能否收集满足条件的文件夹名称做ts类型？-灵析社区

需要用node去读文件夹与文件属性: const fs = require('fs'); const path = require('path'); function getDirectoriesWithIndexVue(directory) { let result = []; const files = fs.readdirSync(directory, { withFileTypes: true }); for (const file of files) { if (file.isDirectory()) { const subDir = path.join(directory, file.name); const indexPath = path.join(subDir, 'index.vue'); if (fs.existsSync(indexPath)) { result.push(file.name); } result = result.concat(getDirectoriesWithIndexVue(subDir)); } } return result; } // 与 root 文件夹同级的下执行, 进入 root const rootDirectory = './src'; const items = getDirectoriesWithIndexVue(rootDirectory); console.log(items); // 输出包含 index.vue 文件的子文件夹名称 输出: