Hi
I have a sed command
sed -e "s/sub_date/=$(date +"%d/%m/%Y")/" sub_create_tmp
I want to substitute with the current date in dd/mm/yyyy format .
But the result is an error " cannot parse " .
Pls help .
sars (2 Replies)
I have a file with some date columns in MM/DD/YYYY format:
SMPBR|DUP-DO NOT USE|NEW YORK||16105|BA5270715|6/6/2007 |MWERNER|109||||JOHN||SMITH|MD|72211118||||||74559|21 WILMINGTON RD||D|11/6/2003|SL# MD CONTACT-LIZ RICHARDS|||0|Y|N||1411458|
And I want to convert the date format to:
... (5 Replies)
Hi all,
I know this may have already been asked but hey ho...
i have two dates in the 'yyyy-mm-dd hh:mm:ss' format.
'2009-01-03 01:00:00'
'2009-04-05 00:00:00'
How can i, in shell script determine their differences?
Please note, the time may not be available, so please suggest both... (4 Replies)
(Attention: Green PHP newbie !)
I have an online inquiry form, delivering a date in the form yyyy/mm/dd to my feedback form. If the content passes several checks, the form sends an e-mail to me. All works fine. I just would like to receive the date in the form dd/mm/yyyy. I tried with some code,... (6 Replies)
could you please help be on the below code ..
Requirement is when i pass the parameter(for below 2) i should get current time -2 hours in the format :wall:..
cur_dt=`$ICEBIN/sqsh -S$DSQUERY -U $BATCHID -P $PASSWD -h -C"select getdate()" | sed '2d'`
pr_dt="`$ICEBIN/sqsh -S$DSQUERY -U $BATCHID... (2 Replies)
Hi,
I am having one log files. contains som data according to date. And it is going to append .Eg:abc.log contains below data
2011-10-19 abjhgj
2011-10-19 gjhgjgj
2011-10-20 hhhjh
2011-10-20 hhhhjj
2011-10-21 gg
.
.
.
2011-11-24 yyy
from log files i want catch only... (2 Replies)
How can I convert any user inputted date into yyyy/mm/dd ?
For example user can input date one of the following 20120121 , 2012-01-21 ,01/21/2012,01/21/2012 etc
But I need to convert any of the date entered by user into yyyy/mm/dd (2012/01/2012). Any suggestion. Thanks in advance
this is... (1 Reply)
I am changing epoch times to dates. I was able to do the following:
echo "$varx" | gawk '{print strftime("%c", $0)}'
Mon Dec 31 16:26:40 2012
This changes the epoch date (which is what varx is) into localtime.
However, my problem is that I only want 12/31/2012 and not the Mon Dec 31... (2 Replies)
I need the date format in YYYY.MM format and I am able to get current month date as well as previous month date with below command
PM=`date +'%Y.%m' -d 'last month'`
CM=`date +'%Y.%m' -d 'now'`
but I need to get YYYY.MM date format for previous 12 months so could you please help me how I get... (2 Replies)
I am getting output of YYYY-MM-DD and want to change this to DD/MM/YYYY.
When am running the query in 'Todd' to_date(column_name,'DD/MM/YYYY') am getting the required o/p of DD/MM/YYYY, But when am executing the same query(Netezza) in linux server(bash) am getting the output of YYYY-MM-DD
file... (3 Replies)
Discussion started by: Roozo
3 Replies
LEARN ABOUT MOJAVE
zipgrep
ZIPGREP(1L)ZIPGREP(1L)NAME
zipgrep - search files in a ZIP archive for lines matching a pattern
SYNOPSIS
zipgrep [egrep_options] pattern file[.zip] [file(s) ...] [-x xfile(s) ...]
DESCRIPTION
zipgrep will search files within a ZIP archive for lines matching the given string or pattern. zipgrep is a shell script and requires
egrep(1) and unzip(1L) to function. Its output is identical to that of egrep(1).
ARGUMENTS
pattern
The pattern to be located within a ZIP archive. Any string or regular expression accepted by egrep(1) may be used. file[.zip] Path
of the ZIP archive. (Wildcard expressions for the ZIP archive name are not supported.) If the literal filename is not found, the
suffix .zip is appended. Note that self-extracting ZIP files are supported, as with any other ZIP archive; just specify the .exe
suffix (if any) explicitly.
[file(s)]
An optional list of archive members to be processed, separated by spaces. If no member files are specified, all members of the ZIP
archive are searched. Regular expressions (wildcards) may be used to match multiple members:
* matches a sequence of 0 or more characters
? matches exactly 1 character
[...] matches any single character found inside the brackets; ranges are specified by a beginning character, a hyphen, and an end-
ing character. If an exclamation point or a caret (`!' or `^') follows the left bracket, then the range of characters within
the brackets is complemented (that is, anything except the characters inside the brackets is considered a match).
(Be sure to quote any character that might otherwise be interpreted or modified by the operating system.)
[-x xfile(s)]
An optional list of archive members to be excluded from processing. Since wildcard characters match directory separators (`/'),
this option may be used to exclude any files that are in subdirectories. For example, ``zipgrep grumpy foo *.[ch] -x */*'' would
search for the string ``grumpy'' in all C source files in the main directory of the ``foo'' archive, but none in any subdirectories.
Without the -x option, all C source files in all directories within the zipfile would be searched.
OPTIONS
All options prior to the ZIP archive filename are passed to egrep(1).
SEE ALSO egrep(1), unzip(1L), zip(1L), funzip(1L), zipcloak(1L), zipinfo(1L), zipnote(1L), zipsplit(1L)URL
The Info-ZIP home page is currently at
http://www.info-zip.org/pub/infozip/
or
ftp://ftp.info-zip.org/pub/infozip/ .
AUTHORS
zipgrep was written by Jean-loup Gailly.
Info-ZIP 20 April 2009 ZIPGREP(1L)