我在使用Beautifulsoup解析带有“class”属性的HTML元素时遇到了麻烦。代码看起来像这样
soup = BeautifulSoup(sdata)
mydivs = soup.findAll('div')
for div in mydivs:
if (div["class"] == "stylelistrow"):
print div
我在脚本完成后的同一行上得到一个错误。
File "./beautifulcoding.py", line 130, in getlanguage
if (div["class"] == "stylelistrow"):
File "/usr/local/lib/python2.6/dist-packages/BeautifulSoup.py", line 599, in __getitem__
return self._getAttrMap()[key]
KeyError: 'class'
如何消除这个错误呢?
当前回答
或者我们可以使用lxml，它支持xpath和非常快!
from lxml import html, etree
attr = html.fromstring(html_text)#passing the raw html
handles = attr.xpath('//div[@class="stylelistrow"]')#xpath exresssion to find that specific class
for each in handles:
print(etree.tostring(each))#printing the html as string
其他回答
下面的方法对我很有效
a_tag = soup.find_all("div",class_='full tabpublist')
使用class_=如果你想在不指定HTML标签的情况下查找元素。
对于单个元素:
soup.find(class_='my-class-name')
对于多个元素:
soup.find_all(class_='my-class-name')
直接的方法是:
soup = BeautifulSoup(sdata)
for each_div in soup.findAll('div',{'class':'stylelist'}):
print each_div
确保你使用了findAll的外壳，它不是findAll
这招对我很管用:
for div in mydivs:
try:
clazz = div["class"]
except KeyError:
clazz = ""
if (clazz == "stylelistrow"):
print div
或者我们可以使用lxml，它支持xpath和非常快!
from lxml import html, etree
attr = html.fromstring(html_text)#passing the raw html
handles = attr.xpath('//div[@class="stylelistrow"]')#xpath exresssion to find that specific class
for each in handles:
print(etree.tostring(each))#printing the html as string
