正确引用bash别名定义 [英] Properly quote bash alias definition

问题描述

我尝试将以下命令放入bash别名中.该命令本身可以正常工作，但是当我尝试使用别名时，出现以下错误:

I have the following command that I am trying to put into a bash alias. The command by itself works fine, but when I try to alias it, I am getting the following errors:

命令

find . -maxdepth 1 -mindepth 1 -type d -exec sh -c 'echo "$(find "{}" -type f | wc -l)" {}' \; | sort -nr

别名

alias csfiles='find . -maxdepth 1 -mindepth 1 -type d -exec sh -c 'echo "$(find "{}" -type f | wc -l)" {}' \; | sort -nr'

错误:

-sh: alias 0: not found
-sh: alias {} \; | sort nr: not found

我认为这意味着我没有正确使用引号，但是在确定正确的组合时遇到了麻烦.帮助吗?

I think this means I am not using quotes right but I am having trouble determining the correct combo. Help?

推荐答案

您的外部 find 不能完成简单的glob无法做到的任何事情.这样就消除了一层引号(以及找到的每个目录的 sh 进程).

Your outer find doesn't do anything you couldn't do with a simple glob. This eliminates a layer of quotes (along with the sh process for each directory found).

# Ignoring the issue of assuming no file name contains a newline
for d in ./*/; do
   echo "$(find "$d" -type f | wc -l) $d"
done

只需定义一个shell函数即可消除对 alias 的参数施加的第二层.

Just define a shell function to eliminate the second layer imposed on the argument to alias.

csfiles () {
  for d in ./*/; do
    echo "$(find "$d" -type f | wc -l) $d"
  done
}

对 find 的其余调用也可以替换为 for 循环，从而消除了每个文件名只占一行的问题:

The remaining call(s) to find can also be replaced with a for loop, eliminating the problematic assumption of one line per file name:

csfiles () {
  for d in ./*/; do
    echo "$(for f in "$d"/*; do [ -f "$f" ] && echo; done | wc -l) $d"
  done
}

如果它支持 -printf 主键，则可以保留 find ，因为您不必关心文件的实际名称，只需要一行就可以了.每个文件的输出数.

You could keep find if it supports the -printf primary, because you don't care about the actual names of the files, just that you get exactly one line of output per file.

csfiles () {
  for d in ./*/; do
    echo "$(find "$d" -type f -printf . | wc -l) $d"
  done
}

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