86 lines
1.8 KiB
HTML
86 lines
1.8 KiB
HTML
<!DOCTYPE html>
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<html lang="en">
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<head>
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<meta charset="UTF-8">
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<meta http-equiv="X-UA-Compatible" content="IE=edge">
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<meta name="viewport" content="width-device-width,initial-scale=1.0">
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<title>选择器权重</title>
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<style>
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#ip {
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color: blue !important;
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}
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div .top p {
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color: salmon !important;
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}
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p {
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color: lightblue !important;
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}
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</style>
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</head>
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<body>
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<div id="app">
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<div class="top">
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<p class="top-p tp" id="ip"> <span>天地不仁</span> ,以万物为刍狗,圣人不仁,以百姓为刍狗。 </p>
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</div>
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<div class="bottom">
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<p> <span>天地不仁</span> ,以万物为刍狗,圣人不仁,以百姓为刍狗。 </p>
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</div>
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<p> <span>天地不仁</span> ,以万物为刍狗,圣人不仁,以百姓为刍狗。 </p>
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</div>
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</body>
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</html>
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<!--之前已证:
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继承的样式的权重 < 通配符选择器的权重 < 标签选择器的权重 < class 选择器的权重 < id选择器的权重 < !important
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!important 给个最高权重
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-->
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<!-- 每种权重 都会有当前权重的值
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继承的样式的权重 0
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通配符选择器的权重 01
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标签选择器的权重 001
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class 选择器的权重 0001
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id选择器的权重 00001
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!important 0000000001
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计算权重:
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1.肯定不能进位
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2.同选择器相加
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3.倒着逐一判断大小
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提一:
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#ip
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div .top p p p p p...n (n>10)
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计算:
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00001
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00(n>10)10
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题二: 权重等于
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A. #app .top-P p
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B. .top-p p #ip
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题三:权重A>B
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A: #app .top-p div p
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B: .top-p p #ip
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--> |