获取所有子目录并返回它们的路径？

huangapple
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英文:

How to get all subdirectories and return their paths?

问题

[
'apps/group1/project1',
'apps/group1/project2',
'apps/group2/project3',
'apps/group2/project4'
]

英文:

Assume there is the following folder structure:

  1. /apps
  2. |- /group1
  3. |  |- /project1
  4. |  |- /project2
  5. |- /group2
  6. |  |- /project3
  7. |  |- /project4

I need to get the path of all projects - which are all two level deep (group folder and then project folder).

So I started with this:

  1. import { readdirSync } from 'fs'
  3. const directoriesInDirectory = readdirSync([rootPath, 'apps'].join('/'), {
  4.   withFileTypes: true
  5. })
  6. .filter((item) => item.isDirectory())
  7. .map((item) => item.name)

This gives me all groups ['group1', 'group2'].

But I need to get this:

  1. [
  2.     'apps/group1/project1',
  3.     'apps/group1/project2',
  4.     'apps/group2/project3',
  5.     'apps/group2/project4'
  6. ]

答案1

得分: 0

你需要多次执行readdirSync，一次用于app目录，其余用于嵌套目录。另外，可以使用 promises 来替代同步操作，如下所示：

  1. import { readdir } from 'fs/promises';
  3. const startPath = [rootPath, 'apps'].join('/');
  5. const parentDirs = (
  6.   await readdir(startPath, {
  7.     withFileTypes: true,
  8.   })
  9. )
  10.   .filter((item) => item.isDirectory())
  11.   .map((item) => item.name);
  13. const projectPaths = (
  14.   await Promise.all(
  15.     parentDirs.map(async (dir) => {
  16.       const projects = (
  17.         await readdir(startPath + '/' + dir, {
  18.           withFileTypes: true,
  19.         })
  20.       )
  21.         .filter((item) => item.isDirectory())
  22.         .map((item) => item.name);
  23.       return { dir, projects };
  24.     })
  25.   )
  26. )
  27.   .map(({ dir, projects }) =>
  28.     projects.map((project) => [startPath, dir, project].join('/'))
  29.   )
  30.   .flat();
  32. console.log(projectPaths); // ['apps/group1/project1', 'apps/group1/project2', ...]
英文:

You need to perform readdirSync multiple times, once for the app directory, and the rest for the nested directories. Also, instead of using sync, if you adapt the non-blocking nature, you can use promises instead:

  1. import { readdir } from 'fs/promises'
  3. const startPath = [rootPath, 'apps'].join('/')
  5. const parentDirs = (
  6.   await readdir(startPath, {
  7.     withFileTypes: true,
  8.   })
  9. )
  10.   .filter((item) => item.isDirectory())
  11.   .map((item) => item.name)
  13. const projectPaths = (
  14.   await Promise.all(
  15.     parentDirs.map(async (dir) => {
  16.       const projects = (
  17.         await readdir(startPath + '/' + dir, {
  18.           withFileTypes: true,
  19.         })
  20.       )
  21.         .filter((item) => item.isDirectory())
  22.         .map((item) => item.name)
  23.       return { dir, projects }
  24.     })
  25.   )
  26. )
  27.   .map(({ dir, projects }) =>
  28.     projects.map((project) => [startPath, dir, projects].join('/'))
  29.   )
  30.   .flat()
  32. console.log(projectPaths) // ['apps/group1/project1', 'apps/group1/project2', ...]

答案2

得分: 0

我相信这段代码应该能够满足您的需求，如果您添加更多子目录，不确定它是否能够正常工作，我进行了有限的测试，但应该可以正常运行。

  1. const fs = require("fs");
  2. const path = require("path");
  4. function getSubdirectories(rootPath) {
  5.   const result = [];
  7.   const directoriesInDirectory = fs
  8.     .readdirSync(rootPath, {
  9.       withFileTypes: true
  10.     })
  11.     .filter(item => item.isDirectory())
  12.     .map(item => item.name);
  14.   directoriesInDirectory.forEach(dir => {
  15.     const subdirectories = getSubdirectories(path.join(rootPath, dir));
  16.     // 对于根目录中的每个目录，检查是否有子目录
  17.     if (subdirectories.length > 0) {
  18.       subdirectories.forEach(subdir => {
  19.         result.push(path.join(dir, subdir));
  20.       });
  21.     } else {
  22.       result.push(dir);
  23.     }
  24.   });
  26.   return result;
  27. }
  29. const subdirectories = getSubdirectories("./app");
  31. const subdirectoryPaths = subdirectories.map(subdir => {
  32.   return path.join("app", subdir);
  33. });
  35. console.log(subdirectoryPaths);

希望这对您有帮助。

英文:

I believe this code should do what you need, I dont know how well it would work if you add more subdirectories, I did limited testing, but it should work fine.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

  1. const fs = require(&quot;fs&quot;);
  2. const path = require(&quot;path&quot;);
  4. function getSubdirectories(rootPath) {
  5.   const result = [];
  7.   const directoriesInDirectory = fs
  8.     .readdirSync(rootPath, {
  9.       withFileTypes: true
  10.     })
  11.     .filter(item =&gt; item.isDirectory())
  12.     .map(item =&gt; item.name);
  14.   directoriesInDirectory.forEach(dir =&gt; {
  15.     const subdirectories = getSubdirectories(path.join(rootPath, dir));
  16.     // for each directory in the root directory, check if it has subdirectories
  17.     if (subdirectories.length &gt; 0) {
  18.       subdirectories.forEach(subdir =&gt; {
  19.         result.push(path.join(dir, subdir));
  20.       });
  21.     } else {
  22.       result.push(dir);
  23.     }
  24.   });
  26.   return result;
  27. }
  29. const subdirectories = getSubdirectories(&quot;./app&quot;);
  31. const subdirectoryPaths = subdirectories.map(subdir =&gt; {
  32.   return path.join(&quot;app&quot;, subdir);
  33. });
  35. console.log(subdirectoryPaths);

<!-- end snippet -->

huangapple
  • 本文由 发表于 2023年2月19日 06:05:46
  • 转载请务必保留本文链接：https://go.coder-hub.com/75496679.html
  • javascript
  • node.js-fs
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