AWS CodeBuild buildspec.yml 以递归方式获取所有文件和子文件夹

Question

I'm trying to use AWS CodeBuild to get all files and subfolders inside a nested public<\/code> folder and deploy to an S3 bucket using CodePipeline. I was able to hook them all together but struggling to configure the buildspec.yml<\/code> file to get the output I want.

<path>/public/

├── 404.html
├── css
│   ├── ...
├── fonts
│   ├── bootstrap
│   │   ├── ...
│   ├── icomoon
│   │   ├── icomoon
│   │   │   ├── ...
│   └── simple-line-icons
│       ├── ...
├── images
│   ├── ...
├── index.html
├── index.xml
├── js
│   ├── ...
└── tags
    └── index.xml

Answer 1

TL;DR You probably don't need to use " discard-paths: yes ". So, drop " discard-paths: yes " and instead use " base-directory " with " **/* " for glob pattern. Read on for more reasoning why.

So yeah, as it usually happens, the culprit was pretty trivial. In my case the directory structure looked like this:

dist/
|-- favicon.ico
|-- index.html
|-- js
|   `-- app.js
`-- revision.txt

So in order for the artifact to contain everything from the " dist " directory whilst preserving the nested directories structure, the buildspec had to look like this:

artifacts:
  files:
    - '**/*'
  base-directory: 'dist'

Essentially " base-directory " parameter behaves as unix's "cd" command, so at first CodeBuild jumps to that directory and only after that resolves the glob pattern, in this case - **/* . If you also happen to use " discard-paths " then what happens is that, presumably, CodeBuild will change current directory to " dist ", resolve the files structure correctly but then remove all path prefixes from the resolved paths, thus the result will be a plain directory structure with no nesting.

Hope that helps!

Answer 2

I've figured out a way to work around it. I don't think it's the best way but it works. I hope you can find this solution useful buildspec.yml :

version: 0.2

phases:
  build:
    commands:
      - mkdir build-output
      - cp -R <path>/public/ build-output
  post_build:
    commands:
      - mv build-output/**/* ./
      - mv build-output/* ./
      - rm -R build-output *.yml LICENSE README* .git*
artifacts:
  files:
    - '**/*'

In words:

• I copy everything inside the nested <path>/public folder out to a folder called ./build-output .

• Then in post_build I move everything out of folder build-output .

• Delete files pulled from my GitHub's repo which aren't needed to host the static website in the S3 bucket

• Then I get the result I want: all files inside public in the root of the S3 bucket with the right folder tree.

Update:

• You can also use my buildspec.yml example file here . Since the example is out of the scope of this question, I don't paste the code here.
• I explain in details here .

Answer 3

I was running into a similar issue and after many permutations below syntax worked for me.

artifacts:
  files:
    - './**/*'
  base-directory: public
  name: websitename

Answer 4

对于那些在 CodePipline 中使用 CodeBuid 并尝试正确配置 buildspec.yml 文件的人来说，罪魁祸首是在部署阶段，我将源工件而不是构建工件设置为输入工件。

Answer 5

This worked for me.

    /
    buildspec.yml
    /public
        index.html
        /js
        /img
        appspec.yml
    /src

in buildspec.yml

    artifacts:
    files:
      - 'public/*'
    discard-paths: yes