为什么 zsh `for` 循环出错时出现 `command not found: filename.txt`?尽管 filename.txt 是参数而不是命令

why does zsh `for` loop error with `command not found: filename.txt`? despite filename.txt being a parameter and not a command

以下 shell 历史记录显示了两次重命名文件的尝试,第二次尝试成功了。但是第一次尝试没有,我不知道为什么...

❯ ls
step-10.txt  step-12.txt  step-14.txt  step-16.txt  step-18.txt  step-1.txt  step-3.txt  step-5.txt  step-7.txt  step-9.txt
step-11.txt  step-13.txt  step-15.txt  step-17.txt  step-19.txt  step-2.txt  step-4.txt  step-6.txt  step-8.txt
❯ for f in *.txt; do
newname=$(echo $f | sed 's/txt/md/')\
mv $f $newname
done
zsh: command not found: step-10.txt
zsh: command not found: step-11.txt
zsh: command not found: step-12.txt
zsh: command not found: step-13.txt
zsh: command not found: step-14.txt
zsh: command not found: step-15.txt
zsh: command not found: step-16.txt
zsh: command not found: step-17.txt
zsh: command not found: step-18.txt
zsh: command not found: step-19.txt
zsh: command not found: step-1.txt
zsh: command not found: step-2.txt
zsh: command not found: step-3.txt
zsh: command not found: step-4.txt
zsh: command not found: step-5.txt
zsh: command not found: step-6.txt
zsh: command not found: step-7.txt
zsh: command not found: step-8.txt
zsh: command not found: step-9.txt
↵ Error. Exit status 127.
❯ for f in *.txt; do
mv $f $(echo $f | sed 's/txt/md/')
done
❯ ls
step-10.md  step-12.md  step-14.md  step-16.md  step-18.md  step-1.md  step-3.md  step-5.md  step-7.md  step-9.md
step-11.md  step-13.md  step-15.md  step-17.md  step-19.md  step-2.md  step-4.md  step-6.md  step-8.md

为什么当我将 $(echo $f | sed 's/txt/md/') 放在 mv 行时它起作用,但当它被分配给变量时却不起作用?当我的循环提供 $f 作为 mv 命令的参数时,为什么它说 command not found: $f

那个反斜杠表示它正在计算

newname=$(echo $f | sed 's/txt/md/')mv $f $newname

这当然将文件名视为 运行 的命令名称(mvnewname 变量的一部分)

这里根本不需要sed;只需使用参数扩展:

for f in *.txt; do
    mv "$f" "${f/%txt/md}"
done

还有几个名为 renameprenamerename.ul 等,取决于 OS)的不同程序可以批量重命名文件;检查您已安装的手册页(如果有)和使用情况。