python非貪婪、多行匹配正則表達式
一些regular的tips：
1 非貪婪flag
>>> re . findall ( r " a( \d +?) " , " a23b " )
[ ' 2 ' ]
>>> re . findall ( r " a( \d +) " , " a23b " )
[ ' 23 ' ]
[ ' 2 ' ]
>>> re . findall ( r " a( \d +) " , " a23b " )
[ ' 23 ' ]
注意比較這種情況：
>>> 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標志
re.S：.將會匹配換行符，默認.不會匹配換行符
>>> re . findall ( r " a( \d +)b.+a( \d +)b " , " a23b \n a34b " )
[]
>>> re . findall ( r " a( \d +)b.+a( \d +)b " , " a23b \n a34b " , re . S )
[( ' 23 ' , ' 34 ' )]
>>>
[]
>>> re . findall ( r " a( \d +)b.+a( \d +)b " , " a23b \n a34b " , re . S )
[( ' 23 ' , ' 34 ' )]
>>>
re.M：^$標志將會匹配每一行，默認^和$只會匹配第一行
>>> re . findall ( r " ^a( \d +)b " , " a23b \n a34b " )
[ ' 23 ' ]
>>> re . findall ( r " ^a( \d +)b " , " a23b \n a34b " , re . M )
[ ' 23 ' , ' 34 ' ]
[ ' 23 ' ]
>>> re . findall ( r " ^a( \d +)b " , " a23b \n a34b " , re . M )
[ ' 23 ' , ' 34 ' ]
但是，如果沒有^標志，
>>> re . findall ( r " a( \d +)b " , " a23b \n a23b " )
[ ' 23 ' , ' 23 ' ]
[ ' 23 ' , ' 23 ' ]
可見，是無需re.M
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