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unixdatetimegrep
Dor Zohar
Dor Zohar

Reputation: 199

Getting the date format for grep -E in Unix

I write the dates on my file with the next format

date +%a-%b-%d-%Y

My goal in my exercise is to get the list of the dates in my file. I know I need to do it with grep -E, but I don't know how to put correctly the format of the date.

Desired input:

"grep -E (the format of the dates I'm looking for)" ~/file1

Desired output:

Tue-Oct-15-2013
Wen-Oct-16-2013
Wen-Oct-16-2013
Thu-Oct-17-2013

Upvotes: 4

Views: 14693

Answers (2)

Thor
Thor

Reputation: 47099

If you want to be explicit about it and locale-aware, you could do it like this with sh, date, grep and coreutils:

days="($(  for i in $(seq  7); do date -d 2013/01/$i +%a; done | paste -sd'|'))" 
months="($(for i in $(seq 12); do date -d 2013/$i/01 +%b; done | paste -sd'|'))"

You can now grep the date format like this: 

grep -E "${days}-${months}-[0-9]{1,2}-[0-9]{4}" infile

Note that you seem to be using a non-standard Wednesday abbreviation, if this is not a locale variation, you need to modify the days line to this:

days="($(for i in $(seq  7); do date -d 2013/01/$i +%a; done | sed s/Wed/Wen/ | paste -sd'|'))"

Upvotes: 1

jkshah
jkshah

Reputation: 11703

Try following:

grep -E '[[:alpha:]]{3}-[[:alpha:]]{3}-[[:digit:]]{2}-[[:digit:]]{4}' ~/file1

Or more concise

grep -E '\w{3}-\w{3}-[0-9]{2}-[0-9]{4}' ~/file1

Upvotes: 6

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