如何在jQuery中使用：not和hasClass（）来获取没有类的特定元素

Question

I have this line of code:

$('#sitesAccordion .groupOfSites').click(function() {
    var lastOpenSite = $(this).siblings().hasClass(':not(.closedTab)');
    console.log(lastOpenSite);
});

I get "false" instead of getting one of the other elements (assuming that there is one - and there must be). I guess the problem is with:

.hasClass(':not(.closedTab)');

What is the problem?

My purpose is to create my own accordion (without using jQuery UI)

and I am trying to write it like this:

$('#sitesAccordion .groupOfSites').click(function() {

    //Get the last opened tab
    var lastOpenSite = $(this).siblings().hasClass(':not(.closedTab)');

    //Close last opened tab and add class
    lastOpenSite.hide().toggleClass('closedTab');

    //Open the current Tab
    $(this).children('.accordionContent').toggle('fast');

    // remove class from open tab
    $(this).toggleClass('closedTab');


});

Is this the best way? thanks, Alon

Answer 1

Use the not function instead:

var lastOpenSite = $(this).siblings().not('.closedTab');

hasClass only tests whether an element has a class, not will remove elements from the selected set matching the provided selector.

Answer 2

It's much easier to do like this:

if(!$('#foo').hasClass('bar')) { 
  ...
}

The ! in front of the criteria means false, works in most programming languages.

Answer 3

jQuery's hasClass() method returns a boolean (true/false) and not an element. Also, the parameter to be given to it is a class name and not a selector as such.

For ex: x.hasClass('error');

Answer 4

你也可以使用jQuery - is（selector）方法 ：

var lastOpenSite = $(this).siblings().is(':not(.closedTab)');

Answer 5

我不知道在原始发布时这是否属实，但兄弟姐妹方法允许选择器，因此减少OP列出的应该起作用。

 $(this).siblings(':not(.closedTab)');