Question

我试图让Python解析Avro模式，例如以下内容......

from avro import schema

mySchema = """
{
    "name": "person",
    "type": "record",
    "fields": [
        {"name": "firstname", "type": "string"},
        {"name": "lastname", "type": "string"},
        {
            "name": "address",
            "type": "record",
            "fields": [
                {"name": "streetaddress", "type": "string"},
                {"name": "city", "type": "string"}
            ]
        }
    ]
}"""

parsedSchema = schema.parse(mySchema)

...我得到以下异常：

avro.schema.SchemaParseException: Type property "record" not a valid Avro schema: Could not make an Avro Schema object from record.

我做错了什么？

Answer 1

根据网络上的其他消息来源，我会重写你的第二个地址定义：

mySchema = """
{
    "name": "person",
    "type": "record",
    "fields": [
        {"name": "firstname", "type": "string"},
        {"name": "lastname", "type": "string"},
        {
            "name": "address",
            "type": {
                        "type" : "record",
                        "name" : "AddressUSRecord",
                        "fields" : [
                            {"name": "streetaddress", "type": "string"},
                            {"name": "city", "type": "string"}
                        ]
                    },
        }
    ]
}"""

Answer 2

每次我们提供类型为命名类型时，该字段需要给出：

"name":"some_name",
"type": {
          "name":"CodeClassName",
           "type":"record/enum/array"
 }

但是，如果命名类型是union，那么我们不需要额外的类型字段，并且可以用作：

"name":"some_name",
"type": [{
          "name":"CodeClassName1",
           "type":"record",
           "fields": ...
          },
          {
           "name":"CodeClassName2",
            "type":"record",
            "fields": ...
}]

希望这进一步澄清！

如何在Avro架构中嵌套记录？

2 个答案: