I have an external table that is built off of a json file.
All of the json keys are columns and are populated as expected except for one key that has a space.
Here is the DDL:
CREATE EXTERNAL TABLE foo.bar
( event ARRAY <STRUCT
value:STRING
,info:STRUCT
<id:STRING
,event_source:STRING>>
ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
WITH SERDEPROPERTIES("mapping.event_source"="event source")
STORED AS INPUTFORMAT 'org.apache.hadoop.mapred.TextInputFormat'
OUTPUTFORMAT 'org.apache.hadoop.hive.ql.io.HiveIgnoreKeyTextOutputFormat'
LOCATION 'foo/bar'
All of the values show up as expected except for event_source, which shows up as NULL. The original form of event_source in the json file is 'event source' without the single quotes.
Is there something I need to do different with the WITH SERDEPROPERTIES setting in order to get the key to work properly?
Thanks
you mean that the json has data like
{ id: "myid", event source: "eventsource" }
If so, there's not much that can be done since it's simply broken JSON.
If not, can you post a sample of the JSON you're trying to read ?
I have encountered a similar problem as above but with a slight variation that the input data is correct json.
I have an external table that is built off of a json file. All of the json keys are populated except one msrp_currency
Here is the DDL:
CREATE EXTERNAL TABLE foo.bar
( id string,
variants array<struct<pid:string, msrp_currency:string>>
)
ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
WITH SERDEPROPERTIES ( "ignore.malformed.json" = "true" ,
'mapping.variants.msrp_currency' = 'variants.msrpcurrency')
LOCATION 'foo/bar'
All of the values show up as expected except for msrp_currency, which shows up as NULL. The reason I need to introduce underscore is because later I need to extract the same field value as msrpCurrecny using brickhouse to_json UDF.
sample values:
{ "pid": "mypid", "msrpCurrency": "USD" }
Related
I'm very new in Hadoop,
I'm using Spark with Java.
I have dynamic JSON, exmaple:
{
"sourceCode":"1234",
"uuid":"df123-....",
"title":"my title"
}{
"myMetaDataEvent": {
"date":"10/10/2010",
},
"myDataEvent": {
"field1": {
"field1Format":"fieldFormat",
"type":"Text",
"value":"field text"
}
}
}
Sometimes I can see only field1 and sometimes I can see field1...field50
And maybe the user can add fields/remove fields from this JSON.
I want to insert this dynamic JSON to hadoop (to hive table) from Spark Java code,
How can I do it?
I want that the user can after make HIVE query, i.e: select * from MyTable where type="Text
I have around 100B JSON records per day that I need to insert to Hadoop,
So what is the recommanded way to do that?
*I'm looked on the following: SO Question but this is known JSON scheme where it isnt my case.
Thanks
I had encountered kind of similar problem, I was able to resolve my problem using this. ( So this might help if you create the schema before you parse the json ).
For a field having a string data type you could create the schema :-
StructField field = DataTypes.createStructField(<name of the field>, DataTypes.StringType, true);
For a field having a int data type you could create the schema :-
StructField field = DataTypes.createStructField(<name of the field>, DataTypes.IntegerType, true);
After you have added all the fields in a List<StructField>,
Eg:-
List<StructField> innerField = new ArrayList<StructField>();
.... Field adding logic ....
Eg:-
innerField.add(field1);
innerField.add(field2);
// One instance can come, or multiple instance of value comes in an array, then it needs to be put in Array Type.
ArrayType getArrayInnerType = DataTypes.createArrayType(DataTypes.createStructType(innerField));
StructField getArrayField = DataTypes.createStructField(<name of field>, getArrayInnerType,true);
You can then create the schema :-
StructType structuredSchema = DataTypes.createStructType(getArrayField);
Then I read the json using the schema generated using the Dataset API.
Dataset<Row> dataRead = sqlContext.read().schema(structuredSchema).json(fileName);
I have a file which contains the data as follows
[{"col1":"col1","col2":1}
,{"col1":"col11","col2":11}
,{"col1":"col111","col2":2}
]
I am trying to load the table in Hive.
