bash - 当管道中的一个进程失败时退出

Question

目标是制作一个简单的非侵入式包装器，将标准输入和标准输出跟踪到标准错误:

#!/bin/bash

tee /dev/stderr | ./script.sh | tee /dev/stderr

exit ${PIPESTATUS[1]}

测试脚本script.sh:

#!/bin/bash

echo asd
sleep 1
exit 4

但是当脚本退出时，它不会终止包装器。可能的解决方案是从管道的第二个命令结束第一个 tee:

#!/bin/bash

# Second subshell will get the PID of the first one through the pipe.
# It will be able to kill the whole script by killing the first subshell.

# Create a temporary named pipe (it's safe, conflicts will throw an error).
pipe=$(mktemp -u)
if ! mkfifo $pipe; then
    echo "ERROR: debug tracing pipe creation failed." >&2
    exit 1
fi

# Attach it to file descriptor 3.
exec 3<>$pipe

# Unlink the named pipe.
rm $pipe

(echo $BASHPID >&3; tee /dev/stderr) | (./script.sh; r=$?; kill $(head -n1 <&3); exit $r) | tee /dev/stderr

exit ${PIPESTATUS[1]}

那是很多代码。还有别的办法吗？

Answer 1

我认为您正在寻找 pipefail 选项。来自 bash 手册页:

pipefail

If set, the return value of a pipeline is the value of the last (rightmost) command to exit with a non-zero status, or zero if all commands in the pipeline exit successfully. This option is disabled by default.

所以如果你开始你的包装脚本

#!/bin/bash

set -e
set -o pipefail

然后包装器将在发生任何错误时退出 (set -e) 并将以您想要的方式设置管道的状态。