Date format change in a csv file


 
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# 15  
Old 07-05-2015
Quote:
Originally Posted by Aia
Why? The GNU utilities are so ubiquity that chances are that it would work for the OP as well as for anyone searching how to do something similar. In fact, you have used the GNU date all this time, without knowing you could have been using another crippled version.
I think this is a fair question, although I risk taking us a long way off topic here Smilie

The thing to remember is that most Unix is still not Linux and that many Unixes that are not Linux do not implement GNU versions of the commands by default. One reason that I could suggest (I don't know for sure) is that not all Unix is free and therefor would clash with the GPL.

If you can solve something for every platform, then wonderful!
If you can solve something for most platforms, then that's still pretty good
If you have to solve something for a small number of platforms, then OK, but you should get to this conclusion as a last resort.
This User Gave Thanks to Smiling Dragon For This Post:
# 16  
Old 07-06-2015
Hi.

A utility, dadd, in package dateutils can do the arithmetic and reformatting. It seems to be in many repositories ( e.g. arch, FreeBSD, Slackware, Mac OS X "brew" etc.), and source is available (github). I compiled it for an earlier version of Debian.

A single command suffices to process all conforming dates:
Code:
#!/usr/bin/env bash

# @(#) s1	Demonstrate date reformat, arithmetic, dateutils, dadd.
# "dadd" is part of dateutils, repository Debian/Jessie, MiNT, # Ubuntu, etc.
# Also: http://www.fresse.org/dateutils.

# Utility functions: print-as-echo, print-line-with-visual-space, debug.
# export PATH="/usr/local/bin:/usr/bin:/bin"
LC_ALL=C ; LANG=C ; export LC_ALL LANG
pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C dadd

FILE=${1-data1}

pl " Input data file $FILE:"
cat $FILE

# In : MM/DD/YYYY HH:MM:SS
# Out: DD/Mon/YYYY HH:MM:SS
pl " Results:"
dadd -S -i '%m/%d/%Y' +1mo -f '%d/%b/%Y' < $FILE

exit 0

producing:
Code:
$ ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian 5.0.8 (lenny, workstation) 
bash GNU bash 3.2.39
dadd 0.2.6

-----
 Input data file data1:
First line, no dates, some blank lines appear.
Today was: Mon Jul  6 08:18:06 CDT 2015

06/23/2015 20:59:12
Now is the time 06/23/2015 20:59:12 for all good men

No dates here.
When was 06/23/2015 20:59:12, another 06/23/2015 20:59:12, not 2015.06.23 20:59:12

Last line, no dates.

-----
 Results:
First line, no dates, some blank lines appear.
Today was: Mon Jul  6 08:18:06 CDT 2015

23/Jul/2015 20:59:12
Now is the time 23/Jul/2015 20:59:12 for all good men

No dates here.
When was 23/Jul/2015 20:59:12, another 23/Jul/2015 20:59:12, not 2015.06.23 20:59:12

Last line, no dates.

Sample input and output are best to supply with the original question, otherwise responders need to create them, which may or may not match what is really needed.

Best wishes ... cheers, drl
This User Gave Thanks to drl For This Post:
# 17  
Old 07-07-2015
Using ksh93 builtin printf is powerful date / time handler. Same date format values as in date command. Input format supported us standard month/day/year or iso YYYY-MM-DD or epoc or ...

Code:
LC_ALL=C
export LC_ALL
fromdate="04/25/2015 12:01:02"
printf "%(%d/%b/%Y %H:%M:%S)T\n" "$fromdate"
year=$(printf "%(%Y)T\n" "$fromdate"
# ...

# current time, builtin value "now"
printf "%(%d/%b/%Y %H:%M:%S)T now" 
# epoc
epoc=$(printf "%(%s)T" now)
# date counting using epoc
((day=60*60*24))
#...

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