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? )
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? )
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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