>>> re.findall(r"a(\d+?)", "a23b") ['2'] >>> re.findall(r"a(\d+)", "a23b") ['23']
注意比较这种情况:
>>> re.findall(r"a(\d+)b", "a23b") ['23'] >>> re.findall(r"a(\d+?)b", "a23b") ['23']
2 如果你要多行匹配,那么加上re.S和re.M标志
re.S:.将会匹配换行符,默认.不会匹配换行符
>>> re.findall(r"a(\d+)b.+a(\d+)b", "a23b\na34b") [] >>> re.findall(r"a(\d+)b.+a(\d+)b", "a23b\na34b", re.S) [('23', '34')] >>>
re.M:^$标志将会匹配每一行,默认^和$只会匹配第一行
>>> re.findall(r"^a(\d+)b", "a23b\na34b") ['23'] >>> re.findall(r"^a(\d+)b", "a23b\na34b", re.M) ['23', '34']
但是,如果没有^标志,
>>> re.findall(r"a(\d+)b", "a23b\na23b") ['23', '23']
可见,是无需re.M
来自 http://www.juyimeng.com/python-multi-line-non-greedy-regular-expression-sample.html
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