It's basically the same as my original suggestion, it's just generating the 'Dec 2' part of the pattern automatically - date | cut -c5-10 gives characters 5-10 from the date command's response. .*: is a regular expression for 'any number of any character, followed by a colon'.
EDIT:
Code:
ls -ltrh | egrep "`date | cut -c5-10`.*:" || echo "No files"
Hi,
I am trying to list names of only today's files OR say, files which are not older than 1 hour and copy them in 'list.txt' file. I know,
:ls > list.txt will list all the files. But, how to list today's files? Any help will be appriciated. (4 Replies)
Hi,
Can any one tell the command to list all the files that are created as of today from all the directories?
The Command "ls -ltR" is listing all the files.
But I want the list of files that has been created as of today along with the directory path:)
Thank you in advance.:)
Regards,... (4 Replies)
Is there a tool that can diff a directory and generate a change list of files in that directory based on a previous snapshot on the directory?
For example
/etc/a.txt:changed
/etc/b.txt:removed
/etc/c.txt:added
Thanks! (1 Reply)
Hi
I am looking for the correct syntax to find all files in the current directory without listing sub-directoris. I was using the following command, but it still returns subdirectoris and files inside them:
$ ls -laR | grep -v ^./
Any idea? Thanks
PS I am in ksh88 (4 Replies)
I am very new to unix as well as shell scripting.
I have to write a script for the following requirement. In have to list all the files in directory and its sub directories along with file path and size of the file
Please help me in this regard and many thanks in advance. (3 Replies)
Hi all,
I'd very grateful for some help with the following:
I have a directory with several subdirectories with files in them. All files are named different, even between different subdirectories. I also have a list with some of those file names in a txt file (without the path, just the file... (5 Replies)
Hi, ALL
thanks in advance,
i listed all files using this command
ls -ltr $(date +%Y%m%d)*.xmlbut i would like to exclude the last one created ;
Best regard
MEROUAN
Use code tags, thanks. (4 Replies)
bash or c-shell On Solaris 10
I need a command that will list the files of a directory, and grep each of those filenames out of a static text file, and then only echo field 2 of the static text file.
Something like this:
#> ls -1 /home/dir/ | each filename grep $filename static.txt |awk... (4 Replies)
Discussion started by: ajp7701
4 Replies
LEARN ABOUT PHP
datetime.setdate
DATETIME.SETDATE(3) 1 DATETIME.SETDATE(3)DateTime::setDate - Sets the date
Object oriented style
SYNOPSIS
public DateTime DateTime::setDate (int $year, int $month, int $day)
DESCRIPTION
Procedural style
DateTime date_date_set (DateTime $object, int $year, int $month, int $day)
Resets the current date of the DateTime object to a different date.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $year
- Year of the date.
o $month
- Month of the date.
o $day
- Day of the date.
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
CHANGELOG
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 5.3.0 | |
| | |
| | Changed the return value on success from NULL to |
| | DateTime. |
| | |
+--------+---------------------------------------------------+
EXAMPLES
Example #1
DateTime.setDate(3) example
Object oriented style
<?php
$date = new DateTime();
$date->setDate(2001, 2, 3);
echo $date->format('Y-m-d');
?>
Procedural style
<?php
$date = date_create();
date_date_set($date, 2001, 2, 3);
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2001-02-03
Example #2
Values exceeding ranges are added to their parent values
<?php
$date = new DateTime();
$date->setDate(2001, 2, 28);
echo $date->format('Y-m-d') . "
";
$date->setDate(2001, 2, 29);
echo $date->format('Y-m-d') . "
";
$date->setDate(2001, 14, 3);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-02-28
2001-03-01
2002-02-03
SEE ALSO DateTime.setISODate(3), DateTime.setTime(3).
PHP Documentation Group DATETIME.SETDATE(3)