1
我一直堅持這一段時間了。我正嘗試使用PHP將數據發送到Java服務器。當我加載Bukkit插件,它停止加載時,我調用這個函數:Java Socket偵聽器問題?
public void SocketListen()
{
String clientSentence;
String capitalizedSentence;
try
{
ServerSocket welcomeSocket = new ServerSocket(25566);
while(true)
{
Socket connectionSocket = welcomeSocket.accept();
BufferedReader inFromClient =
new BufferedReader(new InputStreamReader(connectionSocket.getInputStream()));
DataOutputStream outToClient = new DataOutputStream(connectionSocket.getOutputStream());
clientSentence = inFromClient.readLine();
System.out.println("Received: " + clientSentence);
capitalizedSentence = clientSentence.toUpperCase() + '\n';
outToClient.writeBytes(capitalizedSentence);
}
}
catch(IOException e)
{
getLogger().severe(e.toString());
}
}
它告訴我,我做錯了什麼。我的PHP是:
<?php
$host = "localhost";
$port = 25566;
$data = 'test\n';
if (($socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP)) === FALSE)
echo "socket_create() failed: reason: " . socket_strerror(socket_last_error());
else
{
echo "Attempting to connect to '$host' on port '$port'...<br>";
if (($result = socket_connect($socket, $host, $port)) === FALSE)
echo "socket_connect() failed. Reason: ($result) " . socket_strerror(socket_last_error($socket));
else {
echo "Sending data...<br>";
socket_write($socket, $data, strlen($data));
echo "OK<br>";
echo "Reading response:<br>";
while ($out = socket_read($socket, 2048)) {
echo $out;
}
}
socket_close($socket);
}
?>
我在做什麼錯?基本上,我只需要PHP服務器發送「Hello」,然後Java客戶端使用getLogger()。info(data)將其吐出。
此外,我需要端口轉發25566在客戶端或服務器與PHP?它們都使用相同的外部IP在本地網絡上託管。
我的世界沒有使用端口:25565嗎? @ user2310289 –
你在一行寫東西? – Dru
抱歉需要戴上眼鏡。 –