CP退出了64错误状态 [英] cp exits with a 64 error status

问题描述

我使用的是preflight bash脚本中的 packagemaker

I am using a preflight bash script in packagemaker :

运行CP -pf/文件夹/到/我/ DB/库/ Application Support /应用/数据库

run函数（即我StackOverflow上发现的方式）：

The run function (that I found on StackOverflow by the way) :

的run（）{$ *; code = $ ?; [$$ C $ç-ne 0]放大器;＆安培; echo命令[$ *]，错误code $ $ C $ç\\ n错误失败：$ @ \\ n; }

命令 CP 返回64 code。这是什么64的状态吗？
我该如何解决？

The command cp returns a 64 code. What is this 64 status please? How can I resolve that?

推荐答案

的问题是，你没有一个文件夹支持/应用/数据库的命令复制文件 /文件夹/到/我/ DB 和 /库/应用程序到

The problem is that you don't have a folder Support/app/db for the command to copy files /folder/to/my/db and /Library/Application to.

替换错误的（几乎总是错的） $ * 用正确的$ @：

Replace the misguided (almost always wrong) $* with the correct "$@":

run()
{
    "$@"
    code=$?
    [ $code -ne 0 ] && echo "command [$*] failed with error code $code\nERROR: $@\n"
}

平原 $ * 在空间打破字; $ @ preserves在争论的空间。大多数情况下， $ * 是不正确的符号（虽然这将是在回声，您使用<$罚款C $ C> $ @ ）。为什么该命令的参数错误消息被列出了两次，是我也不清楚。

Plain $* breaks words at spaces; "$@" preserves spaces in arguments. Most often, $* is not the right notation (though it would be fine in the echo where you used $@). It isn't clear to me why the command's arguments are being listed twice in the error message.

错误报告将通过增加改善＆GT;和2 到最后输出到标准错误，这是在错误信息属于重定向。 （而我在这我想删除的重复。）请注意，使用 $ * 的参数内部回声是完全适当的。

The error reporting would be improved by adding >&2 to 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 echo is entirely appropriate.

    [ $code -ne 0 ] && echo "command [$*] failed with error code $code" >&2

其实，的run（）功能可以更加简化;变量 code 真的不需要

In fact, the run() function can be simplified still more; the variable code really isn't needed:

run()
{
    "$@" || echo "command [$*] failed with error code $?" >&2
}

如果您希望脚本退出过，那么你可以使用：

If you want the script to exit too, then you can use:

run()
{
    "$@" || { echo "command [$*] failed with error code $?" >&2; exit 1; }
}

的 {...; } 符号把命令中作为一个单元进行I / O重定向，而无需启动一个子shell。

The { ...; } notation treats the commands within as a unit for I/O redirection without starting a sub-shell.

又见<一个href=\"http://stackoverflow.com/questions/255898/how-to-iterate-over-arguments-in-bash-script/256225#256225\">How迭代变量在Bash脚本。

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