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 (9 Replies)
Hi Friends,
I've a special requirement, even though I know how to implement this using shell scripting, current requirement is PERL, in which I'm not much familiar !!!.
I've a record, which has around 200 fields, out of which I need to extract only one date value from the 97th field (this... (1 Reply)
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)
Hello Group,
I would like to sort the below file by date (first year then month and day) and I used the following command but it does not work
sort -n -t"/" -k3 -k1 -k2
"sample original file"
12/28/2009,1.0353
12/31/2009,1.0357
12/30/2009,1.0364
12/29/2009,1.0366
12/24/2009,1.0386... (6 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)
Discussion started by: Roozo
3 Replies
LEARN ABOUT PHP
basename
BASENAME(3) 1 BASENAME(3)basename - Returns trailing name component of pathSYNOPSIS
string basename (string $path, [string $suffix])
DESCRIPTION
Given a string containing the path to a file or directory, this function will return the trailing name component.
PARAMETERS
o $path
- A path. On Windows, both slash ( /) and backslash ( ) are used as directory separator character. In other environments, it is
the forward slash ( /).
o $suffix
- If the name component ends in $suffix this will also be cut off.
RETURN VALUES
Returns the base name of the given $path.
EXAMPLES
Example #1
basename(3) example
<?php
echo "1) ".basename("/etc/sudoers.d", ".d").PHP_EOL;
echo "2) ".basename("/etc/sudoers.d").PHP_EOL;
echo "3) ".basename("/etc/passwd").PHP_EOL;
echo "4) ".basename("/etc/").PHP_EOL;
echo "5) ".basename(".").PHP_EOL;
echo "6) ".basename("/");
?>
The above example will output:
1) sudoers
2) sudoers.d
3) passwd
4) etc
5) .
6)
NOTES
Note
basename(3) operates naively on the input string, and is not aware of the actual filesystem, or path components such as " ..".
Note
basename(3) is locale aware, so for it to see the correct basename with multibyte character paths, the matching locale must be set
using the setlocale(3) function.
SEE ALSO dirname(3), pathinfo(3).
PHP Documentation Group BASENAME(3)