Parsing JSON file in Snowflake Database
1
Database :SNOWFLAKE
My table contains JSON data for example:
{
"bucket":"IN_Apps",
"bySeqno":56,
"cas":1527639206906626048,
"content":"eyJoaWdoQmluIjoiNTQ4NTA4MDkiLCJkb2N1bWVudFR5cGUiOiJJSU5ETyIsImNhcmRUeXBlIyayI6Ik1BU1RFUkNBUkQifQ==",
"event":"mutation",
"expiration":0,
"flags":33554432,
"key":"iin54850809",
"lockTime":0,
"partition":948,
"revSeqno":1,
"vBucketUuid":137987627737694
}
when i tried to parse it.
select
parse_json:bucket::string as bucket ,
parse_json:bySeqno::string as bySeqno ,
parse_json:cas::INT as cas ,
parse_json:content::string as content ,
parse_json:event::string as event
,parse_json:expiration::string as expiration
,parse_json:flags::string as flags
,parse_json:key::string as key
,parse_json:lockTime::string as lockTime
,parse_json:partition::string as partition
,parse_json:revSeqno::string as revSeqno
,parse_json:vBucketUuid::string as vBucketUuid
from STG_YS_APPS v
but it is throwing error like.
SQL compilation error: error line 2 at position 0 invalid identifier > >'PARSE_JSON'
may someone please help me.
sql json snowflake-datawarehouse
asked Nov 13 '18 at 14:55
OMGOMG
464
add a comment |
1
Database :SNOWFLAKE
My table contains JSON data for example:
{
"bucket":"IN_Apps",
"bySeqno":56,
"cas":1527639206906626048,
"content":"eyJoaWdoQmluIjoiNTQ4NTA4MDkiLCJkb2N1bWVudFR5cGUiOiJJSU5ETyIsImNhcmRUeXBlIyayI6Ik1BU1RFUkNBUkQifQ==",
"event":"mutation",
"expiration":0,
"flags":33554432,
"key":"iin54850809",
"lockTime":0,
"partition":948,
"revSeqno":1,
"vBucketUuid":137987627737694
}
when i tried to parse it.
select
parse_json:bucket::string as bucket ,
parse_json:bySeqno::string as bySeqno ,
parse_json:cas::INT as cas ,
parse_json:content::string as content ,
parse_json:event::string as event
,parse_json:expiration::string as expiration
,parse_json:flags::string as flags
,parse_json:key::string as key
,parse_json:lockTime::string as lockTime
,parse_json:partition::string as partition
,parse_json:revSeqno::string as revSeqno
,parse_json:vBucketUuid::string as vBucketUuid
from STG_YS_APPS v
but it is throwing error like.
SQL compilation error: error line 2 at position 0 invalid identifier > >'PARSE_JSON'
may someone please help me.
sql json snowflake-datawarehouse
asked Nov 13 '18 at 14:55
OMGOMG
464
any clue techie please
– OMG
Nov 13 '18 at 15:42
What is the exact schema of your table? Do you have there a column name called "parse_json" ? That's what your SQL suggests, but that would be weird. Maybe do "DESC TABLE v" and add the output to the question
– Marcin Zukowski
Nov 13 '18 at 22:59
@MarcinZukowski: No i don't have column named as parse_json. it is inbuilt in snowflake for parsing JSON.
– OMG
Nov 14 '18 at 5:05
Well, your query is written like you did. Please run "DESC TABLE v" and provide output
– Marcin Zukowski
Nov 14 '18 at 5:06
add a comment |
1
1
1
Database :SNOWFLAKE
My table contains JSON data for example:
{
"bucket":"IN_Apps",
"bySeqno":56,
"cas":1527639206906626048,
"content":"eyJoaWdoQmluIjoiNTQ4NTA4MDkiLCJkb2N1bWVudFR5cGUiOiJJSU5ETyIsImNhcmRUeXBlIyayI6Ik1BU1RFUkNBUkQifQ==",
"event":"mutation",
"expiration":0,
"flags":33554432,
"key":"iin54850809",
"lockTime":0,
"partition":948,
"revSeqno":1,
"vBucketUuid":137987627737694
}
when i tried to parse it.
select
parse_json:bucket::string as bucket ,
parse_json:bySeqno::string as bySeqno ,
parse_json:cas::INT as cas ,
parse_json:content::string as content ,
parse_json:event::string as event
,parse_json:expiration::string as expiration
,parse_json:flags::string as flags
,parse_json:key::string as key
,parse_json:lockTime::string as lockTime
,parse_json:partition::string as partition
,parse_json:revSeqno::string as revSeqno
,parse_json:vBucketUuid::string as vBucketUuid
from STG_YS_APPS v
but it is throwing error like.
