jQuery: Selecting all elements where attribute is greater than a value

Question

I know that to filter an element with an atttribute called attrName which has value attrValue I do:

filter("[attrName='attrValue']")

but looking at the docs http://api.jquery.com/category/selectors/ I can't see an option to select all elements s.t. attrName>attrValue

Will this work

   filter("[attrName>'attrValue']")

If you are looking for elements with attributes greater than zero, you can use `[attribute!=0]` — Damien, Jun 30 '16 at 05:42

score 113 · Accepted Answer · answered Apr 10 '10 at 14:14

113

You can do this using the function overload of .filter(), like this:

.filter(function() {
  return $(this).attr("attrName") > "someValue";
})

answered Apr 10 '10 at 14:14

Nick Craver

score 29 · Answer 2 · answered Apr 10 '10 at 14:18

29

The solution is jQuery.filter():

$("selector").filter(function() {
    return  $(this).attr("my-attr") > 123;
});

answered Apr 10 '10 at 14:18

Crozin

score 29 · Answer 3 · answered Apr 10 '17 at 19:40

29

Be careful, if you are playing with integers make sure you are using parseInt().

$("selector").filter(function() {
    return parseInt($(this).attr("my-attr")) > 123;
});

answered Apr 10 '17 at 19:40

7

Agreed! Some users may see that "3" > "1" is true and think that they don't need to parse the string to an integer before comparing. This works because the ASCII value of "3" is higher than the ASCII value for "1". But things go south when you are comparing values with more than one digit (e.g. "33" > "4" is false while 33>4 is true). – Arthur Hebert-Ryan Jul 16 '18 at 18:12

score 0 · Answer 4 · answered Mar 05 '21 at 06:31

0

For ES6 arrow function

$("selector").filter((index, el) => {
    return parseInt($(el).attr("attrName")) > 123;
});

answered Mar 05 '21 at 06:31

Fangxing

4 Answers4