I want to grep all dates in August or September (date format mm/dd/yyyy with leading zeroes).
I tried:
grep '0\(8\|9\)/[0-9][0-9]/2012'
But command prompt outputs:
The system cannot find the path specified.
How to use grep with alternating part of the regular expression?
152 Views Asked by ljustin At
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There are 1 best solutions below
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grepwants a regular expression and a file name. The error message looks like you somehow passed an invalid file name. Assuming the file containing the log entries is namedlog, try this;(I also tweaked your regex a bit.)