linux bash - Parse date in custom format

Question

I have a date in a the %c format (could be any other) and I need to use it in the date command. %c is NOT the American format. It is the German one because it's a German server. This also did not work properly on an American server. (Locales set to German or American)

This does not work (error included):

user@server:~$ NOW=$(date +%c); echo $NOW
Do 19 Dez 2013 22:33:28 CET
user@server:~$ date --date="$NOW" +%d/%m/%Y
date: ungültiges Datum ?Do 19 Dez 2013 22:33:28 CET“

(date: ungültiges Datum ?Do 19 Dez 2013 22:33:28 CET“ = date: invalid date ?Do 19 Dez 2013 22:33:28 CET“)

The difficulty is that I don't know which locale or even whci dateformat will be used later since the user can set their own format. So a simple specific parsing solution ist not really going to work!

But how do I do it?

To gerneralize the issue:

If I have a date format format1 (which could be any or at least one that can be reversed) I can use date to get a formatted date. But if I want to format it to another date (format2) how do I do it?
Any solution using anything else than the coreutils is pointless since I am trying to develop a bash script for as many unix machines as possible.

DATE=$(date "+$format1")

date --date="$DATE" "+$format2" # Error in most cases!

This is needed because I have a command which the user can give a date format. This date string is going to be displayed. But in a later step I need to convert this date string into another fixed one. I can manipulate the whcih format the command will get and I can maniplulate the output (or what the user will see).
I cannot run the command twice because it is very time consuming.

---

Update:

I have found something like a solution:

# Modify $user_format so it can be parsed later
user_format="$user_format %s"

# Time consuming command which will print a date in the given format
output=$(time_consuming_command params "$user_format" more params)

# This will only display what $user_format used to be
echo ${output% *}

# A simple unix timestamp parsing ("${output##* }" will only return the timestamp)
new_formated_date=$(date -d "1970-01-01 ${output##* } sec UTC" "+$new_format")

This is working and might be helpful to others. So I will share this with you.

Answer 1

Not possible with --date as of GNU coreutils 8.22. From the date manual:

‘-d datestr’

‘--date=datestr’

Display the date and time specified in datestr instead of the current date and time. datestr can be in almost any common format. It can contain month names, time zones, ‘am’ and ‘pm’, ‘yesterday’, etc. For example, --date="2004-02-27 14:19:13.489392193 +0530" specifies the instant of time that is 489,392,193 nanoseconds after February 27, 2004 at 2:19:13 PM in a time zone that is 5 hours and 30 minutes east of UTC.

Note: input currently must be in locale independent format. E.g., the LC_TIME=C below is needed to print back the correct date in many locales:

date -d "$(LC_TIME=C date)"

http://www.gnu.org/software/coreutils/manual/html_node/Options-for-date.html#Options-for-date

Note it says that the input format cannot be in a locale-specific format.

There may be other libraries or programs that would recognize more date formats, but for a given date format it would not be difficult to write a short program to convert it to something date recognizes (for example, with Perl or awk).