python - if语句的与或运算

Question

最近想提取出特定的URL，遇到问题为预期提取出URL中带有webshell或者phpinfo字段的URL，但是全部URL都匹配出来了：

for url in urls:
    if "webshell" or "phpinfo" in url:
        print url

改成and语句也不符合预期，只提取出了含有phpinfo的url：

for url in urls:
    if "webshell" and "phpinfo" in url:
        print url

Answer 1

for url in urls:
    if "webshell" in url or "phpinfo" in url:
        print url          
               
                   
        

这样才可以, 你原来的是先判断"webshell",如果不为零再判断"phpinfo" in url. "webshell" 和 "phpinfo" in url并列...

Answer 2

if "webshell" or "phpinfo" in url:

這樣做的意思是 if "webshell" or if "phpinfo" in url 而前者恆成立。

if "webshell" and "phpinfo" in url:

這樣做的意思是 if "phpinfo" in url 因為 if "webshell" 恆成立。

解法基本上如 @洛克 所說:

for url in urls:
    if "webshell" in url or "phpinfo" in url:
        print url 

如果今天用來匹配的 word 很多的話:

urls = [
    'https://www.example.com/aaa',
    'https://www.example.com/bbb',
    'https://www.example.com/ccc',
]


def urlcontain(url, lst):
    return any(seg for seg in url.split('/') if seg and seg in lst)

for url in urls:
    if urlcontain(url, ['aaa', 'bbb']):
        print(url)

結果:

https://www.example.com/aaa
https://www.example.com/bbb

urlcontain(url, lst) 可以問 url 裡面是不是有 lst 裡面的任何一個 string

這樣子要比對十個關鍵字也不會寫出太長的 if 述句。

當然要用 re 也可以，只是我個人不太喜歡 re 就是了...

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我回答過的問題: Python-QA