JQuery .find with multiple selectors

Question

I have this problem with Jquery items[c] is well defined as HTMLElement but when I use .find with more than one selector I get this error:

TypeError: X[g].exec is not a function
http://code.jquery.com/jquery-latest.min.js

when i checked some watches i got what is in the image below:

$(items[c]).find('.received') works fine and returns some elements as there is elements with that class

$(items[c]).find('.receive') works fine too and returns zero elements as there is no elements with that class.

but $(items[c]).find('.received.unseen') returns undefined and bugs. so what is happening here?

EDIT: here is what is inside items[c], from the debugger firefox

EDIT: here is the function where i have the bug and i switched to jquery 2.1.1:

function updateUnseenBell(){
    var m;
    for (var c in items)
        if (items.hasOwnProperty(c) && (m = items[c].gEbCN("chat-partner-tab")[0])) {
        if($(items[c]).find('.received.unseen:not(.shown)').length > 0){
            if (!(m.l2_newMsgBell)) {
                m.appendChild(m.l2_newMsgBell = newMsgBell.cloneNode());
                playSound("message");
            }
        } else if (m.l2_newMsgBell) {
            m.removeChild(m.l2_newMsgBell);
            delete m.l2_newMsgBell;
        }
    }
}

and i reduced it to this minimum for debug but still get the same error:

function updateUnseenBell(){
    for (var c in items) {
        if (items.hasOwnProperty(c)) {
            if ($(items[c]).find('.received.unseen:not(.shown)').length > 0) {
                alert(1);
            } else {
                alert(2);
            }
        }
    }
}

can you add html mark up? is element .received and .unseen same or child?if child just add space between them — guradio
– guradio, Commented Sep 19, 2016 at 3:45
further to @guradio - if you are targetting multple selectors (ie not child) you need a comma between them ... $(items[c]).find('.received, .unseen').... — gavgrif
– gavgrif, Commented Sep 19, 2016 at 3:47
I cannot reproduce this (besides, I'm pretty sure .find handles compound selectors just fine). — BoltClock
– BoltClock, Commented Sep 19, 2016 at 3:51
What is items[c]? what does it contains? Please log it and post it — Abhishek Dhanraj Shahdeo
– Abhishek Dhanraj Shahdeo, Commented Sep 19, 2016 at 3:52
@guradio thank you. no they are classes for same element so there should not be a space in fact my finale purpose is to use this $(items[c]).find('.received.unseen:not(.shown)'). to select all elements in items[c] that have received and unseen classes but not shown. but it does not work even with just ('.received.unseen') — fekiri malek
– fekiri malek, Commented Sep 19, 2016 at 3:55

Shubham Khatri · Accepted Answer · 2016-09-19 08:27:11Z

8

Use

$(items[c]).find('.message.received.unseen')

and that should work.

One other way to solve this will be

$(items[c]).find(".received").find(".unseen").find(":not(.sh‌own)")

Its not an elegant approach but works too.

edited Sep 19, 2016 at 8:27

answered Sep 19, 2016 at 4:17

Shubham Khatri

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6 Comments

fekiri malek Over a year ago

i have tried that but it does not work. item misses an s btw.

fekiri malek Over a year ago

yes thank you i managed to get the full functionality (with :not(.show)) too on the site, and it works fine here .. so I don't know what i'm missing in my code. i posted the function above in the edit.

Shubham Khatri Over a year ago

inside theif (items.hasOwnProperty(c)) { if you do console.log(items[c]) do you get what you attached as an image. Also what does items contain.

fekiri malek Over a year ago

yes i get the right object. i even added var a = items[c]; which i use after: ($(a).find(".received.unseen:not(.shown)"); but still the same.

Shubham Khatri Over a year ago

try var a = $(items[c]).not(":visible"); and then $(a).find(".received.unseen:not(.shown)". It also works in the fiddle. jsfiddle.net/y8tfeuy1/3

|

Marcio Mazzucato · Accepted Answer · 2023-10-23 02:22:18Z

0

Another approach you can use:

$(items[c]).find(".received, .unseen, :not(.sh‌own)");

answered Oct 23, 2023 at 2:22

Marcio Mazzucato

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