配置单元无法使用嵌套的avro架构创建表

h9a6wy2h  于 2021-06-26  发布在  Hive
关注(0)|答案(1)|浏览(394)

我正在尝试使用嵌套的avro模式来创建一个配置单元表。但它不起作用。我正在使用cdh5.7.2中的hive1.1。
以下是我的嵌套avro架构:

[
    {
        "type": "record",
        "name": "Id",
        "namespace": "com.test.app_list",
        "doc": "Device ID",
        "fields": [
            {
                "name": "idType",
                "type": "int"
            },{
                "name": "id",
                "type": "string"
            }
        ]
    },

    {
        "type": "record",
        "name": "AppList",
        "namespace": "com.test.app_list",
        "doc": "",
        "fields": [
            {
                "name": "appId",
                "type": "string",
                "avro.java.string": "String"
            },
            {
                "name": "timestamp",
                "type":  "long"
            },

            {
                "name": "idList",
                "type": [{"type": "array", "items": "com.test.app_list.Id"}]
            }

        ]
    }
]

和我的sql来创建表:

CREATE EXTERNAL TABLE app_list
ROW FORMAT SERDE
'org.apache.hadoop.hive.serde2.avro.AvroSerDe'
STORED AS INPUTFORMAT
'org.apache.hadoop.hive.ql.io.avro.AvroContainerInputFormat'
OUTPUTFORMAT
'org.apache.hadoop.hive.ql.io.avro.AvroContainerOutputFormat'
TBLPROPERTIES (
'avro.schema.url'='/hive/schema/test_app_list.avsc');

但是Hive给了我:

FAILED: Execution Error, return code 1 from org.apache.hadoop.hive.ql.exec.DDLTask. java.lang.RuntimeException: MetaException(message:org.apache.hadoop.hive.serde2.avro.AvroSerdeException Schema for table must be of type RECORD. Received type: UNION)
``` `hive` 文件显示: `Supports arbitrarily nested schemas.` 发件人:https://cwiki.apache.org/confluence/display/hive/avroserde#avroserde-概述–从Hive使用AVR
数据样本:

{
"appId":{"string":"com.test.app"},
"timestamp":{"long":1495893601606},
"idList":{
"array":[
{"idType":15,"id":"6c:5c:14:c3:a5:39"},
{"idType":13,"id":"eb297afe56ff340b6bb7de5c5ab09193"}
]
}

}

但我不知道该怎么做。我需要一些帮助来解决这个问题。谢谢!
jfewjypa

jfewjypa1#

avro模式的顶层应该是记录类型,这就是为什么hive不允许这样做的原因。解决方法可以是创建顶层作为记录,内部创建两个字段作为记录类型。

{ 
        "type": "record",
        "name": "myRecord",
        "namespace": "com.test.app_list"
          "fields": [
    {
        "type": "record",
        "name": "Id",
        "doc": "Device ID",
        "fields": [
            {
                "name": "idType",
                "type": "int"
            },{
                "name": "id",
                "type": "string"
            }
        ]
    },

    {
        "type": "record",
        "name": "AppList",
        "doc": "",
        "fields": [
            {
                "name": "appId",
                "type": "string",
                "avro.java.string": "String"
            },
            {
                "name": "timestamp",
                "type":  "long"
            },

            {
                "name": "idList",
                "type": [{"type": "array", "items": "com.test.app_list.Id"}]
            }

        ]
    }
    ]
}

相关问题