combine multiple variables

I tried this but didn't work.

varONE = "td.class1";
varTWO = "td.class2";

$('table').find(varONE,varTWO).css('background', 'red');

AmmarCSE · Accepted Answer · 2015-06-14 20:29:43Z

3

You need the comma in the string itself like

$('table').find('td.class1,td.class2').css('background', 'red');

or

varONE = "td.class1";
varTWO = "td.class2";

$('table').find(varONE+','+varTWO).css('background', 'red');

The only overloads for find() are a selector(string) or an element. However, there isn't an overload which accepts multiple selectors(stings) as separate parameters

edited Jun 14, 2015 at 20:29

answered Jun 14, 2015 at 20:11

AmmarCSE

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Comments

Robert W. Hunter · Accepted Answer · 2015-06-14 20:16:02Z

2

Well, you actually need to set the selectors the jquery way, for example

varONE = $("td.class1");
varTWO = $("td.class2");

$('table').find(varONE,varTWO).css('background', 'red');

Then your script will work.

You can also use a concatenated string

varONE = "td.class1";
varTWO = "td.class2";

$('table').find(varONE + "," + varTWO).css('background', 'red');

And it will also work. I recommend you the first solution as it is cleaner, but you can choose.

answered Jun 14, 2015 at 20:16

Robert W. Hunter

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2 Comments

Oka Over a year ago

@MShack You need to invoke those functions to return the results. $('.two_column_layout').find(getPlayerByeQuery() + ',' + getPlayerQuery()).css('background', 'red'); - Works just fine.

MShack Over a year ago

ty , forgot the () aside each one

Vic · Accepted Answer · 2015-06-14 20:12:24Z

1

same as concatenating strings

$('table').find(varONE + ',' + varTWO).css('background', 'red');

answered Jun 14, 2015 at 20:12

Vic

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Comments

Oka · Accepted Answer · 2015-06-14 20:35:31Z

1

As others have stated, the .find method doesn't take multiple selector strings.

If you want to use .find in that way, you could create your own method.

$.fn.argfind = function () {
  return this.find(Array.prototype.join.call(arguments));
};

var one = "td.class1",
    two = "td.class2";

$('table').argfind(one, two).css('background', 'red');

<table>
  <tr>
    <td class="class1">Example</td>
    <td>Text</td>
  </tr>
  <tr>
    <td>Example</td>
    <td class="class2">Text</td>
  </tr>
</table>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

answered Jun 14, 2015 at 20:35

Oka

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