Socket 函数
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socket_getsockname

(PHP 4 >= 4.1.0, PHP 5)

socket_getsocknameQueries the local side of the given socket which may either result in host/port or in a Unix filesystem path, dependent on its type

说明

bool socket_getsockname ( resource $socket , string &$addr [, int &$port ] )

Note: socket_getsockname() should not be used with AF_UNIX sockets created with socket_connect(). Only sockets created with socket_accept() or a primary server socket following a call to socket_bind() will return meaningful values.

参数

socket

A valid socket resource created with socket_create() or socket_accept().

addr

If the given socket is of type AF_INET or AF_INET6, socket_getsockname() will return the local IP address in appropriate notation (e.g. 127.0.0.1 or fe80::1) in the address parameter and, if the optional port parameter is present, also the associated port.

If the given socket is of type AF_UNIX, socket_getsockname() will return the Unix filesystem path (e.g. /var/run/daemon.sock) in the address parameter.

port

If provided, this will hold the associated port.

返回值

成功时返回 TRUE, 或者在失败时返回 FALSEsocket_getsockname() may also return FALSE if the socket type is not any of AF_INET, AF_INET6, or AF_UNIX, in which case the last socket error code is not updated.

参见


Socket 函数
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用户评论:

not at valid dot com (2007-10-13 12:35:30)

Just a quick note:
I found socket_getsockname() is not IPv6 compatible or may just return some unexpected results?

<?php
Simple Code Example
:

$socket socket_create(AF_INET,SOCK_STREAM,SOL_TCP);
    
socket_bind($socket,'0.0.0.0',150);

socket_getsockname($socket$IP$PORT);

print 
$IP.":".$PORT."\n";

?>

This does not print $IP 127.0.0.1 or 192.168.1.1 or even 0.0.0.0 ... etc ... like you would expect ... in my case I receive 10.0.0.0 witch is not a valid port on my system using IPv5 and IPv6! Thus you should rely on using the initial values of socket_bind() to get the local address. 

Hope this helps...

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