使用Beautiful Soup找到第三个出现的``标签(Find third occurring `` tag using with Beautiful Soup)

`标签(Find third occurring `

` tag using with Beautiful Soup)

正如标题所示，我试图了解如何找到网站的第三个<p> （例如，我使用以下网站： http ： //www.musicmeter.nl/album/31759 ）。

使用这个问题的答案，我尝试了以下代码

但是这段代码会导致third_paragraph出现以下错误：

我试图查找错误，但我无法弄清楚出了什么问题。

As the title suggests, I'm trying to understand how to find the third occurring <p> of a website (as an example, I used the following website: http://www.musicmeter.nl/album/31759).

Using the answer to this question, I tried the following code

But this code results in the following error for the third_paragraph:

I tried to lookup the error, but I couldn't figure out what is wrong.

最满意答案

你正在使用兄弟姐妹，即复数，所以你得到一个ResultSet / list back，你不能调用.find_next_siblings 。

如果你想要每个下一段，你会使用兄弟姐妹而不是兄弟姐妹 ：

哪个可以链接：

一种更简单的方法是使用nth-of-type ：

你还应该知道找到第三个发生的p与找到你在页面上找到的第一个p的第二个兄弟是不一样的，使用nth-of-type实际上确实找到了页面中的第三个p标签，如果是第一个p没有两个兄弟p标签，那么你的逻辑就会失败。

要使用find逻辑真正获得第三个p，请使用find_next ：

如果你想要前三个使用find_all并将限制设置为3：

使用原始逻辑获得前两个：

如果你只想要x标签然后只解析x标签，那么请确保你理解找到第三个发生的<p>标签和第二个兄弟标签到第一个p标签之间的区别，因为它们可能不同。

You are using siblings i.e plural so you are getting a ResultSet/list back which you cannot call .find_next_siblings on.

If you wanted each next paragraph you would use sibling not siblings:

Which can be chained:

A much simpler way is to use nth-of-type:

You should also be aware that finding the third occurring p is not the same as finding the 2nd sibling to the first p you find on the page, using nth-of-type actually does find the third p tag in the page, if the first p does not have two sibling p tags then your logic will fail.

To really get the third occurring p using find logic just use find_next:

Of if you want the first three use find_all with a limit set to 3:

Of using your original logic to get the first two:

If you only want x tags then only parse x tags, just be sure you understand the difference between finding the third occurring <p> tag and the 2nd sibling to the first p tag as they may be different.