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5
于 2013-12-18T13:30:20.273 回答
1
Using a negative-lookahead:
regex = r '"(?!<|\s)'
|
means "or"
\s
means whitespace
You don't need to capture, since you know you're only matching a "
.
Alternatively, you could use a character class instead of the or, ie: [<\s]
.
于 2013-12-18T13:30:51.170 回答