无法在 PHP 中创建客户端服务器套接字程序

Question

我正在为类似聊天的应用程序开发客户端服务器套接字程序，但它给出了以下错误消息

警告：socket_bind()：无法绑定地址 [98]：地址已在使用中......无法绑定到第 15 行的地址

我看过各种教程，但我希望服务器持续监听客户端请求。我已经看到它也打开了端口号，但仍然是相同的消息。我被困在这一点上好几天无法得到正确的解决方案。请帮忙解决谢谢.....

服务器.php：

 <?php
    error_reporting(E_ALL);

    /* Allow the script to hang around waiting for connections. */
    set_time_limit(0);

    /* Turn on implicit output flushing so we see what we're getting
     * as it comes in. */
    ob_implicit_flush();

    $address = 'XXX.XX.XX.XXX';
    $port = 15213;

    if (($sock = socket_create(AF_INET, SOCK_STREAM, SOL_TCP)) === false) {
        echo "socket_create() failed: reason: " . socket_strerror(socket_last_error()) . "\n";
    }

    if (socket_bind($sock, $address, $port) === false) {
        echo "socket_bind() failed: reason: " . socket_strerror(socket_last_error($sock)) . "\n";
    }

    if (socket_listen($sock, 5) === false) {
        echo "socket_listen() failed: reason: " . socket_strerror(socket_last_error($sock)) . "\n";
    }

    do {
        if (($msgsock = socket_accept($sock)) === false) {
            echo "socket_accept() failed: reason: " . socket_strerror(socket_last_error($sock)) . "\n";
            break;
        }
        /* Send instructions. */
        $msg = "\nWelcome to the PHP Test Server. \n" .
            "To quit, type 'quit'. To shut down the server type 'shutdown'.\n";
        socket_write($msgsock, $msg, strlen($msg));

        do {
            if (false === ($buf = socket_read($msgsock, 2048, PHP_NORMAL_READ))) {
                echo "socket_read() failed: reason: " . socket_strerror(socket_last_error($msgsock)) . "\n";
                break 2;
            }
            if (!$buf = trim($buf)) {
                continue;
            }
            if ($buf == 'quit') {
                break;
            }
            if ($buf == 'shutdown') {
                socket_close($msgsock);
                break 2;
            }
            $talkback = "PHP: You said '$buf'.\n";
            socket_write($msgsock, $talkback, strlen($talkback));
            echo "$buf\n";
        } while (true);
        socket_close($msgsock);
    } while (true);

    socket_close($sock);
    ?>

客户端.php

    <?php
error_reporting(E_ALL);

echo "<h2>TCP/IP Connection</h2>\n";

/* Get the port for the WWW service. */
$service_port = getservbyname('www', 'tcp');

/* Get the IP address for the target host. */
$address = gethostbyname('www.example.com');

/* Create a TCP/IP socket. */
$socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);
if ($socket === false) {
    echo "socket_create() failed: reason: " . socket_strerror(socket_last_error()) . "\n";
} else {
    echo "OK.\n";
}

echo "Attempting to connect to '$address' on port '$service_port'...";
$result = socket_connect($socket, $address, $service_port);
if ($result === false) {
    echo "socket_connect() failed.\nReason: ($result) " . socket_strerror(socket_last_error($socket)) . "\n";
} else {
    echo "OK.\n";
}

$in = "HEAD / HTTP/1.1\r\n";
$in .= "Host: www.example.com\r\n";
$in .= "Connection: Close\r\n\r\n";
$out = '';

echo "Sending HTTP HEAD request...";
socket_write($socket, $in, strlen($in));
echo "OK.\n";

echo "Reading response:\n\n";
while ($out = socket_read($socket, 2048)) {
    echo $out;
}

echo "Closing socket...";
socket_close($socket);
echo "OK.\n\n";
?>

Answer 1

在两个 php 文件中，您在同一端口上创建一个 Socket。这就是问题所在。

服务器必须创建套接字，而客户端必须使用套接字（也不创建它）

两个应用程序不能创建同一个套接字并监听同一个端口。

编辑： 我已经在本地主机上使用 xampp 测试了您的脚本。所以IP是127.0.0.1。服务器在端口 15213 上正确侦听，客户端在此端口上正确连接。 我用 xampp 控制面板看到了端口是打开的。

如果你使用你在这里发布的脚本，那么你必须在 Server.php 中替换

$address = 'XXX.XX.XX.XXX';

与

$address = '127.0.0.1';

在Client.php中

$address = gethostbyname('www.example.com');

与

$address = gethostbyname('localhost');
or
$address = '127.0.0.1';