Question

我想使用dhall-kubernetes为服务生成ConfigMap。使用YAML文件配置服务。我可以使用dhall分两步创建配置：

$ dhall-to-yaml < server.dhall > server.yaml
$ dhall-to-yaml < configmap.dhall 
apiVersion: v1
data:
  server.yaml: |
    server:
      port: 8080
kind: ConfigMap
metadata:
  name: my-configmap

server.dhall：

let _config = { server.port = 8080 } in _config

configmap.dhall：

let kubernetes =
      ./package.dhall sha256:d9eac5668d5ed9cb3364c0a39721d4694e4247dad16d8a82827e4619ee1d6188

let _configMap1 =
      kubernetes.ConfigMap::{
      , metadata = kubernetes.ObjectMeta::{ name = "my-configmap" }
      , data = Some
        [ { mapKey = "server.yaml", mapValue = ./server.yaml as Text } ]
      }

in  _configMap1

package.dhall

https://raw.githubusercontent.com/dhall-lang/dhall-kubernetes/master/package.dhall

是否可以通过从dhall内部生成YAML来仅使用一次dhall-to-yaml调用？像这样：

let kubernetes =
      ./package.dhall sha256:d9eac5668d5ed9cb3364c0a39721d4694e4247dad16d8a82827e4619ee1d6188

let _config = { server.port = 8080 }

let _configMap3 =
      kubernetes.ConfigMap::{
      , metadata = kubernetes.ObjectMeta::{ name = "my-configmap" }
      , data = Some
        [ { mapKey = "server.yaml"
          , mapValue = makeYAML _config -- make this work 
          }   
        ]   
      }   

in  _configMap3

我从Prelude中找到了JSON.renderYAML函数，但这似乎并不完全是必需的。

Answer 1

要使用Prelude.JSON.renderYAML生成YAML，首先需要将配置转换为输入Prelude.JSON.Type类型。请参阅下面的我的makeJSON版本中包含的makeYAML函数：

let Prelude = https://prelude.dhall-lang.org/v15.0.0/package.dhall

let J = Prelude.JSON

let Config = { server : { port : Natural } }

let makeJSON
    : Config → J.Type
    =   λ(c : Config)
      → J.object
          ( toMap
              { server = J.object (toMap { port = J.natural c.server.port }) }
          )

let makeYAML
    : Config → Text
    = λ(c : Config) → J.renderYAML (makeJSON c)

let example =
      let _config
          : Config
          = { server.port = 8080 }

      in    assert
          :   makeYAML _config
            ≡ ''
              "server":
                "port": 8080
              ''

in  makeYAML

是否可以通过dwith表达式创建YAML？

1 个答案: