0
我想使用嵌套avro架構來創建配置單元表。但它不起作用。我在cdh5.7.2中使用hive 1.1。配置單元不能創建嵌套avro架構表
這裏是我的嵌套的Avro模式:
[
{
"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"}]
}
]
}
]
而我的SQL創建表:
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');
但是蜂房給我:從Supports arbitrarily nested schemas.
:
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)
數據樣本:
{
"appId":{"string":"com.test.app"},
"timestamp":{"long":1495893601606},
"idList":{
"array":[
{"idType":15,"id":"6c:5c:14:c3:a5:39"},
{"idType":13,"id":"eb297afe56ff340b6bb7de5c5ab09193"}
]
}
}
但我不知道怎麼樣。我需要一些幫助來解決這個問題。謝謝!
已經添加了數據樣本。 –