Python - 递归函数返回无[重复]

Question

我正在尝试清理 URL，并从中仅提取 ID（通过跳过 URL 部分）。通过进行一些调试，我可以看到我正在寻找的值正在打印，但是，没有返回（或返回 None）

这是代码：

def _sanitize_urls(urls=None):
    redact_list = [
        ("abcd.google.com", 3),
        ("xyz.yahoo.com", 4),
    ]
    urls_sanitized = []
    redact_found = [redact for redact in redact_list if redact[0] in urls]
    if redact_found:
        urls = urls.split(" ")
        print(urls)
        urls_sanitized = [
            words.split("/")[redact_found[0][1]] if redact_found[0][0] in words else words for words in urls
        ]
        print(urls_sanitized)
        urls_sanitized = " ".join(urls_sanitized)
        print(urls_sanitized)
        redact_found = [redact for redact in redact_list if redact[0] in urls_sanitized]
        print(redact_found)
        if not redact_found:
            print(urls_sanitized)
            return urls_sanitized
        else:
            _sanitize_urls(urls_sanitized)


def main():
    urls = "https://abcd.google.com/ID-XXXX and https://xyz.yahoo.com/Id/ID-XXXX"
    redact_exists = _sanitize_urls(urls)
    print(redact_exists)


if __name__ == "__main__":
    main()

我期望的输出是 =>“ID-XXXX 和 ID-XXXX”。我现在得到的输出是无。

在我这边进行一些调试 =>

['https://abcd.google.com/ID-XXXX', 'and', 'https://xyz.yahoo.com/Id/ID-XXXX']
['ID-XXXX', 'and', 'https://xyz.yahoo.com/Id/ID-XXXX']
ID-XXXX and https://xyz.yahoo.com/Id/ID-XXXX
[('xyz.yahoo.com', 4)]
['ID-XXXX', 'and', 'https://xyz.yahoo.com/Id/ID-XXXX']
['ID-XXXX', 'and', 'ID-XXXX']
ID-XXXX and ID-XXXX
[]
ID-XXXX and ID-XXXX
None

如您所见，print 打印正确的值，直到最后一刻，但它没有返回到 main 函数，而是返回 None。有什么想法吗？

Answer 1

在_sanitize_urls 函数的最后一行，你有

 _sanitize_urls(urls_sanitized)

在此之后您的函数不再有代码。在 Python 中，如果一个函数在没有明确的 return 语句的情况下用完代码，则会自动返回 None。