SQL compilation error: error line 2 at position 0 invalid identifier > >'PARSE_JSON'
may someone please help me.
sql json snowflake-datawarehouse
asked Nov 13 '18 at 14:55
OMGOMG
464
Database :SNOWFLAKE
My table contains JSON data for example:
{
"bucket":"IN_Apps",
"bySeqno":56,
"cas":1527639206906626048,
"content":"eyJoaWdoQmluIjoiNTQ4NTA4MDkiLCJkb2N1bWVudFR5cGUiOiJJSU5ETyIsImNhcmRUeXBlIyayI6Ik1BU1RFUkNBUkQifQ==",
"event":"mutation",
"expiration":0,
"flags":33554432,
"key":"iin54850809",
"lockTime":0,
"partition":948,
"revSeqno":1,
"vBucketUuid":137987627737694
}
when i tried to parse it.
select
parse_json:bucket::string as bucket ,
parse_json:bySeqno::string as bySeqno ,
parse_json:cas::INT as cas ,
parse_json:content::string as content ,
parse_json:event::string as event
,parse_json:expiration::string as expiration
,parse_json:flags::string as flags
,parse_json:key::string as key
,parse_json:lockTime::string as lockTime
,parse_json:partition::string as partition
,parse_json:revSeqno::string as revSeqno
,parse_json:vBucketUuid::string as vBucketUuid
from STG_YS_APPS v
but it is throwing error like.
SQL compilation error: error line 2 at position 0 invalid identifier > >'PARSE_JSON'
may someone please help me.
sql json snowflake-datawarehouse
sql json snowflake-datawarehouse
asked Nov 13 '18 at 14:55
OMGOMG
464
asked Nov 13 '18 at 14:55
OMGOMG
464
edited Nov 14 '18 at 6:31
OMG
asked Nov 13 '18 at 14:55
OMGOMG
464
asked Nov 13 '18 at 14:55
OMGOMG
464
asked Nov 13 '18 at 14:55
OMGOMG
464
464
any clue techie please
– OMG
Nov 13 '18 at 15:42
What is the exact schema of your table? Do you have there a column name called "parse_json" ? That's what your SQL suggests, but that would be weird. Maybe do "DESC TABLE v" and add the output to the question
– Marcin Zukowski
Nov 13 '18 at 22:59
@MarcinZukowski: No i don't have column named as parse_json. it is inbuilt in snowflake for parsing JSON.
– OMG
Nov 14 '18 at 5:05
Well, your query is written like you did. Please run "DESC TABLE v" and provide output
– Marcin Zukowski
Nov 14 '18 at 5:06
add a comment |
any clue techie please
– OMG
Nov 13 '18 at 15:42
What is the exact schema of your table? Do you have there a column name called "parse_json" ? That's what your SQL suggests, but that would be weird. Maybe do "DESC TABLE v" and add the output to the question
– Marcin Zukowski
Nov 13 '18 at 22:59
@MarcinZukowski: No i don't have column named as parse_json. it is inbuilt in snowflake for parsing JSON.
– OMG
Nov 14 '18 at 5:05
Well, your query is written like you did. Please run "DESC TABLE v" and provide output
– Marcin Zukowski
Nov 14 '18 at 5:06
any clue techie please
– OMG
Nov 13 '18 at 15:42
any clue techie please
– OMG
Nov 13 '18 at 15:42
What is the exact schema of your table? Do you have there a column name called "parse_json" ? That's what your SQL suggests, but that would be weird. Maybe do "DESC TABLE v" and add the output to the question
– Marcin Zukowski
Nov 13 '18 at 22:59
What is the exact schema of your table? Do you have there a column name called "parse_json" ? That's what your SQL suggests, but that would be weird. Maybe do "DESC TABLE v" and add the output to the question
– Marcin Zukowski
Nov 13 '18 at 22:59
@MarcinZukowski: No i don't have column named as parse_json. it is inbuilt in snowflake for parsing JSON.
– OMG
Nov 14 '18 at 5:05
@MarcinZukowski: No i don't have column named as parse_json. it is inbuilt in snowflake for parsing JSON.
– OMG
Nov 14 '18 at 5:05
Well, your query is written like you did. Please run "DESC TABLE v" and provide output
– Marcin Zukowski
Nov 14 '18 at 5:06
Well, your query is written like you did. Please run "DESC TABLE v" and provide output
– Marcin Zukowski
Nov 14 '18 at 5:06
add a comment |
1 Answer
1
active
oldest
votes
1
Answer with a known schema
Update: Since you provided schema, which shows a VAR column of VARIANT type, here's what you need, couldn't be simpler:
select
var:bucket::string as bucket,
var:bySeqno::string as bySeqno,
var:cas::int as cas
...
from STG_YS_APPS v
Below the answer before the schema was known
I'll assume you have a VARCHAR (or similar) column in your table that is called json, and stores the values you presented. You didn't provide the schema, so please adjust the column name as necessary.
