Python:从列表中解析只打印最后一项，而不是全部？

Question

我的代码:

from urllib2 import urlopen
from bs4 import BeautifulSoup

url = "https://realpython.com/practice/profiles.html"

html_page = urlopen(url)
html_text = html_page.read()

soup = BeautifulSoup(html_text)

links = soup.find_all('a', href = True)

files = []
base = "https://realpython.com/practice/"


def page_names():
    for a in links:
        files.append(base + a['href'])

page_names()

for i in files:
    all_page = urlopen(i)

all_text = all_page.read()
all_soup = BeautifulSoup(all_text)
print all_soup

解析的前半部分收集三个链接，后半部分应该打印出它们的所有 html。

遗憾的是，它只打印最后一个链接的 html。

可能是因为

for i in files:
    all_page = urlopen(i)

它以前使用 8 行代码为 for i in files: purpose 服务，但我想清理它并将其归结为这两行。嗯，显然不是，因为它不起作用。

虽然没有错误!

Answer 1

您只在循环中存储最后一个值，您需要在循环内移动所有赋值和打印:

for i in files:
    all_page = urlopen(i)
    all_text = all_page.read()
    all_soup = BeautifulSoup(all_text)
    print all_soup

如果您要使用函数，我会传递参数并创建列表，否则您可能会得到意想不到的输出:

def page_names(b,lnks):
    files = []
    for a in lnks:
        files.append(b + a['href'])
    return files


for i in page_names(base,links):
    all_page = urlopen(i)
    all_text = all_page.read()
    all_soup = BeautifulSoup(all_text)
    print all_s

然后您的函数可以返回一个列表推导式:

def page_names(b,lnks):
    return [b + a['href'] for a in lnks]