Question

I wanna match patterns like: '''abc'''. But there are some noise like ''''''abc''' in the document. I just want the words which are preceded by exact three " ' ". I tried " [^']'''[^'] ". But it does not work. Regular expression " '{3} " will match more than three " ' ".

So can anybody help me?

Answer 1

You basically already gave the solution yourself. If you want to match strings which are surrounded by exactly 3 ', then do this (I use Python for my examples):

In [33]: re.findall("'{3}\w+'{3}", "'''abc'''")
Out[33]: ["'''abc'''"]

If you want avoid those strings that are preceded by more than 3 ', then you best take a negative look-behind assertion (?<!...):

In [59]: re.findall("(?<!')'{3}\w+'{3}", "'''abc'''")
Out[59]: ["'''abc'''"]

In [60]: re.findall("(?<!')'{3}\w+'{3}", "''''''''''abc'''")
Out[60]: []

In [62]: re.findall("(?<!')'{3}\w+'{3}", "'''abc''''''''''")
Out[62]: ["'''abc'''"]

If you also want to avoid strings that are succeeded by more than 3 ', add a negative look-ahead assertion (?!...):

In [63]: re.findall("(?<!')'{3}\w+'{3}(?!')", "'''abc'''")
Out[63]: ["'''abc'''"]

In [64]: re.findall("(?<!')'{3}\w+'{3}(?!')", "''''''''''abc'''")
Out[64]: []

In [65]: re.findall("(?<!')'{3}\w+'{3}(?!')", "'''abc''''''''''")
Out[65]: []

Note: It depends on your programming language whether look-ahead and look-behind are supported. If so, this is the best way to accomplish what you want.