[英]jQuery Selecting Elements That Have Class A or B or C
I working on something where I need two functions. 我在需要两个功能的地方工作。
1 - I need to look at a group of children under the same parent and based on a class name, "active", add that element's ID to an array. 1-我需要查看一组在同一父项下的孩子,并基于类名“ active”,将该元素的ID添加到数组中。
['foo-a','foo-b','foo-d']
2 - I then iterate through all the children in another parent and for each element I want to find out if it has any class name that match the ids in the array. 2-然后,我遍历另一个父级中的所有子级,对于每个元素,我想找出它是否具有与数组中的ID相匹配的类名。
Does this element have class foo-a, foo-b or foo-d? 这个元素是否具有foo-a,foo-b或foo-d类?
For the first part, I'd use map
and get
: 对于第一部分,我将使用map
并get
:
var activeGroups = $('#parent .active').map(function() {
return this.id;
}).get();
This gives you an array of id values (say, ['foo-a', 'foo-d']
). 这为您提供了一组ID值(例如['foo-a', 'foo-d']
)。 You can then make a selector like .foo-a, .foo-b, .foo-c
(the multiple selector ) using join
: 然后,您可以使用join
创建一个.foo-a, .foo-b, .foo-c
之类的选择器 ( 多重选择器 ):
var activeSelector = '.' + activeGroups.join(', .');
This makes a valid jQuery selector string, eg '.foo-a, .foo-d'
. 这将生成一个有效的jQuery选择器字符串,例如'.foo-a, .foo-d'
。 You can then use this selector to find the elements you want using find
: 然后,您可以使用此选择器使用find
查找所需的元素:
var activeEls = $('#secondParent').find(activeSelector);
You can then do whatever you need to with activeEls
. 然后,您可以使用activeEls
做任何您需要做的事情。
var active = $("#foo").find(".active").map(function() {
return this.id;
}).get();
$("#anotherParent *").each(function() {
var that = this;
var classes = $(this).attr("class");
if(classes.indexOf(" ") !== -1) {
classes = classes.split(" ");
} else {
classes = [ classes ];
}
$.each(classes, function(i, val) {
if($.inArray(val, active)) {
// this element has one of 'em, do something with it
$(that).hide();
}
});
});
There's always .is('.foo-a, .foo-b, .foo-d')
. 总是有.is('.foo-a, .foo-b, .foo-d')
。 Or if you actually just wanted to select them, instead of iterating and deciding for each element, $('.foo-a, .foo-b, .foo-d', startingPoint)
. 或者,如果您实际上只是想选择它们,而不是为每个元素进行迭代和决定,则使用$('.foo-a, .foo-b, .foo-d', startingPoint)
。
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