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20090620231013 to date format i am using substr, any simple way in perl?
Post 302327807 by jimmy_y on Monday 22nd of June 2009 03:48:00 PM
06-22-2009
20090620231013 to date format i am using substr, any simple way in perl?
Hi Everyone,
$tmp[2]="20090620231013";
$tmp[2] = substr($tmp[2],0,8)." ".substr($tmp[2],8,2).":".substr($tmp[2],10,2).":".substr($tmp[2],12,2);
So my output is:
20090620 23:10:13.
I only can think substr is easy, any perl can do this just one line very simple efficient one?
Thanks
$tmp[2]="20090620231013";
$tmp[2] = substr($tmp[2],0,8)." ".substr($tmp[2],8,2).":".substr($tmp[2],10,2).":".substr($tmp[2],12,2);
So my output is:
20090620 23:10:13.
I only can think substr is easy, any perl can do this just one line very simple efficient one?
Thanks
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bup-fuse
bup-fuse(1) General Commands Manual bup-fuse(1) NAME
bup-fuse - mount a bup repository as a filesystem SYNOPSIS
bup fuse [-d] [-f] [-o] <mountpoint> DESCRIPTION
bup fuse opens a bup repository and exports it as a fuse(7) userspace filesystem. This feature is only available on systems (such as Linux) which support FUSE. WARNING: bup fuse is still experimental and does not enforce any file permissions! All files will be readable by all users. When you're done accessing the mounted fuse filesystem, you should unmount it with umount(8). OPTIONS
-d, --debug run in the foreground and print FUSE debug information for each request. -f, --foreground run in the foreground and exit only when the filesystem is unmounted. -o, --allow-other permit other users to access the filesystem. Necessary for exporting the filesystem via Samba, for example. EXAMPLE
rm -rf /tmp/buptest mkdir /tmp/buptest sudo bup fuse -d /tmp/buptest ls /tmp/buptest/*/latest ... umount /tmp/buptest SEE ALSO
fuse(7), fusermount(1), bup-ls(1), bup-ftp(1), bup-restore(1), bup-web(1) BUP
Part of the bup(1) suite. AUTHORS
Avery Pennarun <apenwarr@gmail.com>. Bup unknown- bup-fuse(1)