linux - 了解 xargs 选项 -i 和 -t

Question

这个例子中的-i和-t是什么

ls . | xargs -i -t cp ./{} $1

我知道这是将当前目录中的所有内容复制到传入的参数 ($1) 中，但我不明白什么是 -i 和 -t 确实如此。

the example above is a snippet in a bash script file in case you are wondering of the "argument.

Answer 1

来自 man xargs :

   -I replace-str
          Replace occurrences of replace-str in the initial-arguments
          with names read from standard input.  Also, unquoted blanks do
          not terminate input items; instead the separator is the
          newline character.  Implies -x and -L 1.

   -i[replace-str], --replace[=replace-str]
          This option is a synonym for -Ireplace-str if replace-str is
          specified.  If the replace-str argument is missing, the effect
          is the same as -I{}.  This option is deprecated; use -I
          instead.

   -t, --verbose
          Print the command line on the standard error output before
          executing it.

在您的例子中，-i 后面没有 replace-str，因此可以将其视为 -I{}。这意味着 cp ./{} $1 中的 {} 被替换为每个要复制的文件名。