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Date format change in a csv file

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# 1  
Old 07-02-2015
Date format change in a csv file
Hi,

We have csv file where date is coming in MM/DD/YYYY HH:MM:SS (06/23/2015 20:59:12) in multiple places
But we need to change the date format to DD/Mon/YYYY HH:MM:SS (23/Jul/2015 20:59:12) using shell script.

Please let us know how can we achieve the same.
# 2  
Old 07-02-2015
What field(s) in the CSV file contains the date(s) to be changed?

Are you just trying to change a particular date, or are you trying to change all dates in the file?

What operating system and shell are you using?

What have you tried?
# 3  
Old 07-02-2015
I'm going to assume this is some sort of *nix environment due to the requirement for it to be in shell.
The date command can do this for you:
Code:
date -d "$yourolddateformat" +"%d/%b/%Y %H:%M:%S"

If you need to do this via shell (I'll use bourne as it's the only shell you have on all *nix's), you'll just need to do a little lookup:
Code:
month=`echo "$yourolddateformat" | cut -d '/' -f 1`
day=`echo "$yourolddateformat" | cut -d '/' -f 2`
therest=`echo "$yourolddateformat" | cut -d '/' -f 3`
case $month in
  01)
    monthname="Jan"
    ;;
  02)
    monthname="Feb"
    ;;
  03)
    monthname="Mar"
    ;;
  04)
    monthname="Apr"
    ;;
  05)
    monthname="May"
    ;;
  06)
    monthname="Jun"
    ;;
  07)
    monthname="Jul"
    ;;
  08)
    monthname="Aug"
    ;;
  09)
    monthname="Sep"
    ;;
  10)
    monthname="Oct"
    ;;
  11)
    monthname="Nov"
    ;;
  12)
    monthname="Dec"
    ;;
  *)
    exit 1
    ;;
esac
echo "${day}/${monthname}/$therest"

The date command is tested, the script is not - some debug may be required.
Some editing will be required to make it fit you larger needs of course too but the core should be enough to get you 95% there.
# 4  
Old 07-04-2015
Quote:
Originally Posted by Smiling Dragon
I'm going to assume this is some sort of *nix environment due to the requirement for it to be in shell.
The date command can do this for you:
Code:
date -d "$yourolddateformat" +"%d/%b/%Y %H:%M:%S"


That will only work with the GNU date command; -d is not standard.
Quote:
If you need to do this via shell (I'll use bourne as it's the only shell you have on all *nix's), you'll just need to do a little lookup:
Code:
month=`echo "$yourolddateformat" | cut -d '/' -f 1`
day=`echo "$yourolddateformat" | cut -d '/' -f 2`
therest=`echo "$yourolddateformat" | cut -d '/' -f 3`


I doubt that OP is using a system without a POSIX shell, so there's no need for cut:
Code:
IFS=/ read month day year whatever <<.
$yourolddateformat
.

Or, in bash or ksh93:
Code:
IFS=/ read month day year whatever <<< "$yourolddateformat"

---------- Post updated at 10:46 PM ---------- Previous update was at 10:36 PM ----------

Quote:
Originally Posted by dholea
Hi,

We have csv file where date is coming in MM/DD/YYYY HH:MM:SS (06/23/2015 20:59:12) in multiple places
But we need to change the date format to DD/Mon/YYYY HH:MM:SS (23/Jul/2015 20:59:12) using shell script.

Please let us know how can we achieve the same.

I'd strongly recommend changing it to the ISO standard format:
Code:
YYYY-MM-DD_HH:MM:SS

(Strictly speaking, the standard uses T where I used an underscore, but that's a minor variation.)
Last edited by cfajohnson; 07-05-2015 at 02:26 AM..
This User Gave Thanks to cfajohnson For This Post:
Smiling Dragon
# 5  
Old 07-05-2015
Quote:
Originally Posted by cfajohnson
[..]

I doubt that OP is using a system without a POSIX shell, so there's no need for cut:
Code:
IFS/ read month day year <<.
$yourolddateformat
.

Or, in bash or ksh93:
Code:
IFS/ read month day year <<< "$yourolddateformat"

[..]
I think you meant to write:
Code:
IFS=/ read month day therest ...

no?
# 6  
Old 07-05-2015
Quote:
Originally Posted by Scrutinizer
I think you meant to write:
Code:
IFS=/ read month day therest ...

no?

Indeed. Fixed.
# 7  
Old 07-05-2015
Quote:
Originally Posted by cfajohnson
[..]
Code:
IFS=/ read month day year whatever <<.
$yourolddateformat
.

Or, in bash or ksh93:
Code:
IFS=/ read month day year whatever <<< "$yourolddateformat"

[...]
Since the `/' is used as separator, the string can be only divided into three parts, making whatever useless, or in other words, there's no whatever; the year will receive the time as well.
Perhaps a minor modification.
Code:
IFS='/ ' read month day year time <<.
> $yourolddateformat
> .

Last edited by Aia; 07-05-2015 at 03:22 AM..
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