Flask里一个也没有，但我不明白为什么

Question

我一直在尝试建立一个网络刮板，以帮助我跟上在我的行业发表的文章。你知道吗

我束手无策，因为当我试图通过Flask运行我的代码时，我总是遇到这样的错误：

TypeError:view函数未返回有效响应。函数要么返回None，要么在没有return语句的情况下结束。你知道吗

以下是产生错误的代码：

文件1是博客.py上面写着：

import requests
from bs4 import BeautifulSoup


def blog_parser(url) -> 'html':
    import requests
    headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/50.0.2661.102 Safari/537.36'}
    result = requests.get(url, headers=headers)
    return result.content


def html(url) -> 'html':
    website = blog_parser(url)
    html = BeautifulSoup(website, 'html.parser')
    return html

def site_articles(url, element, unique_element) -> 'html':
    sitehtml = html(url)
    article_data = sitehtml.find_all(element, unique_element)
    return article_data

def skillsoft_titles(list_item):
    skillsoft_articles = site_articles('https://www.skillsoft.com/blog', "h1", {"class": "entry-title"})
    entries = skillsoft_articles[list_item]
    title = entries.find('a').get_text()
    return title

def skillsoft_link(list_item):
    skillsoft_articles = site_articles('https://www.skillsoft.com/blog', "h1", {"class": "entry-title"})
    entries = skillsoft_articles[list_item]
    link = entries.find('a').get('href')
    return link

def skillsoft_description(list_item):
    skillsoft_articles = site_articles('https://www.skillsoft.com/blog', "div", {"class": "entry-content"})
    entries = skillsoft_articles[list_item]
    description = entries.select_one("div p:nth-of-type(2)").text
    return description

def opensesame_titles(list_item) -> str:
    opensesame_articles = site_articles('https://www.opensesame.com/site/blog/', "div", {"class": "blog-post-right"})
    entries = opensesame_articles[list_item]
    title = entries.find('a').get_text()
    return title

def opensesame_link(list_item) -> str:
    opensesame_articles = site_articles('https://www.opensesame.com/site/blog/', "div", {"class": "blog-post-right"})
    entries = opensesame_articles[list_item]
    link = entries.find('a').get('href')
    return link

def opensesame_description(list_item):
    opensesame_articles = site_articles('https://www.opensesame.com/site/blog/', "section", {"class": "entry-content"})
    entries = opensesame_articles[list_item]
    description = entries.find('p').text
    return description

def cornerstone_titles(list_item) -> str:
    cornerstone_articles = site_articles('https://www.cornerstoneondemand.com/rework', "h2", {"class": "text-blue"})
    entries = cornerstone_articles[list_item]
    title = entries.find('a').get_text()
    return title

def cornerstone_link(list_item) -> str:
    cornerstone_articles = site_articles('https://www.cornerstoneondemand.com/rework', "h2", {"class": "text-blue"})
    entries = cornerstone_articles[list_item]
    link = entries.find('a').get('href')
    return link

def cornerstone_description(list_item) -> str:
    cornerstone_articles = site_articles('https://www.cornerstoneondemand.com/rework', "div", {"class": "col3-teaser-cont"})
    entries = cornerstone_articles[list_item]
    description = entries.find('p').text
    return description


def print_values(list_item, title_func, link_func, desc_func):
    return (print('Title:', title_func(list_item), '\n' 'Link:', link_func(list_item), '\n' 'Description:', desc_func(list_item)))

它本身工作得很好，在pycharm范围内，它返回的正是我想要的。你知道吗

Doc 2是我的烧瓶Doc，代码是：

import blogscraper
from flask import Flask

app = Flask(__name__)

skillsoft_titles = blogscraper.skillsoft_titles
skillsoft_link = blogscraper.skillsoft_link

skillsoft_description = blogscraper.skillsoft_description

@app.route('/', methods = ['GET'])



def skillsoft():
    output = blogscraper.print_values(1, skillsoft_titles, skillsoft_link, skillsoft_description)
    return output

skillsoft()

app.debug = True
app.run()
app.run(debug = True)

这会产生错误。出于某种原因，这会产生一个无回报或无回报，这对我来说没有意义后，谷歌的无回报。非常感谢您的帮助！你知道吗

Answer 1

函数print_values返回print(...)的结果，即None。这就是我所抱怨的。你知道吗

如果删除print语句，将返回一个元组： 'Title:', title_func(list_item), ...是一个元组，因为在Python中，用逗号分隔的几个值是一个元组。你知道吗

如果flask函数返回一个元组，flask会假定它是一个包含某些元素的元组，例如（response，status），请参见about responses

函数应该返回一个字符串，例如：

def print_values(list_item, title_func, link_func, desc_func):
    value = ''.join(('Title: ', title_func(list_item), '<br>\n',
                     'Link: <a href=', link_func(list_item), '>', 
                     link_func(list_item), '</a><br>\n',
                     'Description: ', desc_func(list_item)))
    # print (value)
    return value

或者一个列表也应该是有效的，这取决于你想要实现什么。你知道吗

请注意，同时将所有答案打印到控制台并不是一个好的解决方案，因此您可能希望在了解了print语句的工作原理之后删除它。。。你知道吗

Answer 2

您的print_values函数返回print的返回值，它恰好是None

def print_values(list_item, title_func, link_func, desc_func):
    return (print('Title:', title_func(list_item), '\n' 'Link:', link_func(list_item)

您需要更改此方法以返回您想要返回的内容。你知道吗

像这样：

def print_values(list_item, title_func, link_func, desc_func):
    return 'Title:' + title_func(list_item) + '\n' + 'Link:' + link_func(list_item)