没有class属性美丽的汤提取元素 [英] beautiful soup extract element with no class attribute

问题描述

我需要导航到特定类型的HTML元素，但有一些类型的页面上的有许多不同类类型的许多这样的元件。我需要它不具有任何类属性。我应该找一台带班=''或者是有一些其他的方式？

I need to navigate to an html element of a particular type, however there are many such elements of that type on the page with many different class types. I need one which does NOT have any class attribute. SHould I look for one with class = '' or is there some other way?

推荐答案

使用

soup.findAll(attrs={'class': None})

从文档报价：

如果你需要把进口的属性，其名称是Python的保留字，像类的限制，或者您可以使用ATTRS;或属性，其名称都是非关键字参数来美丽的汤搜索方式：名称，递归限制，文本或ATTRS本身。

You can use attrs if you need to put restrictions on attributes whose names are Python reserved words, like class, for, or import; or attributes whose names are non-keyword arguments to the Beautiful Soup search methods: name, recursive, limit, text, or attrs itself.

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