I am using following Hive serde
CREATE EXTERNAL TABLE my_table (
my_array ARRAY<struct<col1:string,col2:int>>
)ROW FORMAT SERDE 'org.apache.hive.hcatalog.data.JsonSerDe'
WITH SERDEPROPERTIES ( "ignore.malformed.json" = "true")
LOCATION "MY_LOCATION";
I am getting error when I try to run select * after running the create command -
['*org.apache.hive.service.cli.HiveSQLException:java.io.IOException: org.apache.hadoop.hive.serde2.SerDeException: java.io.IOException: Start token not found where expected:25:24', 'org.apache.hive.service.cli.operation.SQLOperation:getNextRowSet:SQLOperation.java:499', 'org.apache.hive.service.cli.operation.OperationManager:getOperationNextRowSet:OperationManager.java:307', 'org.apache.hive.service.cli.session.HiveSessionImpl:fetchResults:HiveSessionImpl.java:878', 'sun.reflect.GeneratedMethodAccessor29:invoke::-1', 'sun.reflect.DelegatingMethodAccessorImpl:invoke:DelegatingMethodAccessorImpl.java:43', 'java.lang.reflect.Method:invoke:Method.java:498', 'org.apache.hive.service.cli.session.HiveSessionProxy:invoke:HiveSessionProxy.java:78', 'org.apache.hive.service.cli.session.HiveSessionProxy:access$000:HiveSessionProxy.java:36', 'org.apache.hive.service.cli.session.HiveSessionProxy$1:run:HiveSessionProxy.java:63', 'java.security.AccessController:doPrivileged:AccessController.java:-2', 'javax.security.auth.Subject:doAs:Subject.java:422', 'org.apache.hadoop.security.UserGroupInformation:doAs:UserGroupInformation.java:1698', 'org.apache.hive.service.cli.session.HiveSessionProxy:invoke:HiveSessionProxy.java:59', 'com.sun.proxy.$Proxy35:fetchResults::-1', 'org.apache.hive.service.cli.CLIService:fetchResults:CLIService.java:559', 'org.apache.hive.service.cli.thrift.ThriftCLIService:FetchResults:ThriftCLIService.java:751', 'org.apache.hive.service.rpc.thrift.TCLIService$Processor$FetchResults:getResult:TCLIService.java:1717', 'org.apache.hive.service.rpc.thrift.TCLIService$Processor$FetchResults:getResult:TCLIService.java:1702', 'org.apache.thrift.ProcessFunction:process:ProcessFunction.java:39', 'org.apache.thrift.TBaseProcessor:process:TBaseProcessor.java:39', 'org.apache.hive.service.auth.TSetIpAddressProcessor:process:TSetIpAddressProcessor.java:56', 'org.apache.thrift.server.TThreadPoolServer$WorkerProcess:run:TThreadPoolServer.java:286', 'java.util.concurrent.ThreadPoolExecutor:runWorker:ThreadPoolExecutor.java:1149', 'java.util.concurrent.ThreadPoolExecutor$Worker:run:ThreadPoolExecutor.java:624', 'java.lang.Thread:run:Thread.java:748', '*java.io.IOException:org.apache.hadoop.hive.serde2.SerDeException: java.io.IOException: Start token not found where expected:29:4', 'org.apache.hadoop.hive.ql.exec.FetchOperator:getNextRow:FetchOperator.java:521', 'org.apache.hadoop.hive.ql.exec.FetchOperator:pushRow:FetchOperator.java:428', 'org.apache.hadoop.hive.ql.exec.FetchTask:fetch:FetchTask.java:147', 'org.apache.hadoop.hive.ql.Driver:getResults:Driver.java:2207', 'org.apache.hive.service.cli.operation.SQLOperation:getNextRowSet:SQLOperation.java:494', '*org.apache.hadoop.hive.serde2.SerDeException:java.io.IOException: Start token not found where expected:30:1', 'org.apache.hive.hcatalog.data.JsonSerDe:deserialize:JsonSerDe.java:184', 'org.apache.hadoop.hive.ql.exec.FetchOperator:getNextRow:FetchOperator.java:502', '*java.io.IOException:Start token not found where expected:30:0', 'org.apache.hive.hcatalog.data.JsonSerDe:deserialize:JsonSerDe.java:170'], statusCode=3), results=None, hasMoreRows=None)
I tried several things, none of which worked as expected. I can't change the input data format as it is someone else who is providing the data.
This is a malformed JSON issue. A JSON file will always have "curly braces" at the beginning and the end. So change your JSON file to look something like below.
{"my_array":[{"col1":"col1","col2":1},{"col1":"col11","col2":11},{"col1":"col111","col2":2}]}
Create your table in the exact same way as you are doing it already.
CREATE EXTERNAL TABLE my_table
(
my_array ARRAY<struct<col1:string,col2:int>>
)
ROW FORMAT SERDE 'org.apache.hive.hcatalog.data.JsonSerDe'
WITH SERDEPROPERTIES ( "ignore.malformed.json" = "true")
LOCATION "MY_LOCATION";
Now fire a select * on your newly created table to see following results.
[{"col1":"col1","col2":1},{"col1":"col11","col2":11},{"col1":"col111","col2":2}]
Use select my_array.col1 from my_table; to see the values for col1 from your array.
["col1","col11","col111"]
PS - Not the most efficient way to store the data. Consider transforming the data and storing it as ORC/Parquet.
Hope that helps!
