如何在Python中将JSON字符串转换为Avro？

Question

有没有办法在不使用Python的模式定义的情况下将JSON字符串转换为Avro？还是只有Java才能处理的事情？

Answer 1

我最近遇到了同样的问题，我最终开发了一个 python 包，它可以采用任何 python 数据结构，包括解析的 JSON 并将其存储在 Avro 中，而无需专用模式。

我为python 3测试了它。

您可以将其安装为pip3 install rec-avro或在https://github.com/bmizhen/rec-avro 上查看代码和文档

用法示例：

from fastavro import writer, reader, schema
from rec_avro import to_rec_avro_destructive, from_rec_avro_destructive, rec_avro_schema

def json_objects():
    return [{'a': 'a'}, {'b':'b'}]

# For efficiency, to_rec_avro_destructive() destroys rec, and reuses it's
# data structures to construct avro_objects 
avro_objects = (to_rec_avro_destructive(rec) for rec in json_objects())

# store records in avro
with open('json_in_avro.avro', 'wb') as f_out:
    writer(f_out, schema.parse_schema(rec_avro_schema()), avro_objects)

#load records from avro
with open('json_in_avro.avro', 'rb') as f_in:
    # For efficiency, from_rec_avro_destructive(rec) destroys rec, and 
    # reuses it's data structures to construct it's output
    loaded_json = [from_rec_avro_destructive(rec) for rec in reader(f_in)]

assert loaded_json == json_objects()

要将 JSON 字符串转换为 json 对象，请使用 json.loads('{"a":"b"}')

Answer 2

Apache Avro\xe2\x84\xa2 1.7.6 入门 (Python)：

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import avro.schema\navro.schema.parse(json_schema_string)\n

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