I want to change a date from format dd-mmm-yyyy to mm/dd/yyyy. Is there a way to do this with sed or do you have to write a case statement to convert JAN to 01? Thanks
How to convert the date field from dd/mm/yyyy to yyyy/mm/dd in unix
my script will generate text file which have two fields
one is date and another is name of the server for example this is sample date which I have to sort based on older to newer date the problem is when I found out sort will... (4 Replies)
I have a file which has 100k+ records like this
abc,05-JUN-1974,def,lkj,aaa
def,11-SEP-1975,ghj,dis,dea
I want to convert ex 05-JUN-1974 to 06/05/1974
Please help me with awk script to convert the whole file into MM-DD-YYYY
Thank you! (2 Replies)
Hi all
I have some pipe-separated data in the form:
5/12/2008 00:00:00|31/1/2009 00:00:00|SOMESTUFF|OTHERSTUFF
12/31/2008 00:00:00|15/1/2009 00:00:00|MORESTUFF|REMAININGSTUFF
1/1/1023 00:00:00|16/5/2047 00:00:00|THEREST|YETMORE
I need to zero-pad the single-digit days and months, using... (3 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)
Hello,
I am writing a script that parses different logs and produces one. In the source files, the date is in DD MM YYYY HH24:MI:SS format. In the output, it should be in DD MON YYY HH24:MI:SS (ie 25 Jan 2010 16:10:10)
To extract the dates, I am using shell substrings, i.e.:
read line
... (4 Replies)
Hi I have a problem with Date format in my code.
1st I am trying to convert today's date to yesterday's using
YESTERDAY3=`perl -e '@y=localtime(time()-86400); printf "%04d/%02d/%02d",$y+1900,$y+1,$y;$y;'`
And once it is done I am trying to using the yesterday date in a grep command to... (3 Replies)
I've seen a lot of posts on this and have tried the following:
echo 1257000000| perl -e '($d,$m,$y)=(localtime(time-86400));$m+=1;$y+=1900;printf "$y/$m/$d\n";'
But I am unable to convert a past Epoch date into a format such as YYYY/MM/DD or MM/DD/YYYY.
I am using bash and don't know... (4 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)
Hi All,
I have line
,A,FDRM0002,12/21/2017,,0.961751583,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
it contains date in mm/dd/yyyy format i want to change this to yyyymmdd format using perl.
Use code tags, thanks. (8 Replies)
How to convert mmm-yy to mm/dd/yyyy format in unix ?
example:
Jan-99 to 01/01/1999
Jan-00 to 01/01/2000
Jan-25 to 01/01/2025
Dec-99 to 01/12/1999
Dec-00 to 01/12/2000
Dec-25 to 01/12/2025
YY anything between 00-50 should be 2000-2050
YY anything between 51-99 should be 1951-1999
... (2 Replies)
Discussion started by: gksenthilkumar
2 Replies
LEARN ABOUT PHP
datefmt_set_pattern
DATEFMT_SET_PATTERN(3) 1 DATEFMT_SET_PATTERN(3)IntlDateFormatter::setPattern - Set the pattern used for the IntlDateFormatter
Object oriented style
SYNOPSIS
public bool IntlDateFormatter::setPattern (string $pattern)
DESCRIPTION
Procedural style
bool datefmt_set_pattern (IntlDateFormatter $fmt, string $pattern)
Set the pattern used for the IntlDateFormatter.
PARAMETERS
o $fmt
- The formatter resource.
o $pattern
- New pattern string to use. Possible patterns are documented at http://userguide.icu-project.org/formatparse/datetime.
RETURN VALUES
Returns TRUE on success or FALSE on failure. Bad formatstrings are usually the cause of the failure.
EXAMPLES
Example #1
datefmt_set_pattern(3) example
<?php
$fmt = datefmt_create(
'en_US',
IntlDateFormatter::FULL,IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'MM/dd/yyyy'
);
echo 'pattern of the formatter is : ' . datefmt_get_pattern($fmt);
echo 'First Formatted output with pattern is ' . datefmt_format($fmt, 0);
datefmt_set_pattern($fmt, 'yyyymmdd hh:mm:ss z');
echo 'Now pattern of the formatter is : ' . datefmt_get_pattern($fmt);
echo 'Second Formatted output with pattern is ' . datefmt_format($fmt, 0);
?>
Example #2
OO example
<?php
$fmt = new IntlDateFormatter(
'en_US',
IntlDateFormatter::FULL,IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'MM/dd/yyyy'
);
echo 'pattern of the formatter is : ' . $fmt->getPattern();
echo 'First Formatted output is ' . $fmt->format(0);
$fmt->setPattern('yyyymmdd hh:mm:ss z');
echo 'Now pattern of the formatter is : ' . $fmt->getPattern();
echo 'Second Formatted output is ' . $fmt->format(0);
?>
The above example will output:
pattern of the formatter is : MM/dd/yyyy
First Formatted output with pattern is 12/31/1969
Now pattern of the formatter is : yyyymmdd hh:mm:ss z
Second Formatted output with pattern is 19690031 04:00:00 PST
SEE ALSO datefmt_get_pattern(3), datefmt_create(3).
PHP Documentation Group DATEFMT_SET_PATTERN(3)