Looks like the issue is with your json data. Can you try with below example?
Create employee json with below content and place it in hdfs.
[root#quickstart spark]# hadoop fs -cat /user/cloudera/spark/employeejson/*
{"Name":"Vinayak","age":35}
{"Name":"Nilesh","age":37}
{"Name":"Raju","age":30}
{"Name":"Karthik","age":28}
{"Name":"Shreshta","age":1}
{"Name":"Siddhish","age":2}
Add below jar(execute only if you get any error. )
hive> ADD JAR /usr/lib/hive-hcatalog/lib/hive-hcatalog-core.jar;
hive>
CREATE TABLE employeefromjson(name string, age int)
ROW FORMAT SERDE 'org.apache.hive.hcatalog.data.JsonSerDe'
STORED AS TEXTFILE
LOCATION '/user/cloudera/hive/employeefromjson'
;
hive> LOAD DATA INPATH '/user/cloudera/spark/employeejson' OVERWRITE INTO TABLE employeefromjson;
hive> select * from employeefromjson;
OK
Vinayak 35
Nilesh 37
Raju 30
Karthik 28
Shreshta 1
Siddhish 2
Time taken: 0.174 seconds, Fetched: 6 row(s)
A JSON should always start with '{' and not with '['. That is the problem. As you know, JSON has a structure of {'key':'value'}. What you have given in your file is a value which does not have any key. So, change your JSON to the below formmat
{"my_array":[{"col1":"col1","col2":1},{"col1":"col11","col2":11},{"col1":"col111","col2":2}]}
Your Create table statement should work fine.
If you want to get the data for each column for all the rows, use the below query.
select my_array.col1, my_array.col2 from my_table;
The above command will give you the below result.
OK
["col1","col11","col111"] [1,11,2]
If you want to get the result column wise for each row seperately, use the below query.
select a.* from my_table m lateral view outer inline (m.my_array) a;
The above command will give you the below result.
OK
col1 1
col11 11
col111 2
Hope you this helps!
I've created a crawler that looks at a PostgreSQL 9.6 RDS table with a JSONB column but the crawler identifies the column type as "string". When I then try to create a job that loads data from a JSON file on S3 into the RDS table I get an error.
How can I map a JSON file source to a JSONB target column?
It's not quite a direct copy, but an approach that has worked for me is to define the column on the target table as TEXT. After the Glue job populates the field, I then convert it to JSONB. For example:
alter table postgres_table
alter column column_with_json set data type jsonb using column_with_json::jsonb;
Note the use of the cast for the existing text data. Without that, the alter column would fail.
Crawler will identify JSONB column type as "string" but you can try to use Unbox Class in Glue to convert this column to json
let's check the following table in PostgreSQL
create table persons (id integer, person_data jsonb, creation_date timestamp )
There is an example of one record from person table
ID = 1
PERSON_DATA = {
"firstName": "Sergii",
"age": 99,
"email":"Test#test.com"
}
CREATION_DATE = 2021-04-15 00:18:06
The following code need to be added in Glue
# 1. create dynamic frame from catalog
df_persons = glueContext.create_dynamic_frame.from_catalog(database = "testdb", table_name = "persons", transformation_ctx = "df_persons ")
# 2.in path you need to add your jsonb column name that need to be converted to json
df_persons_json = Unbox.apply(frame = df_persons , path = "person_data", format="json")
# 3. converting from dynamic frame to data frame
datf_persons_json = df_persons_json.toDF()
# 4. after that you can process this column as a json datatype or create dataframe with all necessary columns , each json data element can be added as a separate column in dataframe :
final_df_person = datf_persons_json.select("id","person_data.age","person_data.firstName","creation_date")
You can also check the following link:
https://docs.aws.amazon.com/glue/latest/dg/aws-glue-api-crawler-pyspark-transforms-Unbox.html
I am trying to load a huge volume json data with nested structure to hive using a Json serde. some of the field names start with $ in nested structure. I am mapping hive filed names Using SerDeproperties, but how ever when i query the table, getting null in the field starting with $, tried with different syntax,but no luck.
Sample JSON:
{
"_id" : "319FFE15FF90",
"SomeThing" :
{
"$SomeField" : 22,
"AnotherField" : 2112,
"YetAnotherField": 1
}
. . . etc . . . .
Using a schema as follows:
create table testSample
(
`_id` string,
something struct
<
$somefield:int,
anotherfield:bigint,
yetanotherfield:int
>
)
row format serde 'org.openx.data.jsonserde.JsonSerDe'
with serdeproperties
(
"mapping.somefield" = "$somefield"
);
This schema builds OK, however, somefield(starting with $) in the above table is always returning null (all the other values exist and are correct).
We've been trying a lot of syntax combinations, but to no avail.
