Python正则表达式匹配多次

Question

我正在尝试将模式与可能具有多个模式实例的字符串进行匹配.我需要单独的每个实例.re.findall() 应该这样做,但我不知道我做错了什么.

pattern = re.compile('/review: (http://url.com/(\d+)\s?)+/', re.IGNORECASE)
match = pattern.findall('this is the message. review: http://url.com/123 http://url.com/456')

我需要" http://url.com/123 ",http://url.com/456和两个数123 456是不同的元素match列表.

我也尝试'/review: ((http://url.com/(\d+)\s?)+)/'过这种模式,但没有运气.

Answer 1

Use this. You need to place 'review' outside the capturing group to achieve the desired result.

pattern = re.compile(r'(?:review: )?(http://url.com/(\d+))\s?', re.IGNORECASE)

This gives output

>>> match = pattern.findall('this is the message. review: http://url.com/123 http://url.com/456')
>>> match
[('http://url.com/123', '123'), ('http://url.com/456', '456')]

Answer 2

你在正则表达式中有额外的东西.在python中,模式应该只是一个字符串.例如,而不是这样:

pattern = re.compile('/review: (http://url.com/(\d+)\s?)+/', re.IGNORECASE)

它应该是:

pattern = re.compile('review: (http://url.com/(\d+)\s?)+', re.IGNORECASE)

通常在python中你实际上使用这样的"原始"字符串:

pattern = re.compile(r'review: (http://url.com/(\d+)\s?)+', re.IGNORECASE)

字符串前面的额外r可以避免你不得不做大量的反斜杠转义等.