The above script works perfectly fine. But I'm really not sure how the above script works. I'm just an amateur when it comes to RegEx. A search on google did not help me either.
My question, how the above script works? Can someone trace an example for me?
Hi,
I want to know how to change this string to date format
20061102122042 to 02-11-2006 12:20:42
or 02-Nov-2006 12:20:42
Please let me know at the earliest.Thanks in advance.
Regards,
Preetham R. (3 Replies)
I know the command date +"%Y%m%d" can change today's date to digit format as below .
$date +"%Y%m%d"
20071217
it works fine .
now I want to do it back . If I have a file like below, (in the file , there are three lines, and each line have ; sign , after the ; sign is the date ) , I... (4 Replies)
Hi All,
this is my second post, last post reply was very helpful.
I have a data that has date in DD/MM/YYYY (07/11/2008) format i want to replace the backslash by a dot(.) so that my awk script can read it inside the C shell script that i have written.
i want to change 07/11/2008 to... (3 Replies)
dear members,
ls -l
drwxr-xr-x 40 root sys 1024 Jul 11 22:19 usr
drwxr-xr-x 43 root sys 1024 Feb 1 2009 var
i am using solaris 10
is that possibe to do
drwxr-xr-x 40 root sys 1024 25-08-2009 22:19 usr
drwxr-xr-x 43 root sys ... (4 Replies)
Hi guys,
I have a text file with lots of lines like this:
MCOGT23R27815 27/07/07 27/05/09
SO733AM0235 30/11/07 30/11/10
NL123403N 04/03/08 04/03/11
0747AM7474 04/04/08 04/04/11
I want to change each line so the date format looks like this:
MCOGT23R27815 07/07/27 09/05/27 ... (7 Replies)
Dear Friends,
Need your help once again,
I have a variable ( e.g. ${i}) whoch has date in MM/DD/YYYY (E.g. 12/31/2011) format.
I want to change it to DD/MM/YYYY (e.g. 31/12/2011) format.
Request you to guide me as we are unable to do the same.
Thanks in advance
Anu. (1 Reply)
Hi,
I have the variable "$date_update" in that form:
2011-12-31T13:00:09Z and I would like to change it to
31/12/2011 13:00:09 (Date and Time separated by a blank).
Does anyone has a simple solution for that? (using Korn Shell)
Cheers
Jurgen (4 Replies)
Hi all,
I have a file that every line starts with the date and time. The format is like YYYYMMDDHHMM and I woulk like to change it to MM/DD/YY<space>HH:MM.
I tried to figure out a way to do it with sed, but I don't know how I could reorganize the digits of the first format. Does anyone have any... (1 Reply)
Discussion started by: geovas
1 Replies
LEARN ABOUT DEBIAN
dp
DP(8) [nmh-1.5] DP(8)NAME
dp - parse dates 822-style
SYNOPSIS
/usr/lib/mh/dp [-form formatfile] [-format string] [-width columns] [-version] [-help] dates ...
DESCRIPTION
Dp is a program that parses dates according to the ARPA Internet standard. It also understands many non-standard formats, such as those
produced by TOPS-20 sites and some UNIX sites using ctime(3). It is useful for seeing how nmh will interpret a date.
The dp program treats each argument as a single date, and prints the date out in the official 822-format. Hence, it is usually best to
enclose each argument in quotes for the shell.
To override the output format used by dp, the -format string or -format file switches are used. This permits individual fields of the
address to be extracted with ease. The string is simply a format string and the file is simply a format file. See mh-format(5) for the
details.
Here is the default format string used by dp:
%<(nodate{text})error: %{text}%|%(putstr(pretty{text}))%>
which says that if an error was detected, print the error, a `:', and the date in error. Otherwise, output the 822-proper format of the
date.
FILES
$HOME/.mh_profile The user profile
PROFILE COMPONENTS
None
SEE ALSO ap(8), Standard for the Format of ARPA Internet Text Messages (RFC-822)
DEFAULTS
`-format' default as described above
`-width' default to the width of the terminal
CONTEXT
None
BUGS
The argument to the -format switch must be interpreted as a single token by the shell that invokes dp. Therefore, one must usually place
the argument to this switch inside quotes.
MH.6.8 11 June 2012 DP(8)