python多行匹配正则表达式
1 非贪婪flag
>>> re.findall(r"a(\d+?)", "a23b")
['2']
>>> re.findall(r"a(\d+)", "a23b")
['23']
['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']
['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.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']
['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']
['23', '23']
可见，是无需re.M
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