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Q&A How to convert json to yaml?

With yq: $ yq --output-format yaml . file.json > file.yaml . is a filter which is applied to the data, but since . just stands for the document root, this means that the data is passed throu...

posted 1mo ago by Iizuki‭  ·  edited 17d ago by Iizuki‭

Answer
#2: Post edited by user avatar Iizuki‭ · 2024-04-29T11:27:22Z (17 days ago)
Got the two yqs mixed up
  • With [`yq`](https://mikefarah.gitbook.io/yq):
  • ```shell
  • $ yq --yaml-output . file.json > file.yaml
  • ```
  • `.` is a filter which is applied to the data, but since `.` just stands for the document root, this means that the data is passed through without modification.
  • `yq` is capable of rich document manipulation, but here we are doing just the most simple of operations.
  • With [`yq`](https://mikefarah.gitbook.io/yq):
  • ```shell
  • $ yq --output-format yaml . file.json > file.yaml
  • ```
  • `.` is a filter which is applied to the data, but since `.` just stands for the document root, this means that the data is passed through without modification.
  • `yq` is capable of rich document manipulation, but here we are doing just the most simple of operations.
  • Note that this `yq` here is implemented in go. There's also a python project with the same name.
#1: Initial revision by user avatar Iizuki‭ · 2024-04-11T07:42:30Z (about 1 month ago)
With [`yq`](https://mikefarah.gitbook.io/yq):

```shell
$ yq --yaml-output . file.json > file.yaml
```

`.` is a filter which is applied to the data, but since `.` just stands for the document root, this means that the data is passed through without modification.

`yq` is capable of rich document manipulation, but here we are doing just the most simple of operations.