我的服务器在23999端口上打开了一个telnet端口,当我输入
telnet localhost 23999
时,它会显示以下内容:< BP-SAS ==> bplin19 !>telnet 0 23999
Trying 0.0.0.0...
Connected to 0.
Escape character is '^]'.
Please enter password to authenticate:
(here i give password for example abc123)
Enter 'help' at any point to get a listing of all registered commands...
BAS> log set-info 1 ( commad i have entered and it does somthing )
现在,我不再像这样直接打开文件,而是必须编写Java代码来完成此操作。
- 连接到23999端口的主机
- 输入密码
- 输入命令
Socket soc=new Socket("192.168.9.7",23999);
while(true){
//create buffered writer
BufferedReader bwin = new BufferedReader(new InputStreamReader(soc.getInputStream()));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(soc.getOutputStream()));
BufferedWriter bw1 = new BufferedWriter(new OutputStreamWriter(soc.getOutputStream()));
String readFir = bwin.readLine();
System.out.println(readFir);
if(readFir.startsWith("Please")){
System.out.println("Password Entered");
bw.write("abc123");
bw.flush();
bw.close(); //close buffered Reader
}
readFir = bwin.readLine();
if(readFir.startsWith("Enter")){
System.out.println("Enter command");
bw1.write("log set-info 1");
bw1.flush();
bw1.close(); //close buffered Reader
}
//readFir = bwin.readLine();
}
这不起作用。实际上,我有点困惑应该采取什么方法。在读者/写者之间非常困惑。
请帮忙。