ZSH:如何禁用`make`的默认完成? [英] ZSH: How to disable default completion for `make`?

问题描述

我为 make 命令编写了我的自动完成功能，并将其放置在 ~/.zsh 中:

I wrote my auto-completion function for make command, and placed it in ~/.zsh:

function __comp_make {
    # ... function body ....
}
compctl -K __comp_make make

不幸的是，它不会工作，因为 make 的完成已经在

Unfortunately, it will not work because completion for make is already defined in

/usr/share/zsh/functions/Completion/Unix/_make

显然它优先于我的规则.

and apparently it takes precedence over my rule.

我不得不将 _make 重命名为 unused_make 以便它不会在 zsh 初始化时加载.它有效，但它是相当丑陋的解决方案.

I had to rename _make to unused_make so it is not loaded at zsh initialization. It works, but it is rather ugly solution.

我的问题是:我应该如何设置完成规则，使其优先于加载的默认值?

My question is: how should I set my completion rule so it takes precedence over the loaded defaults?

相关:

• 如何覆盖现有的 zsh 键盘完成?

• zsh 4.3.17

推荐答案

你需要设置你的fpath.请参阅此处.

在你的 ~/.zshrc 中，类似:

fpath=( ~/.zshfunctions $fpath )

其中 ~/.zshfunctions 包含您的自定义完成文件，应该为您解决这个问题.请注意，您的功能在系统功能之前加载.这将不起作用:

where ~/.zshfunctions contains your custom completion files, should solve this for you. Note that your functions are loaded before the system ones. This will not work:

fpath=( $fpath ~/.zshfunctions ) 

顺便说一句，您正在使用 compctl.它老了.我建议改用 compsys ，尽管一个很好的链接解释了为什么现在让我逃脱.

As an aside, you're using compctl. It's old. I'd recommend using compsys instead, although a decent link explaining why escapes me at the moment.

这些是关于编写完成函数的一些细节，包括 fpath.您可能会发现它们可供参考.

These go into some detail about writing completion functions, including the fpath. You might find them useful for reference.

• Z-shell 补全函数: 介绍
• 编写 z-shell 补全函数

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