我有python脚本,它像这样调用shell脚本:
p = subprocess.Popen([/path/to/script.sh], stdout=subprocess.PIPE,
stderr=subprocess.STDOUT)
output, error = p.communicate()
[expect
脚本:
#!/usr/bin/env expect
set username "user"
set password "pass"
cd /my/repo
spawn /usr/bin/git pull origin master
expect "Username"
send "$username\r"
expect "Password"
send "$password\r"
interact
当我手动运行py-script时,它工作正常,我看到如下输出:
Username for repo: user
Password for repo:
* branch master -> FETCH_HEAD
Already up to date.
但是当我通过cron运行脚本时,会看到以下内容:
Username for repo: user
Password for repo:
没有任何反应。我在做什么错?
我的/ etc / crontab文件:
SHELL=/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
10 * * * * my_username /usr/bin/python3 /path/to/python/script.py
您需要将interact
更改为expect -timeout -1 eof
。
[interact
仅在Expect脚本在tty上运行但cron作业不在tty上运行时才起作用。
您可以从交互式外壳验证这一点:
# this works
expect -c 'spawn sleep 5; interact'
# this does not work
true | expect -c 'spawn sleep 5; interact'
# this works
true | expect -c 'spawn sleep 5; expect -timeout -1 eof'