python解析页类名选择错误
使用beautifulsoup写入python程序以解析html页面:python解析页类名选择错误,python,Python,使用beautifulsoup写入python程序以解析html页面: soup_content = BeautifulSoup(u_f_page, "html.parser") if soup_content.find("li", attrs={"class": "_698"}) == None: u_f_name = "" u_f_c_unit = "" else: for u_f_c_li in soup_content.find_
soup_content = BeautifulSoup(u_f_page, "html.parser")
if soup_content.find("li", attrs={"class": "_698"}) == None:
u_f_name = ""
u_f_c_unit = ""
else:
for u_f_c_li in soup_content.find_all("li",attrs={"class":"_698"}):
print(u_f_c_li)
当我检查打印结果时,我发现所有的
<li class="_698">
...
...
</li>
因为
<li class="followListItem _1fic _698">
在类名中还包括“_698”,因此也可以获取其内容,但我不需要它的内容,如何仅获取class=“_698”,而不获取class=“followListItem”\u 1fic\u 698”
你能帮我解决这个问题吗?这个代码有点乱,但它会给你想要的结果
soup_content = BeautifulSoup(html, "html.parser")
if soup_content.find("li", attrs={"class": "_698"}) is None :
u_f_name = ""
u_f_c_unit = ""
else :
li_list = [ li for li in soup_content.find_all("li", {"class":"_698"}) if all(x == '_698' for x in li.get('class')) ]
for u_f_c_li in li_list :
print(u_f_c_li)
好的,但是如果有其他
<li class="followListItem _1fic _698">....</li>
for u_f_c_li in soup_content.find_all("li",attrs={"class":"_698"}):
<li class="followListItem _1fic _698">
soup_content = BeautifulSoup(html, "html.parser")
if soup_content.find("li", attrs={"class": "_698"}) is None :
u_f_name = ""
u_f_c_unit = ""
else :
li_list = [ li for li in soup_content.find_all("li", {"class":"_698"}) if all(x == '_698' for x in li.get('class')) ]
for u_f_c_li in li_list :
print(u_f_c_li)