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이 문서에서는 PHP 중국어 웹사이트 매뉴얼 풀어 주다
(PHP 4, PHP 5, PHP 7)
ftp_site — 向服务器发送 SITE 命令
$ftp_stream
, string $cmd
) ftp_site() 函数向 FTP 服务器发送由参数 cmd
指定的命令。
SITE 命令是非标准化的,不同的服务器不尽相同。主要用于处理文件权限以及组成员等事情。
ftp_stream
FTP 连接资源。
command
SITE 命令。注意本参数没有经过处理,在文件名有存在空格或其它特殊字符的情况下可能会有问题。
成功时返回 TRUE
, 或者在失败时返回 FALSE
。
Example #1 向一个 FTP 服务器发送一条 SITE 命令
<?php
$conn = ftp_connect ( 'ftp.example.com' );
if (! $conn ) die( 'Unable to connect to ftp.example.com' );
if (! ftp_login ( $conn , 'user' , 'pass' )) die( 'Error logging into ftp.example.com' );
if ( ftp_site ( $conn , 'CHMOD 0600 /home/user/privatefile' )) {
echo "Command executed successfully.\n" ;
} else {
die( 'Command failed.' );
}
?>
[#1] www.bossftp.com [2008-11-05 18:30:22]
To send a SITE command and get the result, you can try this:
<?php
function ftpsite($conn, $cmd) {
return ftp_raw($conn, "SITE $cmd");
}
?>
[#2] Marc [2003-08-05 07:02:46]
Use ftp_raw() (PHP 5+) instead of ftp_site() if you need the results of the remote command.
[#3] webmaster at beezm dot net [2003-06-17 23:52:47]
I wrote some functions for an ftp server for windows called GuildFTPd.
Here they are, they are pretty much so self explanitory:
<?php
function guildftpd_adduser($ftp_stream,$group,$username,$password){
ftp_site($ftp_stream,"SITE ADDUSER $group $username $password");
}
function guildftpd_addgroup($ftp_stream,$groupname){
ftp_site($ftp_stream,"SITE ADDGROUP $groupname");
}
function guildftpd_deluser($ftp_stream,$username){
ftp_site($ftp_stream,"SITE DELUSER $username");
}
function guildftpd_addpath($ftp_stream,$username,$path,$vpath){
ftp_site($ftp_stream,"SITE ADDPATH $username $path $vpath");
}
function guildftpd_delpath($ftp_stream,$username,$path,$vpath){
ftp_site($ftp_stream,"SITE DELPATH $username $path $vpath");
}
function guildftpd_anon($ftp_stream){
ftp_site($ftp_stream,"SITE ANON ALLOW");
}
function guildftpd_anond($ftp_stream){
ftp_site($ftp_stream,"SITE ANON DENY");
}
function guildftpd_chpass($ftp_stream,$user,$pass){
ftp_site($ftp_stream,"SITE CHPASS $user $pass");
}
function guildftpd_disable($ftp_stream,$user){
ftp_site($ftp_stream,"SITE DISABLE $user");
}
function guildftpd_enable($ftp_stream,$user){
ftp_site($ftp_stream,"SITE ENABLE $user");
}
function guildftpd_who($ftp_stream){
$who = ftp_site($ftp_stream,"SITE WHO");
}
function guildftpd_kick($ftp_stream,$user);
ftp_site($ftp_stream,"SITE KICK $user");
}
function guildftpd_msg($ftp_stream,$text);
ftp_site($ftp_stream,"SITE MSG $text");
}
?>
[#4] to at mail dot isec dot pt [2002-11-21 05:20:21]
[#5] daemorhedron [2002-11-17 05:32:40]
I'd like to follow up with Ivan's observation there. Unless I am missing the obvious, which is entirely possible of course, this command is quite useless unless it returns the data from the SITE <foo> request.
Besides the obvious SITE HELP command, another one that would require catching the feedback would be the SITE LOCATE | SITE SEARCH commands. I completely recognize that SITE <foo> commands are indeed insanely proprietary, but the fact still remains that changing this to return the results, or at least having the option to do so (such as exec() and the like) would be vastly preferred to the current method of returning true/false values.
Thanks.
[#6] [2001-06-29 09:36:27]
site commands are extra-functionality provided by some ftp servers, and depend on the server :
e.g. trying "ftp> site help" (with an interactive ftp client) on a proftp server would give :
214-The following SITE extensions are recognized:
RATIO -- show all ratios in effect
The following SITE commands are recognized (* =>'s unimplemented).
HELP CHMOD
214 Direct comments to admin@ftp.blahblah.fr
while with wu-ftpd on can typically get :
214-The following SITE commands are recognized (* =>'s unimplemented).
UMASK GROUP INDEX GROUPS
IDLE GPASS EXEC CHECKMETHOD
CHMOD NEWER ALIAS CHECKSUM
HELP MINFO CDPATH
214 blabla
Unfortunately, ftp_site only returns true or false. It would be nice to have a way to get the server's reply in full : ftp_site('help') return 1, which pretty much useless.
Hope that helps,
Ivan