Find elements using part of ID

Question

I have, the div's where id looks like this_div_id_NUMBER, all div's has the different NUMBER part. How I find all div's just using this_div_id part of id ?

Answer 1

you can use querySelectorAll to hit partial attribs, including ID:

document.querySelectorAll("[id^='this_div_id']")

the ^ next to the equal sign indicates "starts with", you can use * instead, but it's prone to false-positives.

you also want to make sure to use quotes (or apos) around the comapare value in attrib selectors for maximum compatibility on querySelectorAll; in jQuery and evergreen browsers it doesn't matter, but in vanilla for old browsers it does matter.

EDIT: late breaking requirement needs a more specific selector:

document.querySelectorAll("[id^='this_div_id']:not([id$='_test_field'])");

the not() segment prevents anything ending with "_test_field" from matching.

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proof of concept / demo: http://pagedemos.com/partialmatch/

Answer 2

querySelectorAll

querySelectorAll takes CSS selectors and returns a HTMLNodeList (a kind of array) of all the elements matching that css selector.

The css selector ^ (starts with) can be used for your purpose. Learn more about it in this article.

For your case, it would be document.querySelectorAll("[id^='this_div_id']");

Note that querySelectorAll doesn't return a live node list. That means, any changes to the dom would not be updated live in the return value of the function.

Another method would to assign all your elements a specific class and then you can use getElementsByClassName (which is much faster than querySelector).

var divs = document.getElementsByClassName("divClass");

Answer 3

Try this selector:

[id^="this_div_id_"]

Pure JavaScript: (reference)

document.querySelectorAll('[id^="this_div_id_"]')

jQuery: (reference)

$('[id^="this_div_id_"]')

CSS: (reference)

[id^="this_div_id_"] {
    /* do your stuff */
}

Why is this working?

With the [] in the selector, you can select attributes. Use '=' to match exactly the value or use the '^=' to check if the value starts with. Read more about this here.