php socket编程 发送json字符串接不到

2024-11-22 05:19:59
推荐回答(1个)
回答1:

代码如下:

// 设置一些基本的变量
$host =
"192.168.1.99";
$port = 1234;
// 设置超时时间
set_time_limit(0);
//
创建一个Socket
$socket = socket_create(AF_INET, SOCK_STREAM, 0) or die("Could
not create
socket\n");
//绑定Socket到端口
$result = socket_bind($socket,
$host, $port) or die("Could not bind to
socket\n");
// 开始监听链接

$result = socket_listen($socket, 3) or die("Could not set up socket

listener\n");
// accept incoming connections
// 另一个Socket来处理通信

$spawn = socket_accept($socket) or die("Could not accept incoming

connection\n");
// 获得客户端的输入
$input = socket_read($spawn, 1024) or
die("Could not read input\n");
// 清空输入字符串
$input = trim($input);

//处理客户端输入并返回结果
$output = strrev($input) . "\n";
socket_write($spawn,
$output, strlen ($output)) or die("Could not write
output\n");
//
关闭sockets
socket_close($spawn);
socket_close($socket);
?>