Does anyone know the trick to hap a nested field with a leading $ in its name?
You almost got it right. Try creating the table like this.
The mistake you're making is that when mapping in the serde properties (mapping.somefield ="$somefield") you're saying "when looking for the hive column named 'somefield', look for the json field '$somefield', but in hive you defined the column with the dollar sign, which if not outright illegal it's for sure not the best practice in hive.
create table testSample
(
`_id` string,
something struct
<
somefield:int,
anotherfield:bigint,
yetanotherfield:int
>
)
row format serde 'org.openx.data.jsonserde.JsonSerDe'
with serdeproperties
(
"mapping.somefield" = "$somefield"
);
I tested it with some test data:
{ "_id" : "123", "something": { "$somefield": 12, "anotherfield":13,"yetanotherfield":100}}
hive> select something.somefield from testSample;
OK
12
I am suddenly starting to see this problem as well but for normal column names as well (no special characters such as $)
I am populating an external table (Temp) from another internal table (Table2) and want the output of Temp table in JSON format. I want column names in camel case in the output JSON file and so am also using the Serdepoperties in the Temp table to specify correct names. However, I am seeing that when I do Select * from the Temp table, it gives NULL values for the columns whose names have been used in the mapping.
I am running Hive 0.13. Here are the commands:
Create table command:
CREATE EXTERNAL TABLE Temp (
data STRUCT<
customerId:BIGINT, region:STRING, marketplaceId:INT, asin:ARRAY<STRING>>
)
ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
WITH SERDEPROPERTIES (
'mapping.customerid' = 'customerId',
'mapping.marketplaceid' = 'marketplaceId'
)
LOCATION '/output';
INSERT INTO TABLE Temp
SELECT
named_struct ('customerId',customerId, 'region', region, 'marketplaceId', marketplaceId, 'asin', asin)
FROM Table2;
Select * from Temp:
{"customerid":null,"region":"EU","marketplaceid":null,"asin":["B000FC1PZC"]}
{"customerid":null,"region":"EU","marketplaceid":null,"asin":["B000FC1C9G"]}
See how "customerid" and "marketplaceid" are null. Generated JSON file is:
{"data":{"region":"EU","asin":["B000FC1PZC"]}}
{"data":{"region":"EU","asin":["B000FC1C9G"]}}
Now, if I remove the with serdeproperties, the table starts getting all values:
{"customerid":1,"region":"EU","marketplaceid":4,"asin":["B000FC1PZC"]}
{"customerid":2,"region":"EU","marketplaceid":4,"asin":["B000FC1C9G"]}
And then the JSON file so generated is:
{"data":{"region":"EU","marketplaceid":4,"asin":["B000FC1PZC"],"customerid":1}}
{"data":{"region":"EU","marketplaceid":4,"asin":["B000FC1C9G"],"customerid":2}}
We are attempting to create a schema to load a massive JSON structure into Hive. We are having a problem, however, in that some fields have leading underscores for names--at the root level, this is fine, but we have not found a way to make this work for nested fields.
Sample JSON:
{
"_id" : "319FFE15FF908EDD86B7FDEADBEEFBD8D7284128841B14AA6A966923C268DF39",
"SomeThing" :
{
"_SomeField" : 22,
"AnotherField" : 2112,
"YetAnotherField": 1
}
. . . etc . . . .
Using a schema as follows:
create table testSample
(
id string,
something struct
<
somefield:int,
anotherfield:bigint,
yetanotherfield:int
>
)
row format serde 'org.openx.data.jsonserde.JsonSerDe'
with serdeproperties
(
"mapping.id" = "_id",
"mapping.somefield" = "_somefield"
);
This schema builds OK--however, after loading the in above sample, the value of "somefield" (the nested + leading underscore one) is always null (all the other values exist and are correct).
We've been trying a lot of syntax combinations, but to no avail.
Does anyone know the trick to hap a nested field with a leading underscore in its name?
Cheers!
Answering my own question here: there is no trick because you can't.
However, there's an easy work-around: you can tell Hive to treat the names as literals upon creating the schema. If you do this, you will also need to query using the same literal syntax. In the above example, it would look like:
`_something` struct<rest_of_definitions>
without any special serde properties for it.
Then use again in query:
select stuff.`_something` from sometable;
e.g., schema:
create table testSample
(
id string,
something struct
<
`_somefield`:int,
anotherfield:bigint,
yetanotherfield:int
>
)
row format serde 'org.openx.data.jsonserde.JsonSerDe'
with serdeproperties("mapping.id" = "_id");
for an input JSON like:
{
"_id": "someuid",
"something":
{
"_somefield": 1,
"anotherfield": 2,
"yetanotherfield": 3
}
}
with a query like:
select something.`_somefield`
from testSample
where something.anotherfield = 2;