bash cp 以 64 错误状态退出
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原文地址: http://stackoverflow.com/questions/25790736/
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cp exits with a 64 error status
提问by arnaud
I am using a preflight bash script in packagemaker:
我在packagemaker 中使用了预检 bash 脚本：
run cp -pf "/folder/to/my/db" "/Library/Application Support/app/db
run cp -pf "/folder/to/my/db" "/Library/Application Support/app/db
The run function (that I found on StackOverflow by the way) :
运行函数（我在 StackOverflow 上找到的）：
run() { $*; code=$?; [ $code -ne 0 ] && echo "command [$*] failed with error code $code\nERROR: $@\n"; }
run() { $*; code=$?; [ $code -ne 0 ] && echo "command [$*] failed with error code $code\nERROR: $@\n"; }
The command cpreturns a 64 code. What is this 64 status please?
How can I resolve that?
该命令cp返回 64 代码。请问这个64状态是什么？我该如何解决？
回答by Jonathan Leffler
The problem is that you don't have a folder Support/app/dbfor the command to copy files /folder/to/my/dband /Library/Applicationto.
问题是，你没有一个文件夹，Support/app/db该命令复制文件/folder/to/my/db和/Library/Application到。
Replace the misguided (almost always wrong) $*with the correct "$@":
用$*正确的替换被误导的（几乎总是错误的）"$@"：
run()
{
"$@"
code=$?
[ $code -ne 0 ] && echo "command [$*] failed with error code $code\nERROR: $@\n"
}
Plain $*breaks words at spaces; "$@"preserves spaces in arguments. Most often, $*is not the right notation (though it would be fine in the echowhere you used $@). It isn't clear to me why the command's arguments are being listed twice in the error message.
平原$*在空格处断词；"$@"保留参数中的空格。大多数情况下，$*不是正确的表示法（尽管echo在您使用的地方会很好$@）。我不清楚为什么命令的参数在错误消息中列出了两次。
The error reporting would be improved by adding >&2to the end to redirect the output to standard error, which is where error messages belong. (I'd remove the repetition while I'm at it.) Note that using $*inside the argument to echois entirely appropriate.
通过添加>&2到末尾以将输出重定向到标准错误，错误报告将得到改进，这是错误消息所属的地方。（当我在做的时候，我会删除重复。）请注意，$*在参数中使用toecho是完全合适的。
[ $code -ne 0 ] && echo "command [$*] failed with error code $code" >&2
In fact, the run()function can be simplified still more; the variable codereally isn't needed:
事实上，run()功能还可以进一步简化；code确实不需要变量：
run()
{
"$@" || echo "command [$*] failed with error code $?" >&2
}
If you want the script to exit too, then you can use:
如果您也希望脚本退出，则可以使用：
run()
{
"$@" || { echo "command [$*] failed with error code $?" >&2; exit 1; }
}
The { ...; }notation treats the commands within as a unit for I/O redirection without starting a sub-shell.
该{ ...; }符号将其中的命令视为 I/O 重定向的一个单元，而无需启动子 shell。
See also How to iterate over arguments in a Bash script.
另请参阅如何在 Bash 脚本中迭代参数。
