在bash中将命令作为超时参数

Question

我可以使用超时：

timeout 5s sleep 6s

但是我怎样才能传递给超时更复杂的命令呢？我试过跟随，但都给了我错误：

timeout 5s sleep 6s && echo "You did not fit in timeout"
timeout 5s 'sleep 6s && echo "You did not fit in timeout"'
timeout 5s {sleep 6s && echo "You did not fit in timeout"}
timeout 5s (sleep 6s && echo "You did not fit in timeout")

Answer 1

看看这个：

$ timeout 5 sleep 2 && echo OK || echo "You did not fit in timeout"
OK
$ timeout 1 sleep 2 && echo OK || echo "You did not fit in timeout"
You did not fit in timeout

Answer 2

几乎在发布后我立即想出来：

timeout 5s sh -c 'sleep 6s && echo "You did not fit in timeout"'

有没有更好的解决方案？