Bash 打开 Shell 的脚本，运行 Shell 中的命令，并保持 Shell 运行

Question

I'm trying to automate these steps using a bash script

1. Run a command to access the shell of a program (bash shell of a Kube.netes pod)

    kubectl exec --stdin --tty hello-node-7567d9fdc9-zvz55 -- bash

2. In this shell, run a second command (eg. python )

    root@hello-node-7567d9fdc9-zvz55:/# python Python 3.7.4 (default, Aug 13 2019, 15:17:50) [Clang 4.0.1 (tags/RELEASE_401/final)]:: Anaconda, Inc. on linux Type "help", "copyright", "credits" or "license" for more information. >>>

3. Keep the shell open/running in the foreground (so more commands can immediately be entered by a user into the python shell started in step 2)

I'm able to do the first step described above using the script below. How do we do the second and third step?

#! /bin/bash
bash -c 'kubectl exec --stdin --tty hello-node-7567d9fdc9-zvz55 -- bash'

Answer 1

For step 2 you can use the nohup command to run a process and even if the user logs off the process does not get killed. Afterwards the shell is kept running for you to perform step 3