Question

我是套接字连接的新手，我想向socket服务器发送一些命令，但我只能发送第一条消息而不是其他消息。

客户端：

$fp = stream_socket_client("tcp://127.0.0.1:1000", $errno, $errstr, 30);
if (!$fp) {
    echo "$errstr ($errno)<br />\n";
} else {
    $cmd_1 = "hello";
    fwrite($fp, $cmd_1);
    while (!feof($fp)) {
        print_r(fgets($fp, 1024));
    }
    $cmd_2 = "second message";
    fwrite($fp, $cmd_2);
    while (!feof($fp)) {
        print_r(fgets($fp, 1024));
    }
    fclose($fp);
}

如何发送多条消息，例如第一个$cmd_1，并且取决于结果，我必须发送$cmd_2？

服务器

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 = '127.0.0.1';
$port = 1000;

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);

Answer 1

问题出在下面一行：

while (!feof($fp)) {

它的作用是：它等待该套接字上的EOF（通常关闭套接字）。它会为您提供第一个响应，并等待进一步响应而不发送任何进一步的请求（循环）。您想要读取您发送的每条请求消息的响应消息，对吗？

像这样修改你的代码（未经测试，但应该有效）：

$cmd_1 = "hello";
fwrite($fp, $cmd_1);
print_r(fgets($fp, 1024));

$cmd_2 = "second message";
fwrite($fp, $cmd_2);
print_r(fgets($fp, 1024));

fclose($fp);

一旦从套接字读取一行，

fgets(..)将返回。你实际上并不需要一个while循环。

此外，开始为您的通信定义结构/协议是一个很好的观点。在客户端，您逐行阅读，但不是服务器。插件（在服务器上）读取可以返回部分消息。如果您决定通信结构，则可以将读取设计为在读取整个结构之前不返回。

Answer 2

问题在于

if (false === ($buf = socket_read($msgsock, 2048, PHP_NORMAL_READ))) {
    echo "socket_read() failed: reason: " . socket_strerror(socket_last_error($msgsock)) . "\n";
    break 2;
}

而不是将其与false === ''

进行比较

if ('' === ($buf = socket_read($msgsock, 2048, PHP_NORMAL_READ))) {
    echo "socket_read() failed: reason: " . socket_strerror(socket_last_error($msgsock)) . "\n";
    break 2;
}

否则连接过早关闭 - 请参阅此处 - http://www.php.net/manual/en/function.socket-read.php#89133

套接字客户端发送消息

2 个答案: