如何配置指定名称的组件才在components.d.ts文件中生成类型声明

[JackieCheung]

例如，有如下组件目录结构：

--component-a
----child-a.vue
----child-b.vue
----index.vue

这样index.vue、child-a.vue、child-b.vue都会生成类型声明，但实际上child-a和child-b组件作为component-a组件的子组件，并不需要单独对外暴露的，而且child-a.vue、child-b.vue还有可能与其它组件重名，有没有办法只匹配到名字为index的组件才生成类型声明，尝试了include属性，好像没效果

[azaleta]

@JackieCheung

currently you can't do this
Glob Scan
does not filter files by include/exclude (they are settings for transform)

[kingyue737]

I also want this feature because I don't want type declarations provided by Vue2 UI library (all the component types are any), I define the types of these components myself. While I still want unplugin-vue-components to generate other components' type for me.

[paul-xia]

I also want this feature too.

[cx741041875]

您好，请问有解决办法了吗？我也有这个困惑

[nigiwen]

@lishaobos 大佬你这个pr怎么使用啊

[lishaobos]

就是和 glob 规则一样，你最开始加个 ! , 例如：!src/xx.vue

[nigiwen]

@lishaobos

你有遇到这种情况嘛，我的插件是最新的，也重启重装过依赖

[lishaobos]

这个 globs 里面可以写多个，你写了一个排除，可以再写几个命中，类似于下面

      globs: [
        'src/components/*.vue',
        '!src/components/ComponentA.vue',
      ],

[AprilLemon]

globs: [
    '!src/components/**/modules/*.vue',  //排除components下的所有modules文件夹里的组件
    'src/components/*.vue',
    'src/components/**/index.vue',
  ],

使用这种方式来排除