bash - timeout命令 - shell超时退出




使用超时重试Bash命令 (2)

重试功能来自:

http://fahdshariff.blogspot.com/2014/02/retrying-commands-in-shell-scripts.html

#!/bin/bash

# Retries a command on failure.
# $1 - the max number of attempts
# $2... - the command to run
retry() {
    local -r -i max_attempts="$1"; shift
    local -r cmd="[email protected]"
    local -i attempt_num=1

    until $cmd
    do
        if (( attempt_num == max_attempts ))
        then
            echo "Attempt $attempt_num failed and there are no more attempts left!"
            return 1
        else
            echo "Attempt $attempt_num failed! Trying again in $attempt_num seconds..."
            sleep $(( attempt_num++ ))
        fi
    done
}

# example usage:
retry 5 ls -ltr foo

如果你想在你的脚本中重试一个函数,你应该这样做:

# example usage:
foo()
{
   #whatever you want do.
}

declare -fxr foo
retry 3 timeout 60 bash -ce 'foo'

如何重试bash命令直到其状态正常或达到超时?

我最好的镜头(我正在寻找更简单的东西):

NEXT_WAIT_TIME=0
COMMAND_STATUS=1
until [ $COMMAND_STATUS -eq 0 || $NEXT_WAIT_TIME -eq 4 ]; do
  command
  COMMAND_STATUS=$?
  sleep $NEXT_WAIT_TIME
  let NEXT_WAIT_TIME=NEXT_WAIT_TIME+1
done

您可以通过在测试中放置command并稍微改变一点来简化一些事情。 否则脚本看起来很好:

NEXT_WAIT_TIME=0
until command || [ $NEXT_WAIT_TIME -eq 4 ]; do
   sleep $(( NEXT_WAIT_TIME++ ))
done






timeout