bash 使用verbose和xtrace（set -vx）记录bash脚本并自定义PS4
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logging bash scripts with verbose and xtrace (set -vx) and customizing PS4
提问by Jason
I am trying to configure PS4 for better xtrace output when I also have verbose set for verbose mode. Several of the things I would like to do spawn subshells but I want to see any -x or -v output for those operations because they would print for every line.
当我还为详细模式设置了详细设置时，我正在尝试配置 PS4 以获得更好的 xtrace 输出。我想做的几件事会产生子外壳，但我想看到这些操作的任何 -x 或 -v 输出，因为它们会为每一行打印。
I also wanted to get rid of the nesting/indirection depth level indicator (+) because I want the trace output to be aligned.
我还想摆脱嵌套/间接深度级别指示器 (+)，因为我希望跟踪输出对齐。
My initial idea was:
我最初的想法是：
BACKSPACES=$'\b\b\b\b\b\b\b\b'
PS4='+`printf "%s[\t] %-16.16s:%03d (%-16.16s) $ " ${BACKSPACES:0:$BASH_SUBSHELL} $(basename $BASH_SOURCE) $LINENO ${FUNCNAME[0]:+${FUNCNAME[0]}}`'
The first problem I ran into was that $BASH_SUBSHELL doesn't always seem to be the same as the number of +'s that would print. Looking into $SHLVL didn't help either.
我遇到的第一个问题是 $BASH_SUBSHELL 似乎并不总是与要打印的 + 的数量相同。查看 $SHLVL 也无济于事。
echo $X
+ 0 2 $ echo x <-- toplevel prints one + and $BASH_SUBSHELL is 0, as expected
x
( echo subshell )
+ 1 2 $ echo subshell <-- $BASH_SUBSHELL increments to 1 as expected, but why only one + instead of two ++?
source ./test2.sh
+ 0 2 $ source ./test2.sh
echo $X$X$X$X
++ 0 2 $ echo xxxx <-- ??? is $BASH_SUBSHELL relative to the current file or something but the + indicators are not???
xxxx
subshell
I think I have gotten around this by forgetting about using $BASH_SUBSHELL and instead making the first character in PS4 an unprintable one but I'd still like to know why $BASH_SUBSHELL isn't what I expect.
我想我已经通过忘记使用 $BASH_SUBSHELL 而是使 PS4 中的第一个字符成为不可打印的字符来解决这个问题的，但我仍然想知道为什么 $BASH_SUBSHELL 不是我所期望的。
To work around the issue with subshells being created within PS4, I looked an equivalent of PS1's PROMPT_COMMAND but didn't find anything except some indications on how to implement it myself.
为了解决在 PS4 中创建子外壳的问题，我查看了 PS1 的 PROMPT_COMMAND 的等价物，但除了有关如何自己实现它的一些指示外，没有找到任何东西。
The best way I thought I found was to trap the DEBUG signal.
我认为我发现的最好方法是捕获 DEBUG 信号。
trap 'debugfun $BASH_SOURCE' DEBUG
Of course, the -vx options also apply to calls to debugfun. My solution to suppress that output was to wrap the function's steps in set +vx and set -vx.
当然，-vx 选项也适用于对 debugfun 的调用。我抑制该输出的解决方案是将函数的步骤包装在 set +vx 和 set -vx 中。
debugfun() {
set +vx
VAR=`basename .sh`
set -vx
}
Now... I had to deal with the matter of every call printing:
现在...我不得不处理每次调用打印的问题：
debugfun $BASH_SOURCE <-- from -v
++ debugfun ./test.sh <-- from -x
++ set +vx <-- from -x
I'm not sure why -v doesn't cause a print on the "set +vx" line either. I thought -T option might do that but it didn't.
我不确定为什么 -v 也不会在“set +vx”行上打印。我认为 -T 选项可能会这样做，但事实并非如此。
Anyway, I thought the output was always going to consistent so all I had to do was erase those 3 lines from within debugfun. My ugly solution was to add this line after the "set +vx":
无论如何，我认为输出总是一致的，所以我所要做的就是从 debugfun 中擦除这 3 行。我丑陋的解决方案是在“set +vx”之后添加这一行：
printf "\b\r3[K\b\r3[K\b\r3[K" # clear previous 3 lines
It worked!
有效！
... Except when I pipe more than once.
......除非我不止一次地吸管。
I'll remove the printf line to show why:
我将删除 printf 行以说明原因：
echo $X$X | sed 's/x/y/' # pipe once... debugfun is called every time before each pipe component
debugfun $BASH_SOURCE
++ debugfun ./test.sh
++ set +vx
debugfun $BASH_SOURCE
++ debugfun ./test.sh
++ set +vx
+ echo xx
+ sed s/x/y/
echo $X$X$X | sed 's/x/y/' | sed 's/y/x/' # pipe twice
debugfun $BASH_SOURCE
++ debugfun ./test.sh
++ set +vx
debugfun $BASH_SOURCE
++ debugfun ./test.sh
++ set +vx
+ echo xxx
+ sed s/x/y/ # because this line is here, clearing 3 previous lines from the next "set +vx" would clear it
debugfun $BASH_SOURCE
++ debugfun ./test.sh
++ set +vx
+ sed s/y/x/
xxx
I did, finally, come up with this solution but I think it's really ugly and, while close, isn't exactly what I want:
最后，我确实想出了这个解决方案，但我认为它真的很丑，虽然很接近，但并不是我想要的：
SPACES=" "
PS4='^@\[\e[31m\]+ [\t] \
${SPACES:0:$((${#BASH_SOURCE} > 24 ? 0 : 24 - ${#BASH_SOURCE}))}${BASH_SOURCE:$((${#BASH_SOURCE} < 24 ? 0 : -24))}:\
${SPACES:0:$((3-${#LINENO}))}$LINENO \
${FUNCNAME:-${SPACES:0:20}}${FUNCNAME:+${SPACES:0:$((20 - ${#FUNCNAME}))}} \
$ \[\e[0m\]'
The main thing I don't like is that I can't nest the bash string manipulations to not only reverse truncate $BASH_SOURCE, but to get it's basename without a subshell (i.e. combine with echo ${BASH_SOURCE##*/})
我不喜欢的主要事情是我不能嵌套 bash 字符串操作，不仅要反向截断 $BASH_SOURCE，还要在没有子shell 的情况下获得它的基本名称（即与 echo ${BASH_SOURCE##*/} 结合）
I'm out of ideas and would appreciate and clarification or tips on how to accomplish what I'm trying to do.
我没有想法，希望得到关于如何完成我正在尝试做的事情的澄清或提示。
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
回答by spbnick
I'd suggest forgoing PS4 and doing all trace output in a DEBUG trap. Most, if not all the information -x prints is available elsewhere. See BASH_SOURCE, BASH_LINENO, FUNCNAME and BASH_COMMAND.
我建议放弃 PS4 并在调试陷阱中进行所有跟踪输出。-x 打印的大多数（如果不是全部）信息都可以在其他地方获得。请参见 BASH_SOURCE、BASH_LINENO、FUNCNAME 和 BASH_COMMAND。
Like this:
像这样：
function trace()
{
echo "TRACE" \
"${BASH_SOURCE[1]}:${BASH_LINENO[0]}:${FUNCNAME[1]}:" \
"$BASH_COMMAND"
}
set -o functrace
shopt -s extdebug
trap trace DEBUG
