C 统计一个目录下的文件数
[code]$userDir = "$c:/users";
chdir($userDir);
while (<*.txt>){
my $filename = $_;
$count = 0;
open(ORIGINALORDERFILE, $_);
@file=
by shijiang1130 - Perl - 2009-01-06 17:11:39 阅读（1872） 回复（3）
$ls -R|grep '^\.\/.*\/.*\/2' ... ./Alarms/2008122908: ./Alarms/2008122909: ./Alarms/2008122910: ... ./Messages/2008122908: ./Messages/2008122909: ./Messages/2008122910: ... ... .user1/Alarms/2008122908: .user1/Alarms/2008122909: .user1/Alarms/2008122910: ... .user1/Messages/2008122908: .user1/Messages/2008122909: .user1/Messages/2008122910: ... .user2/Alarms/2008122908: .user2/Alarms/2008122909: ....
#!/usr/bin/perl $dir=0; if (!(-d "$dir")){mkdir "$dir";} for ($i=0;$i<=6666666;$i++) { open F,">>$dir/$i"; close F; } 随着文件的数目增加,cpu的占用率越来越高。怎么办。 系统是FreeBSD 6.0-RELEASE FreeBSD 6.0-RELEASE #0: Thu Nov 3 09:36:13 UTC 2005 root@x64.samsco.home:/usr/obj/usr/src/sys/GENERIC i386 # sysctl -a | grep mem memdesc 1 4K - 1 4096 vm.kmem_size: 3...
如题,求救,系统从Windows迁移至HP_UX,原Windows下一个逻辑盘里有12万个小文件,能直接移到HP_UX 11i 系统上的一个目录里吗? 如不能,有什么办法吗? 请大侠指点,谢谢!
[code] #!/bin/sh ## filename: dfdir.sh ## script for getting file list of specified directory, ## then output them in order by size. ## Powered by Muddyboot, last modified: 2005-04-26 usage(){ echo -e "\nUsage: `basename $0` DIRECTORY [MIN_SIZE]\n" echo "Get file list of DIRECTORY, then output them in order by size." echo "If MIN-SIZE is specified, only print those greater than MIN-S...
