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python - 尝试将抓取的数据添加到列表时仅返回一个值，Python

问题描述

我正在尝试从网站上抓取数据，我的代码的第一部分似乎有效。但是，当我尝试将该数据添加到列表中时，它仅从抓取的数据中返回一个值。我想也许它被识别为字符串，尝试了分隔符，但没有用。请帮忙！

我使用了 Goodgle Colab

提前致谢！

import requests
from bs4 import BeautifulSoup
url = "https://athletics.baruch.cuny.edu/sports/mens-swimming-and-diving/roster"
page = requests.get(url)

page.content  ##content of classes, for the problem we need "sidearm-roster-player-height" class
soup = BeautifulSoup(page.content, "html.parser")
height_swimmers = soup.findAll('span', class_ = 'sidearm-roster-player-height')
print(height_swimmers)

for text in height_swimmers:
  height = text.get_text()
  print(height)
height_list = []
height_list.append(height)

print(height_list)

标签： pythonweb-scraping

解决方案

在 for 循环之后，您只需添加heightto的最后一个值height_list。

为了在列表中打印所有抓取的结果，请使用

height_list = []

for text in height_swimmers:
    height = text.get_text()
    print(height)
    height_list.append(height)

代替

for text in height_swimmers:
    height = text.get_text()
    print(height)
height_list = []
height_list.append(height)

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