php (client socket) java (server socket) cant receive second data

Question

This Is my code to connect java socket:-

$socket         = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);
socket_connect($socket, '127.0.0.1', 12345);

while(true) 
{
    // read a line from the socket
    $line = socket_read($socket, 1024, PHP_NORMAL_READ);

    var_dump($line);

    $someArray  = json_decode($line, true);
    $otp         = $someArray["otp"];

    if($someArray["msg"] == "otp_generation")
    {           
        $myObj      = new \stdClass();          
        $myObj->msg = "OTP RECEIVED NEED TO CONNECT";       
        $send       = json_encode($myObj);

        socket_send($socket, $send, strlen($send), 0);
    }
    exit;       

}

My Question is -

When connection is established successfully server send one OTP to client and received successfully in client. Then i send data to server OTP RECEIVED acknowledgement, it also received in server. After OTP RECEIVED acknowledgement server send welcome msg to client. I cant get the welcome message. if i remove the "exit" code browser is still loading, finally crashing. Why i didn't receive the second data. anyone solve my issue. what i need to modify. am beginner for socket.

I need to display Welcome msg. What can i do?

Answer 1

You need to continue looping and read the next message, then break out of the loop.

while(true) 
{
    // read a line from the socket
    $line = socket_read($socket, 1024, PHP_NORMAL_READ);

    var_dump($line);

    $someArray  = json_decode($line, true);

    if($someArray["msg"] == "otp_generation")
    {           
        $otp         = $someArray["otp"];
        $myObj      = new \stdClass();          
        $myObj->msg = "OTP RECEIVED NEED TO CONNECT";       
        $send       = json_encode($myObj);

        socket_send($socket, $send, strlen($send), 0);
    } elseif ($someArray["msg"] == "welcome") {
        // do whatever you need to do with the rest of the message
        break; // then get out of the loop
    } else {
        echo "Unknown message received";
        var_dump($someArray);
        break;
    }
}

I had to make a guess about how the welcome message is formatted, but this should give you the general idea.

Answer 2

Without new line cmd data is not send. This is the mistake i done. Finally i got the answer from my friend.

I just add the line below;-

socket_send($socket, $send."\r\n", strlen($send."\r\n"), 0);

Thanks @hemanth kumar and @Barmar