Question

我下面有html代码

<div class = "conf">
Brazil vs. Colombia
</ div>
<div class = "targetHour"> 08:00 pm </ div>
</ div>
</ div>
<div class = "matches">
<div class = "conf">
Chilex Argentina
</ div>
<div class = "targetHour"> 08:00 pm </ div>
</ div>
</ div>

我需要获取父div的值和子div的值，而不必重复结果。将每个游戏的时间表与各自的父代联系起来。

这是我的代码pyhton

for nc in soup.find_all('div', attrs={'class': 'league-data'}):
    campeonato = nc.text
    for hr in soup.find('div', attrs={'class': 'match row cf'}).findAll("div",recursive=False):
        print(campeonato + "|" + hr.text)

Answer 1

您可以使用zip()函数将匹配项与相应的时间表绑定：

from bs4 import BeautifulSoup

data = '''<div class = "conf">
Brazil vs. Colombia
</div>
<div class = "targetHour"> 08:00 pm </div>
</div>
</div>
<div class = "matches">
<div class = "conf">
Chilex Argentina
</div>
<div class = "targetHour"> 08:00 pm </div>
</div>
</div>'''

soup = BeautifulSoup(data, 'lxml')

for match, hour in zip( soup.select('div.conf'), soup.select('div.targetHour') ):
    print(match.text.strip(), hour.text.strip())

打印：

Brazil vs. Colombia 08:00 pm
Chilex Argentina 08:00 pm

Answer 2

另一种选择（假设长度列表为偶数）

from bs4 import BeautifulSoup

data = '''<div class = "conf">
Brazil vs. Colombia
</div>
<div class = "targetHour"> 08:00 pm </div>
</div>
</div>
<div class = "matches">
<div class = "conf">
Chilex Argentina
</div>
<div class = "targetHour"> 08:00 pm </div>
</div>
</div>'''

soup = BeautifulSoup(data, 'lxml')
items = [item.text.strip() for item in soup.select('.conf, .targetHour')]
for i in range(0, len(items), 2):
       print(items[i],items[i+1])

在div中获取节点而不重复

2 个答案: