漂亮的刮擦交替跳水

Question

我正在努力把我作为学习实验写的一个文件刮下来。看起来是这样的：

<div class="container">
    <div class="date">1st</div>
    <div class="events">
        <div class="meeting">
            <span class="name">Bob</span>
        </div>
    </div>
    <div class="date">2nd</div>
    <div class="event">
        <div class="meeting">
            <span class="name">Emma</span>
            <span class="name">Frank</span>
            <span class="name">Charlie</span>
        </div>
    </div>
    <div class="date">3rd</div>
    <div class="event">
        <div class="meeting">
            <span class="name">Lisa</span>
            <span class="name">Tony</span>
        </div>
    </div>
</div>

我想报废的数据，所以它返回的跨度与关联日期。例如：

data = [['1st', 'bob'], ['2nd', 'Emma', 'Frank' 'Charlie'], ['3rd', 'Lisa', 'Tony']]

我遇到的问题是，当我使用以下命令勉强完成时，Div的date和event在同一级别上：

for data in schedule_soup.find_all('div', 'container'):
    for date in data.find_all('div', 'date'):
        print(date)
    for name in data.find_all('span', 'name'):
        print(name)

我明白了：

<div class="date">1st</div>
<div class="date">2nd</div>
<div class="date">3rd</div>
<span class="name">Bob</span>
<span class="name">Emma</span>
<span class="name">Frank</span>
<span class="name">Charlie</span>
<span class="name">Lisa</span>
<span class="name">Tony</span>

Answer 1

您可以使用zip函数：

final_list=[]
dates = soup.find_all('div', 'date')
meetings = soup.find_all('div', 'meeting')
for date1, meeting in zip(dates, meetings):
    temp_list=[]
    temp_list.append(date1.text)
    [temp_list.append(x.text) for x in meeting.find_all('span')]
    final_list.append(temp_list)
print (final_list)

Answer 2

试着使用下面的代码，它为我工作

final_list=[]
dates = soup.find_all('div', 'date')
for c in range(len(dates)):
    temp_list=[]
    temp_list.append(dates[c].text)
    meeting = soup.find_all('div', 'meeting')
    meeting = BeautifulSoup(str(meeting[c]),'html.parser')
    for name in meeting.find_all('span','name'):
        temp_list.append(name.text)
    final_list.append(temp_list)
print(final_list)

输出

[['1st', 'Bob'], ['2nd', 'Emma', 'Frank', 'Charlie'], ['3rd', 'Lisa', 'Tony']]