Python语法错误：'return'在函数外及referer_list函数代码解析求助

Hey there! Let's tackle your problems one by one:

1. Resolving the SyntaxError: 'return' outside function #

This error pops up when you have a return statement that's not tucked inside any function definition (either def or lambda). The referer_list function you shared doesn't include a return, so you’ll need to scan your entire codebase for any stray return lines that aren’t wrapped within a function block. That’s the immediate fix for this syntax error.

2. Issues in Your referer_list Function #

Looking at the code you provided, there are several problems with the URLs and invalid snippets that need addressing:

def referer_list():
    global headers_referers
    headers_referers.append('http://www.google.com/?q=')
    headers_referers.append('http://yandex.ru/yandsearch?text=%D1%%D2%?=g.sql()81%..')
    headers_referers.append('http://vk.com/profile.php?redirect=')
    headers_referers.append('http://www.usatoday.com/search/results?q=')
    headers_referers.append('http://engadget.search.aol.com/search?q=query?=query=..')
    headers_referers.append('https://www.google.ru/#hl=ru&newwindow=1?&saf..,or.r_gc.r_pw=?.r_cp.r_qf...')

Key Problems to Fix: #

• Missing initialization for headers_referers: Before using this global list, you need to define it outside the function first, like:

  headers_referers = []
  

  Without this, you’ll get a NameError when trying to call append().

• Invalid URL formatting & broken snippets:

  • The Yandex URL (http://yandex.ru/yandsearch?text=%D1%%D2%?=g.sql()81%..) has:
    • Incomplete URL encoding (%D1%%D2% is invalid—URL encoded characters are always %XX where XX is a hexadecimal pair)
    • A misplaced SQL-like snippet g.sql() that doesn’t belong in a referer URL, plus invalid parameter syntax (?= instead of a valid key=value pair)
  • The AOL search URL (http://engadget.search.aol.com/search?q=query?=query=..) has two ? characters—URLs only use one ? to separate the path from parameters; additional parameters should use &. The query?=query part is also invalid parameter syntax.
  • The Google.ru URL is incomplete (ends with ...) and uses HTML entities (&) instead of raw & characters for parameters, plus an extra ? that breaks the URL structure.

Cleaned-Up Example #

Here’s a revised version of your function with valid URLs and proper setup:

headers_referers = []  # Initialize the global list first

def referer_list():
    global headers_referers
    headers_referers.append('http://www.google.com/?q=')
    headers_referers.append('http://yandex.ru/yandsearch?text=')
    headers_referers.append('http://vk.com/profile.php?redirect=')
    headers_referers.append('http://www.usatoday.com/search/results?q=')
    headers_referers.append('http://engadget.search.aol.com/search?q=')
    headers_referers.append('https://www.google.ru/?hl=ru&newwindow=1')

This removes invalid snippets, fixes URL structure, and ensures the global list is properly initialized.

内容的提问来源于stack exchange，提问作者Ahmed Shady Elsadany