Apply selector to elements that have already been selected

up vote
0
down vote

favorite

I feel like I am asking something really dumb but perhaps it's not my day. If I have a selected element already, e.g:

let tables = $('table');

But now I want to apply another selector like .some-class on top of those tables, but without creating a new jQuery object like this

$('table.some-class')

How would I do it?

edited Nov 22 at 13:03

Mohammad

14.1k93259

asked Nov 22 at 12:56

dabadaba

3,06564086

May be like $(tables+'.some-class');
– Ayaz Shah
Nov 22 at 12:57

1

What about find()
– Ankit Agarwal
Nov 22 at 12:58

@AnkitAgarwal $('table').find('.some-class') is not the same than $('table.some-class').
– dabadaba
Nov 22 at 13:16

add a comment |

up vote
0
down vote

favorite

I feel like I am asking something really dumb but perhaps it's not my day. If I have a selected element already, e.g:

let tables = $('table');

But now I want to apply another selector like .some-class on top of those tables, but without creating a new jQuery object like this

$('table.some-class')

How would I do it?

edited Nov 22 at 13:03

Mohammad

14.1k93259

asked Nov 22 at 12:56

dabadaba

3,06564086

May be like $(tables+'.some-class');
– Ayaz Shah
Nov 22 at 12:57

1

What about find()
– Ankit Agarwal
Nov 22 at 12:58

@AnkitAgarwal $('table').find('.some-class') is not the same than $('table.some-class').
– dabadaba
Nov 22 at 13:16

add a comment |

up vote
0
down vote

favorite

I feel like I am asking something really dumb but perhaps it's not my day. If I have a selected element already, e.g:

let tables = $('table');

But now I want to apply another selector like .some-class on top of those tables, but without creating a new jQuery object like this

$('table.some-class')

How would I do it?

edited Nov 22 at 13:03

Mohammad

14.1k93259

asked Nov 22 at 12:56

dabadaba

3,06564086

I feel like I am asking something really dumb but perhaps it's not my day. If I have a selected element already, e.g:

let tables = $('table');

But now I want to apply another selector like .some-class on top of those tables, but without creating a new jQuery object like this

$('table.some-class')

How would I do it?

javascript jquery html jquery-selectors

edited Nov 22 at 13:03

Mohammad

14.1k93259

asked Nov 22 at 12:56

dabadaba

3,06564086

edited Nov 22 at 13:03

Mohammad

14.1k93259

asked Nov 22 at 12:56

dabadaba

3,06564086

edited Nov 22 at 13:03

Mohammad

14.1k93259

edited Nov 22 at 13:03

Mohammad

14.1k93259

edited Nov 22 at 13:03

Mohammad

14.1k93259

asked Nov 22 at 12:56

dabadaba

3,06564086

asked Nov 22 at 12:56

dabadaba

3,06564086

asked Nov 22 at 12:56

dabadaba

3,06564086

May be like $(tables+'.some-class');
– Ayaz Shah
Nov 22 at 12:57

1

What about find()
– Ankit Agarwal
Nov 22 at 12:58

@AnkitAgarwal $('table').find('.some-class') is not the same than $('table.some-class').
– dabadaba
Nov 22 at 13:16

add a comment |

May be like $(tables+'.some-class');
– Ayaz Shah
Nov 22 at 12:57

1

What about find()
– Ankit Agarwal
Nov 22 at 12:58

@AnkitAgarwal $('table').find('.some-class') is not the same than $('table.some-class').
– dabadaba
Nov 22 at 13:16

May be like $(tables+'.some-class');
– Ayaz Shah
Nov 22 at 12:57

What about find()
– Ankit Agarwal
Nov 22 at 12:58

@AnkitAgarwal $('table').find('.some-class') is not the same than $('table.some-class').
– dabadaba
Nov 22 at 13:16

add a comment |

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

You need to use .filter() to adding filter to variable selector.

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

answered Nov 22 at 13:00

Mohammad

14.1k93259

add a comment |

up vote
0
down vote

With the following code you would find all elements with class .some-class under the elements with tag name table.

let tables = $('table').find('.some-class');

Of course, it is just for the example, otherwise you can simply do:

let tables = $('table .some-class');

In your question it is not clear if you want children elements or just filter the elements. If you want the tables with a given class, you would do:

let tables = $('table').filter('.some-class');

answered Nov 22 at 12:58

LaurentG

7,86573553

I never mentioned children elements. I am trying to apply the selector on the original element.
– dabadaba
Nov 22 at 13:14

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

You need to use .filter() to adding filter to variable selector.

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

answered Nov 22 at 13:00

Mohammad

14.1k93259

add a comment |

up vote
1
down vote

accepted

You need to use .filter() to adding filter to variable selector.

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

answered Nov 22 at 13:00

Mohammad

14.1k93259

add a comment |

up vote
1
down vote

accepted

You need to use .filter() to adding filter to variable selector.

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

answered Nov 22 at 13:00

Mohammad

14.1k93259

You need to use .filter() to adding filter to variable selector.

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

let tables = $('table');

tables = tables.filter('.some-class');

tables.css('color', 'red');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<table>

  <tr>

    <td>table</td>

  </tr>

</table>

<table class="some-class">

  <tr>

    <td>table has .some-class</td>

  </tr>

</table>

answered Nov 22 at 13:00

Mohammad

14.1k93259

answered Nov 22 at 13:00

Mohammad

14.1k93259

answered Nov 22 at 13:00

Mohammad

14.1k93259

answered Nov 22 at 13:00

Mohammad

14.1k93259

add a comment |

up vote
0
down vote

With the following code you would find all elements with class .some-class under the elements with tag name table.

let tables = $('table').find('.some-class');

Of course, it is just for the example, otherwise you can simply do:

let tables = $('table .some-class');

In your question it is not clear if you want children elements or just filter the elements. If you want the tables with a given class, you would do:

let tables = $('table').filter('.some-class');

answered Nov 22 at 12:58

LaurentG

7,86573553

I never mentioned children elements. I am trying to apply the selector on the original element.
– dabadaba
Nov 22 at 13:14

add a comment |

up vote
0
down vote

With the following code you would find all elements with class .some-class under the elements with tag name table.

let tables = $('table').find('.some-class');

Of course, it is just for the example, otherwise you can simply do:

let tables = $('table .some-class');

In your question it is not clear if you want children elements or just filter the elements. If you want the tables with a given class, you would do:

let tables = $('table').filter('.some-class');

answered Nov 22 at 12:58

LaurentG

7,86573553

I never mentioned children elements. I am trying to apply the selector on the original element.
– dabadaba
Nov 22 at 13:14

add a comment |

up vote
0
down vote

With the following code you would find all elements with class .some-class under the elements with tag name table.

let tables = $('table').find('.some-class');

Of course, it is just for the example, otherwise you can simply do:

let tables = $('table .some-class');

In your question it is not clear if you want children elements or just filter the elements. If you want the tables with a given class, you would do:

let tables = $('table').filter('.some-class');

answered Nov 22 at 12:58

LaurentG

7,86573553

With the following code you would find all elements with class .some-class under the elements with tag name table.

let tables = $('table').find('.some-class');

Of course, it is just for the example, otherwise you can simply do:

let tables = $('table .some-class');

In your question it is not clear if you want children elements or just filter the elements. If you want the tables with a given class, you would do:

let tables = $('table').filter('.some-class');

answered Nov 22 at 12:58

LaurentG

7,86573553

answered Nov 22 at 12:58

LaurentG

7,86573553

answered Nov 22 at 12:58

LaurentG

7,86573553

answered Nov 22 at 12:58

LaurentG

7,86573553

I never mentioned children elements. I am trying to apply the selector on the original element.
– dabadaba
Nov 22 at 13:14

add a comment |

I never mentioned children elements. I am trying to apply the selector on the original element.
– dabadaba
Nov 22 at 13:14

I never mentioned children elements. I am trying to apply the selector on the original element.
– dabadaba
Nov 22 at 13:14

add a comment |

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