如何在 Linux 中使用管道指定命令行参数？

Question

我是 shell 编程的新手，不知道如何解决这个问题。

我刚刚从网上下载了一个文件到默认目录~/Downloads。我想将该文件移动到另一个目录~/Documents.

由于我不知道下载文件的确切名称，我想我可以使用以下命令来实现我的目标：

ls -t ~/Downloads | head -1 | mv [source] [destination]

如何指定要替换的形式参数。就我而言，我想省略[source]并填写我自己的[destination]参数~/Documents。

Answer 1

ls -t ~/Downloads | head -1 | xargs -I  {} mv ~/Downloads/{} ~/Documents

这将适用于名称中包含空格的文件。

Answer 2

您还可以使用 bash 的命令替换运算符（反引号）作为

mv `ls -t ~/Downloads | head -1` ~/Documents

如果您不想一次性移动多个文件，作为一次性解决方案。请参阅 bash 手册页：

Command Substitution
   Command  substitution  allows  the output of a command to replace the command name.  There
   are two forms:

          $(command)
   or
          `command`

   Bash performs the expansion by executing command and replacing  the  command  substitution
   with  the  standard  output  of the command, with any trailing newlines deleted.  Embedded
   newlines are not deleted, but they may be removed during word splitting.  The command sub?
   stitution $(cat file) can be replaced by the equivalent but faster $(< file).

   When  the  old-style backquote form of substitution is used, backslash retains its literal
   meaning except when followed by $, `, or \.  The first backquote not preceded by  a  back?
   slash terminates the command substitution.  When using the $(command) form, all characters
   between the parentheses make up the command; none are treated specially.

   Command substitutions may be nested.  To nest when using the backquoted form,  escape  the
   inner backquotes with backslashes.

   If  the  substitution  appears within double quotes, word splitting and pathname expansion
   are not performed on the results.

Answer 3

你要xargs。

echo "foo" | xargs touch
ls -l foo