如何为复杂的 json 文档定义 avro 模式？

Question

我有一个 JSON 文档，我想将其转换为 Avro，并且需要为此目的指定一个架构。这是我要为其定义 avro 架构的 JSON 文档：

{
 "uid": 29153333,
 "somefield": "somevalue",
 "options": [
   {
     "item1_lvl2": "a",
     "item2_lvl2": [
       {
         "item1_lvl3": "x1",
         "item2_lvl3": "y1"
       },
       {
         "item1_lvl3": "x2",
         "item2_lvl3": "y2"
       }
     ]
   }
 ]
}

我可以为非复杂类型定义架构，但不能为复杂的 "options" 字段定义架构：

{
  "namespace" : "my.com.ns",
  "type" :  "record",
  "fields" : [
     {"name": "uid", "type": "int"},
     {"name": "somefield", "type": "string"}
     {"name": "options", "type": .....}
  ]
}

感谢您的帮助！

Answer 1

您需要使用 Avro complex types, specifically arrays and records。然后将它们嵌套在一起：

{
  "namespace" : "my.com.ns",
  "name": "myrecord",
  "type" :  "record",
  "fields" : [
     {"name": "uid", "type": "int"},
     {"name": "somefield", "type": "string"},
     {"name": "options", "type": {
        "type": "array",
        "items": {
            "type": "record",
            "name": "lvl2_record",
            "fields": [
                {"name": "item1_lvl2", "type": "string"},
                {"name": "item2_lvl2", "type": {
                    "type": "array",
                    "items": {
                        "type": "record",
                        "name": "lvl3_record",
                        "fields": [
                            {"name": "item1_lvl3", "type": "string"},
                            {"name": "item2_lvl3", "type": "string"}
                        ]
                    }
                }}
            ]
        }
     }}
  ]
}

此外，为了提高可读性，您可以 split the schema into multiple files。

Answer 2

这个在线工具(http://avro4s-ui.landoop.com/)非常实用，可以通过给定的有效json.

生成AVRO schema

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