将python代码与Visual Studio c＃集成

Question

我已经在Visual Studio 2019中创建了GuI。

[用户将输入用户名和密码，我必须将其传递给python脚本。当用户单击登录按钮时，将触发python脚本并显示输出。

我尝试过的python代码是：

import paramiko
import time

ssh = paramiko.SSHClient()
ssh.set_missing_host_key_policy(paramiko.AutoAddPolicy())

try:

    hostname = input("Enter host IP address: ")
    username = input("Enter SSH Username: ")
    password = input("Enter SSH Password: ")
    port = 22


    ssh.connect(hostname, port, username, password, look_for_keys=False)
    print("ssh login successfully")



#stdin,stdout,stderr = ssh.exec_command('show version')
#output = stdout.readlines()
#print(output)


    Device_access = ssh.invoke_shell()
    Device_access.send(b'environment no more \n')
    Device_access.send(b'show version\n')
    time.sleep(2)
    output = Device_access.recv(65000)
    print (output.decode('ascii'))

except:
    print("error in connection due to wrong input entered")

但是在此我没有得到如何与输入链接的方法，请使用python脚本输入Gui c＃。请让我知道我该怎么做。

提前感谢！

Answer 1

更改python脚本：

import paramiko
import time
import sys # Used to get arguments

ssh = paramiko.SSHClient()
ssh.set_missing_host_key_policy(paramiko.AutoAddPolicy())

try:

   hostname = sys.args[1] # Skip 0th arg, since it is just the filename
   username = sys.args[2]
   password = sys.args[3]
   port = 22

   ssh.connect(hostname, port, username, password, look_for_keys=False)
   print("ssh login successfully")

   #stdin,stdout,stderr = ssh.exec_command('show version')
   #output = stdout.readlines()
   #print(output)


   Device_access = ssh.invoke_shell()
   Device_access.send(b'environment no more \n')
   Device_access.send(b'show version\n')
   time.sleep(2)
   output = Device_access.recv(65000)
   print (output.decode('ascii'))

except:
       print("error in connection due to wrong input entered")

然后将调用脚本的C＃代码更改为如下代码：

Process pythonScript = new Process();
pythonScript.StartInfo.FileName = "Your python script";
pythonScript.StartInfo.Arguments = $"{YouHostnameVar} {YouUsernameVar} {YourPasswordVar}"; // Start the script with the credentials as arguments
pythonScript.Start();