对于多个命令，是否有类似于pipefail的东西，比如'try'语句，但在bash中。我想这样做:
echo "trying stuff"
try {
command1
command2
command3
}
在任何时候，如果任何命令失败了，就退出并回显该命令的错误。我不想做这样的事情:
command1
if [ $? -ne 0 ]; then
echo "command1 borked it"
fi
command2
if [ $? -ne 0 ]; then
echo "command2 borked it"
fi
等等……或者类似的事情:
pipefail -o
command1 "arg1" "arg2" | command2 "arg1" "arg2" | command3
因为我相信每个命令的参数(如果我错了请纠正我)会相互干扰。这两种方法对我来说似乎非常冗长和讨厌，所以我在这里呼吁一种更有效的方法。
当前回答
假设
alias command1='grep a <<<abc'
alias command2='grep x <<<abc'
alias command3='grep c <<<abc'
要么
{ command1 1>/dev/null || { echo "cmd1 fail"; /bin/false; } } && echo "cmd1 succeed" &&
{ command2 1>/dev/null || { echo "cmd2 fail"; /bin/false; } } && echo "cmd2 succeed" &&
{ command3 1>/dev/null || { echo "cmd3 fail"; /bin/false; } } && echo "cmd3 succeed"
or
{ { command1 1>/dev/null && echo "cmd1 succeed"; } || { echo "cmd1 fail"; /bin/false; } } &&
{ { command2 1>/dev/null && echo "cmd2 succeed"; } || { echo "cmd2 fail"; /bin/false; } } &&
{ { command3 1>/dev/null && echo "cmd3 succeed"; } || { echo "cmd3 fail"; /bin/false; } }
收益率
cmd1 succeed
cmd2 fail
这是乏味的。但是可读性还不错。
其他回答
你说的“退出并回复错误”是什么意思?如果您的意思是希望脚本在任何命令失败时立即终止，那么就这样做
set -e # DON'T do this. See commentary below.
在脚本的开头(但注意下面的警告)。不要回显错误消息:让失败的命令处理它。换句话说，如果你这样做:
#!/bin/sh
set -e # Use caution. eg, don't do this
command1
command2
command3
而command2失败，同时打印一个错误消息到stderr，那么似乎您已经实现了您想要的。(除非我误解了你想要的!)
因此，您编写的任何命令都必须表现良好:它必须向stderr而不是stdout报告错误(问题中的示例代码将错误输出到stdout)，并且在失败时必须以非零状态退出。
However, I no longer consider this to be a good practice. set -e has changed its semantics with different versions of bash, and although it works fine for a simple script, there are so many edge cases that it is essentially unusable. (Consider things like: set -e; foo() { false; echo should not print; } ; foo && echo ok The semantics here are somewhat reasonable, but if you refactor code into a function that relied on the option setting to terminate early, you can easily get bitten.) IMO it is better to write:
#!/bin/sh
command1 || exit
command2 || exit
command3 || exit
or
#!/bin/sh
command1 && command2 && command3
就我个人而言，我更喜欢使用轻量级方法，如下所示;
yell() { echo "$0: $*" >&2; }
die() { yell "$*"; exit 111; }
try() { "$@" || die "cannot $*"; }
asuser() { sudo su - "$1" -c "${*:2}"; }
使用示例:
try apt-fast upgrade -y
try asuser vagrant "echo 'uname -a' >> ~/.profile"
不管怎样，编写代码来检查每个命令是否成功的更简单的方法是:
command1 || echo "command1 borked it"
command2 || echo "command2 borked it"
它仍然很乏味，但至少是可读的。
你可以在非redhat系统上使用@john-kugelman的解决方案，在他的代码中注释掉这一行:
. /etc/init.d/functions
然后，在最后粘贴下面的代码。完全披露:这只是从Centos 7中直接复制和粘贴上述文件的相关位。
在MacOS和Ubuntu 18.04上测试。
BOOTUP=color
RES_COL=60
MOVE_TO_COL="echo -en \\033[${RES_COL}G"
SETCOLOR_SUCCESS="echo -en \\033[1;32m"
SETCOLOR_FAILURE="echo -en \\033[1;31m"
SETCOLOR_WARNING="echo -en \\033[1;33m"
SETCOLOR_NORMAL="echo -en \\033[0;39m"
echo_success() {
[ "$BOOTUP" = "color" ] && $MOVE_TO_COL
echo -n "["
[ "$BOOTUP" = "color" ] && $SETCOLOR_SUCCESS
echo -n $" OK "
[ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
echo -n "]"
echo -ne "\r"
return 0
}
echo_failure() {
[ "$BOOTUP" = "color" ] && $MOVE_TO_COL
echo -n "["
[ "$BOOTUP" = "color" ] && $SETCOLOR_FAILURE
echo -n $"FAILED"
[ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
echo -n "]"
echo -ne "\r"
return 1
}
echo_passed() {
[ "$BOOTUP" = "color" ] && $MOVE_TO_COL
echo -n "["
[ "$BOOTUP" = "color" ] && $SETCOLOR_WARNING
echo -n $"PASSED"
[ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
echo -n "]"
echo -ne "\r"
return 1
}
echo_warning() {
[ "$BOOTUP" = "color" ] && $MOVE_TO_COL
echo -n "["
[ "$BOOTUP" = "color" ] && $SETCOLOR_WARNING
echo -n $"WARNING"
[ "$BOOTUP" = "color" ] && $SETCOLOR_NORMAL
echo -n "]"
echo -ne "\r"
return 1
}
以功能的方式检查状态
assert_exit_status() {
lambda() {
local val_fd=$(echo $@ | tr -d ' ' | cut -d':' -f2)
local arg=$1
shift
shift
local cmd=$(echo $@ | xargs -E ':')
local val=$(cat $val_fd)
eval $arg=$val
eval $cmd
}
local lambda=$1
shift
eval $@
local ret=$?
$lambda : <(echo $ret)
}
用法:
assert_exit_status 'lambda status -> [[ $status -ne 0 ]] && echo Status is $status.' lls
输出
Status is 127
