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JQuery CSS 选择器 e.fn.e.init
原标题:JQuery CSS Selector e.fn.e.init

有时我用jquery来测试自己的 cs 。 通常我会使用简单的选择器( 在 铬 Web Developer 控制台) 。

我会键入 $(.topclasddiv) ,然后返回

e.fn.e.init[125]

作为含有节点的javacampic object 。 (它是一个相当复杂的选择器,但有时是简单的)。

为什么我得到这个?它的意思是标签不见了还是我的选手错了?

编辑: html基本上是嵌套的 divs (可能是因为它得到了孩子的divs吗? )

最佳回答

这意味着您的选择包含太多匹配, 开发者的工具缩短了他们在控制台的代表权。

在控制台尝试这个 :

var a = []
for (var i = 0; i < 10; i++) { a.push( foobar ); }

如果您试图检查 a , 您将会看到 [“ foobar”, "foobar", "foobar", "foobar", "foobar", "foobar", "foobar", "foobar", "foobar", "foobar", "foobar", "foobar"," 。 现在, 请添加一些元素 :

for (var i = 0; i < 100; i++) { a.push( foobar ); }

如果您试图检查它, 请看到 < code> Array[110] 。

并不意味着你的选择者错了, 但如果你期望的元素更少, 那可能是错的。

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