硒：CSS查詢

Question

<b> 
<s> 
    <b>---</b> (1) 
    <b>---</b> (2) 
</s> 
</b> 
<s> 
    <b>---</b> (3) 
    <b>---</b> (4) 
</s> 

1. 我想可能標籤，但僅是標籤的孩子的？ 如何在這種情況下指定CSS？ 這是好的：self.selenium.get_css_count("css=s b") 這會計算所有的b關係，s作爲父母，b作爲孩子。 4在上面的例子？如果沒有，你能幫助解決這個問題嗎？

2. 我想單擊標籤下的第四個（4）標籤。我會怎麼做？ 如果我這樣做：self.selenium.click("css=s b:nth(1)) 它會選擇（2），那麼我如何指定(s b):nth(3)？ 因爲s是b的父親。我想點擊第四種關係。 我希望這是有道理的。

3. 就是第n（）和第n-的類型（）之間的區別

感謝 陽光

Answer 1

你可以使用XPath的，而不是CSS選擇器來做到這一點。

您1.

get_xpath_count("//s/b") 

和2

click("//s[2]/b[2]") 

Answer 2

什麼：

B> S：第一個孩子> b

要點擊第4個孩子：

：n-child（4）

第3個問題：nth-of-type vs nth-child