Bash:当指定文件夹为空时，通过find命令退出脚本 [英] Bash: exit a script when a specified folder is empty via the find command

问题描述

我要检查文件夹是否为空.如果是，则编写一个日志文件并退出脚本.

I want to check whether a folder is empty. If yes, write a log file and exit the script.

我的代码如下:

find $folder -maxdepth 0 -empty -exec echo [`date`] $folder is empty > log.txt & exit 0 \;

但是，我收到了

find: missing argument to `-exec'

您能帮我解决问题吗?

谢谢！

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谢谢大家！

我只是试图只退出没有输出日志的脚本:

I just tried only to exit the script without output log:

find $folder -maxdepth 0 -empth -exec exit 0 \;

或

find $folder -type d -empty -exec exit 0 \;

我收到错误:

find: ‘exit’: No such file or directory

即使我将 set -e 放在脚本顶部，该脚本仍能继续运行.我不知道如何解决它.

And the script keeps running even I have put set -e at the top of the script. I have no idea how to fix it.

推荐答案

如果我正确理解了这个问题，这应该可以工作:

If I understand the question correctly, this should work:

if find "$folder" -maxdepth 0 -empty | grep -q "."; then
    echo "[$(date)] $folder is empty" > log.txt    # Should this be >> to append?
    exit 0
fi

说明:如果目录 $ folder 为空，则 find 命令将打印其路径； grep 检查是否有任何东西("."是匹配任何字符的模式)，如果匹配则退出并显示成功状态；否则，返回成功.然后 if 语句根据 grep 的退出状态运行或跳过其内容.

Explanation: if the directory $folder is empty, the find command prints its path; grep checks to see whether it got anything ("." is a pattern that matches any character), and exits with a success status if there was a match; the if statement then runs or skips its content based on grep's exit status.

说明原始代码为何不起作用的原因: -exec 中的命令不是完整的shell命令，只是将要传递给 execl() 系统调用(或相关功能).也就是说，它是可执行文件的文件名(可以在常见的 PATH 中找到)，后跟要传递给它的参数列表.

Explanation of why the original didn't work: the command in -exec is not a full shell command, it's just a series of words that're going to be passed to the execl() system call (or a related function). That is, it's a filename of an executable (to be found in the usual PATH) followed by a list of arguments to be passed to it.

这有两个重要的含义:您不能在命令中使用任何任何 shell功能，因此不进行任何重定向(> log.txt )，不加入任何多个命令通过& 或; 或 | 或类似的方法，没有变量替换等.就您的命令而言，所有这些将通过外壳 before 进行解析，然后将结果传递给 find -因此，重定向到log.txt的过程将发生在整个 find 命令中， date 替换发生在运行 find 之前，& 将find命令置于后台，并将之后的部分视为完全不同的命令.

This has two important implications: you can't use any shell features in the command, so no redirects (> log.txt), no multiple commands joined by & or ; or | or anything like that, no variable substitutions, etc. In the case of your command, all of those things will get parsed by the shell before passing the result to find -- so the redirect to log.txt happens to the entire find command, the date substitution happens before find is run, the & puts the find command in the background and treats the part after as a completely different command.

此外， exit 不是可以与 execl()或类似代码一起运行的常规命令；它是内置的shell(请尝试 type -a exit )，因此它不能由 -exec 执行.即使可以，它也不会做您想要的事情，因为由 -exec 运行的命令是作为子进程运行的，它只会退出该子进程，而不是整个脚本(甚至是退出) find 命令).

Furthermore, exit is not a regular command that can be run with execl() or such; it's a shell builtin (try type -a exit), and so it cannot be executed by -exec. And even if it could, it wouldn't do what you want, because the command run by -exec is run as a subprocess, and it would just exit that subprocess, not the entire script (or even the find command).

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