Question

我一直在尝试构建网络抓取工具，以帮助我掌握行业内发表的文章。

我精打细算，因为当我尝试通过Flask运行代码时，我会不断收到此错误：

TypeError：视图函数未返回有效响应。该函数返回None或不返回return语句结束。

以下是产生错误的代码：

文档1是blogscraper.py，其内容为：

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)))

这本身就可以很好地工作，在pycharm中，它完全返回我想要的东西。

Doc 2是我的烧瓶文档，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“无”或“无”回报，对谷歌搜索无回报后，这对我来说毫无意义。非常感谢您提供任何帮助！

Answer 1

您的print_values函数返回print的返回值-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要返回的内容。

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)

Answer 2

您的print_values函数返回print(...)的结果None。这就是烧瓶所抱怨的。

如果仅删除print语句，将返回一个元组： 'Title:', title_func(list_item), ...是一个元组，因为用逗号分隔的多个值是Python中的一个元组。

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

您的函数应返回例如字符串，就像这样：

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

或列表也应该有效，具体取决于您要实现的目标。

请注意，将所有答案打印出控制台并不是一个好的解决方案，因此您可能希望在理解所有打印方式后将其删除...

不返回烧瓶，但我不明白为什么

2 个答案: