jQuery show element如果不为空

I have the following jQuery code:

我有以下jQuery代码:

var agencyToolbar = $('#agency-header-toolbar');
if ($(":not(:empty)", agencyToolbar)) {
    agencyToolbar.show();
}

The code above works, but I wasn't sure if it's a proper way to use a jQuery selector.

上面的代码有效,但我不确定它是否是使用jQuery选择器的正确方法。

2 个解决方案

#1

Your code will always fire the show as $(":not(:empty)", agencyToolbar) creates a jQuery object which is never null so is always "truthy".

您的代码将始终以$(“:not(:empty)”,agencyToolbar)创建一个永远不为null的jQuery对象来激活该节目,因此始终是“真实的”。

You wanted to use is (which returns a boolean result) like

你想使用is(返回一个布尔结果)之类的

if (agencyToolbar.is(':not(:empty))')){

or use a single selector like

或使用单个选择器

$("#agency-header-toolbar:not(:empty)").show();

If you need it to be a tad faster, do the id selector first as its own operation, then filter it with not()

如果你需要它更快一点,首先将id选择器作为自己的操作,然后用not()过滤它

$("#agency-header-toolbar").not(":empty").show();

#2

Assuming that you want to select #agency-header-toolbar which are not empty. It is more simpler to use

假设您要选择非空的#agent-header-toolbar。它使用起来更简单

var agencyToolbar = $("#agency-header-toolbar:not(:empty)");
agencyToolbar.show();