CODEWITHSUNDEEP
linuxdateformat
pistach
pistach

Reputation: 401

give the input format when converting a date

I read the date from a file to a variable. The date has the format ddmmyyyy. It has to be converted to yyyy-mm-dd I already searched this forum and got this far : date -d '$DATE' +%F The problem is the input format is not recognised. Is there any way I can specify the input date format? On an other forum I found : date -d "${OLD_DATE}" -D "%d%m%Y" +%F where -D should specify the input format but this doesn't work. But -D is unknown.

thanks for the help and sorry for my English.

Upvotes: 1

Views: 2020

Answers (2)

Lev Levitsky
Lev Levitsky

Reputation: 65791

Yes, date understands a lot of formats for -d, but when it's just 8 digits in a row, it interprets it as YYYYmmdd. I'm not sure if you can force it to read it differently, but in this case you can use a simple editor such as awk or sed instead:

$ OLD_DATE='08032011'
$ echo $OLD_DATE | sed -r 's/(.{2})(.{2})(.{4})/\3-\2-\1/'
2011-03-08

This will work on GNU sed. Note that it doesn't check the input (for brevity).

Upvotes: 1

dstronczak
dstronczak

Reputation: 2454

You could to it like this:

echo "DDMMYYYY" | awk 'BEGIN {OFS="-"} {print substr($1,5,4), substr($1,3,2), substr($1,1,2)}'

Output:

YYYY-MM-DD

Upvotes: 1

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