在bash中快速引用标准输出（即上一条命令的输出）？ [英] Reference stdout (i.e. output of previous command) quickly in bash?

问题描述

有没有一种方法可以快速（例如通过键盘快捷键等）引用前一条命令输出到stdout的输出？

Is there a way to quickly (e.g. via a keyboard shortcut, etc.) to reference the output of the previous command's output that it wrote to stdout?

对于例如，如果我这样做：

For example, if I did this:

which rails

，它返回 / usr / local / bin / rails ，然后我想在textmate中打开该文件，我可以-像这样输入：

and it returned /usr/local/bin/rails and then I wanted to open that file in textmate, I could re-type the output like this:

mate /usr/local/bin/rails

但是有没有一种方法可以快速引用输出而不必重新输入？

but is there a way to quickly reference the output without having to re-type it?

注意：我知道我可以做 mate $（哪个轨道），但是我特别希望引用stdout。

NOTE: I am aware I can just do mate $(which rails), but I am specifically looking to reference stdout.

推荐答案

我在历史记录中使用反引号：

I use backticks with history reference:

$ which rails
/usr/local/bin/rails
$ mate `!!`

实际上，我的编辑器（以gvim开头的脚本）的别名为 e ，因此看起来较短：

Actually, my editor (a script starting gvim) is aliased to e, so it looks even shorter:

$ e `!!`

，您始终可以绑定到热键（有关 bind 命令和 readline支持，请参见bash手册页）。

and you can always bind to a hotkey (see bash man page for bind command and readline support).

此外，如果您可以使用剪切缓冲区（在X应用程序中用鼠标选择），则可能会出现如下所示的热键有用：

Also, if you can use cut buffers (select with a mouse in an X application), a hotkey for something like the below might be useful:

$ e $(xclip -out)

该命令将使用命令行剪切缓冲区中的所有内容如上所述启动编辑器。只需双击即可选择许多路径，则可以非常快速地编辑所选路径。

The command will start the editor as above with whatever was in the cut buffer on command line. Given that many paths are selectable with just a double click, a selected path can be edited very quickly.

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