Using Regex to search for a string unless it finds another string first

Question

Hello I'm trying to use regex to search through a markdown file for a date and only get a match if it finds an instance of a specific string before it finds another date.

This is what I have right now and it definitely doesn't work. (\d{2}\/\d{2}\/\d{2})(string)?(^(\d{2}\/\d{2}\/\d{2}))

So in this instance It would throw a match since the string is before the next date:

01/20/20

string

01/21/20

Here it shouldn't match since the string is after the next date:

01/20/20

this isn't the phrase you're looking for

01/21/20

string

Any help on this would be greatly appreciated.

Do you mean like this? \d{2}\/\d{2}\/\d{2}(?:(?!\d{2}\/\d{2}\/\d{2}).)*string.*?\d{2}\/\d{2}\/\d{2} regex101.com/r/FREPRt/1 — The fourth bird
– The fourth bird, Commented Jan 12, 2020 at 14:58

The fourth bird · Accepted Answer · 2020-01-12 15:13:09Z

1

You could match a date like pattern. Then use a tempered greedy token approach (?:(?!\d{2}\/\d{2}\/\d{2}).)* to match string without matching another date first.

If you have matched the string, use a non greedy dot .*? to match the first occurrence of the next date.

\d{2}\/\d{2}\/\d{2}(?:(?!\d{2}\/\d{2}\/\d{2}).)*string.*?\d{2}\/\d{2}\/\d{2}

Regex demo | Python demo

For example (using re.DOTALL to make the dot match a newline)

import re

regex = r"\d{2}\/\d{2}\/\d{2}(?:(?!\d{2}\/\d{2}\/\d{2}).)*string(?:(?!string|\d{2}\/\d{2}\/\d{2}).)*\d{2}\/\d{2}\/\d{2}"

test_str = """01/20/20\n\n"
    "string\n\n"
    "01/21/20\n\n"
    "01/20/20\n\n"
    "this isn't the phrase you're looking for\n\n"
    "01/21/20\n\n"
    "string"""

print(re.findall(regex, test_str, re.DOTALL))

Output

['01/20/20\n\n"\n\t"string\n\n"\n\t"01/21/20']

If the string can not occur 2 times between the date, you might use

\d{2}\/\d{2}\/\d{2}(?:(?!\d{2}\/\d{2}\/\d{2}|string).)*string(?:(?!string|\d{2}\/\d{2}\/\d{2}).)*\d{2}\/\d{2}\/\d{2}

Regex demo

Note that if you don't want the string and the dates to be part of a larger word, you could add word boundaries \b

edited Jan 12, 2020 at 15:13

answered Jan 12, 2020 at 15:00

The fourth bird

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8 Comments

The fourth bird Over a year ago

@TimBiegeleisen It is just the answer for my comment, which looks like your answer afterwards :) But you will get my vote anyway

Captain Po-Po Over a year ago

@The Fourth Bird That does match correctly but two things. How do I get it to return all instances of this match? and two How do I get it to return only the initial date after it has confirmed that it is a match.

Tim Biegeleisen Over a year ago

@CaptainPo-Po Fourth's answer already matches all instances of the match +1. For the second requirement, you only need to change the capture group in the call to re.findall.

The fourth bird Over a year ago

@CaptainPo-Po You could indeed use a capturing group for the first date regex101.com/r/khGQ8G/1 Using re.findall will return only the capturing group. If you want the match and the groups, you could use re.finditer. This is an example of the auto generated code by regex101 ideone.com/2M19fp

Captain Po-Po Over a year ago

Okay yeah I see that now. I've been tinkering and I've run into an error when running this: TypeError: findall() missing 1 required positional argument: 'string' i've replaced the phrase "string" with the actual string I need, is one of those a command of some sort?

|

Tim Biegeleisen · Accepted Answer · 2020-01-12 15:00:37Z

1

One approach here would be to use a tempered dot to ensure that the regex engine does not cross over the ending date while trying to find the string after the starting date. For example:

inp = """01/20/20

string                  # <-- this is matched

01/21/20

01/20/20

01/21/20

string"""               # <-- this is not matched

matches = re.findall(r'01/20/20(?:(?!\b01/21/20\b).)*?(\bstring\b).*?\b01/21/20\b', inp, flags=re.DOTALL)
print(matches)

This prints string only once, that match being the first occurrence, which legitimately sits in between the starting and ending dates.

answered Jan 12, 2020 at 15:00

Tim Biegeleisen

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