How to construct this in regular expression

Question

I have file names that end in yyyymmdd, eg: myFile.20090601, myFile20090708 , etc

I want to grep for a pattern in all files from June 08 to July 07 of 2009, ie: 20090609 to 20090707

How can I do a regex in one go?

I tried:

grep 'myPattern' *20090(6(09|[1-3][0-9])|70[1-7])

Answer 1

20090(6(09|[1-3][0-9])|70[1-7])$

or

20090(6(0[89]|[1-3][0-9])|70[1-7])$

depending on whether you meant 8th or 9th of July (your question seems contradictory there).

Answer 2

The range of valid dates is 06–30 for June and 01—07 for July. Because the ranges of days are dissimilar, we should use separate regexes for each month. These are

/2009 06 (09 | [12][0-9] | 30)/x

(Notice how the day ranges are divided into cases depending on the tens place, because there are different conditions on what is valid for the units place depending.)

And

/2009 07 0[1-7]/x

and then we can join them into

/(2009 06 (09 | [12][0-9] | 30)) | (2009 07 0[1-7])/x

and then factor out the common points (may not be the best for readabilty) and add the end-of-line assertion:

/2009 0 (6 (09 | [12][0-9] | 30)) | (7 0[1-7]) $/x

Answer 3

I'd suggest a perl/python script (or any other scripting language) that takes 3 parameters:

1. The pattern
2. Start date as yyyymmdd
3. End date as yyyymmdd

It would :

1. decode start and end date.
2. loop through the files in a folder
3. decode any dates in the filename
4. check if it's between the dates, and grep the pattern

Answer 4

grep 'myPattern' `ls | grep -E "20090(6(09|[1-3][0-9])|70[1-7])"`

This works roughly as follows. Take a list of files in the current directory (ls), filter that using the date regex (ls | grep ...), then perform a grep search using your pattern, on the list of files that is produced (grep 'myPattern' ...). The back-ticks surrounding the ls | grep ... executes that part of the command and substitutes in the output of that command into the surrounding command. So if it produced output like "file1 file2 file3", then it would result in a command like grep 'myPattern' file1 file2 file3.