参考了文档 然后根据他的例子做了
其中他文档中的例子是
from BeautifulSoup import BeautifulSoup
doc = ['<html><head><title>Page title</title></head>',
'<body><p id="firstpara" align="center">This is paragraph <b>one</b>.',
'<p id="secondpara" align="blah">This is paragraph <b>two</b>.',
'</html>']
soup = BeautifulSoup(''.join(doc))
print soup.head.nextSibling.name
然后输出了body
但是如果我改成下面这个样子
from BeautifulSoup import BeautifulSoup
html = '''
<html>
<head>
<title>
Page title
</title>
</head>
<body>
<p id="firstpara" align="center">
This is paragraph
<b>
one
</b>
.
</p>
<p id="secondpara" align="blah">
This is paragraph
<b>
two
</b>
.
</p>
</body>
</html>
'''
soup = BeautifulSoup(html)
print soup.head.nextSibling.name
结果会出错，出错信息是
File "/Library/Python/2.7/site-packages/BeautifulSoup.py", line 473, in __getattr__
raise AttributeError, "'%s' object has no attribute '%s'" % (self.__class__.__name__, attr)
AttributeError: 'NavigableString' object has no attribute 'name'
但是如果把上面代码中最后一句使用nextSibling的代码改成下面的形式
print soup.head.nextSibling.nextSibling.name
就又可以正确输出结果了，输出了body
但是一般我们爬取网页返回的都是第二种情况，然后我看了一些开源的抓取其他网页的webservice开源代码，其中也是连续用了两次nextSibling才获取下一个同级元素，想请问下各位大大为什么第二种情况就一定要连续用两次nextSibling才能获得下一个同级dom元素呢，nextSibling不是本意就是下一个同级元素，为什么此处需要用两次才能获取下一个，只用一次就会出现上面那个错误。
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
soup.head.nextSiblingIt should be that the text node behind the head tag has been obtained, right?After testing, beautifulsoup4 only needs one
.nextSiblingto get the<body>element.