You're not using PARSE_JSON as a function in your SQL. You should write something like
select
parse_json(json):bucket::string as bucket,
parse_json(json):bySeqno::string as bySeqno,
parse_json(json):cas::int as cas
...
from STG_YS_APPS v
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Answer with a known schema
Update: Since you provided schema, which shows a VAR column of VARIANT type, here's what you need, couldn't be simpler:
select
var:bucket::string as bucket,
var:bySeqno::string as bySeqno,
var:cas::int as cas
...
from STG_YS_APPS v
Below the answer before the schema was known
I'll assume you have a VARCHAR (or similar) column in your table that is called json, and stores the values you presented. You didn't provide the schema, so please adjust the column name as necessary.
You're not using PARSE_JSON as a function in your SQL. You should write something like
select
parse_json(json):bucket::string as bucket,
parse_json(json):bySeqno::string as bySeqno,
parse_json(json):cas::int as cas
...
from STG_YS_APPS v
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
add a comment |
1
Answer with a known schema
Update: Since you provided schema, which shows a VAR column of VARIANT type, here's what you need, couldn't be simpler:
select
var:bucket::string as bucket,
var:bySeqno::string as bySeqno,
var:cas::int as cas
...
from STG_YS_APPS v
Below the answer before the schema was known
I'll assume you have a VARCHAR (or similar) column in your table that is called json, and stores the values you presented. You didn't provide the schema, so please adjust the column name as necessary.
You're not using PARSE_JSON as a function in your SQL. You should write something like
select
parse_json(json):bucket::string as bucket,
parse_json(json):bySeqno::string as bySeqno,
parse_json(json):cas::int as cas
...
from STG_YS_APPS v
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
add a comment |
1
1
1
Answer with a known schema
Update: Since you provided schema, which shows a VAR column of VARIANT type, here's what you need, couldn't be simpler:
select
var:bucket::string as bucket,
var:bySeqno::string as bySeqno,
var:cas::int as cas
...
from STG_YS_APPS v
Below the answer before the schema was known
I'll assume you have a VARCHAR (or similar) column in your table that is called json, and stores the values you presented. You didn't provide the schema, so please adjust the column name as necessary.
You're not using PARSE_JSON as a function in your SQL. You should write something like
select
parse_json(json):bucket::string as bucket,
parse_json(json):bySeqno::string as bySeqno,
parse_json(json):cas::int as cas
...
from STG_YS_APPS v
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
Answer with a known schema
Update: Since you provided schema, which shows a VAR column of VARIANT type, here's what you need, couldn't be simpler:
select
var:bucket::string as bucket,
var:bySeqno::string as bySeqno,
var:cas::int as cas
...
from STG_YS_APPS v
Below the answer before the schema was known
I'll assume you have a VARCHAR (or similar) column in your table that is called json, and stores the values you presented. You didn't provide the schema, so please adjust the column name as necessary.
You're not using PARSE_JSON as a function in your SQL. You should write something like
select
parse_json(json):bucket::string as bucket,
parse_json(json):bySeqno::string as bySeqno,
parse_json(json):cas::int as cas
...
from STG_YS_APPS v
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
edited Nov 15 '18 at 0:24
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
answered Nov 14 '18 at 5:12
Marcin ZukowskiMarcin Zukowski
1,9351820
1,9351820
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
add a comment |
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
i have attached the snapshot of schema above. when i tried select parse_json(json):bucket::string as bucket from STG_YS_APPS, i got error: invalid identifier 'JSON . please help if i am missing something
– OMG
Nov 14 '18 at 6:35
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
may someone please help me if possible, Snowflake datawarehouse is little new to me.
– OMG
Nov 14 '18 at 14:38
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
am i need to add lateral flatten
– OMG
Nov 14 '18 at 14:42
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
I'm sorry, but you clearly should do some reading and read/watch some tutorials about SQL in general and Snowflake VARIANT processing... Your questions suggest lack of very basic SQL knowledge.
– Marcin Zukowski
Nov 15 '18 at 0:25
add a comment |
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any clue techie please
– OMG
Nov 13 '18 at 15:42
What is the exact schema of your table? Do you have there a column name called "parse_json" ? That's what your SQL suggests, but that would be weird. Maybe do "DESC TABLE v" and add the output to the question
– Marcin Zukowski
Nov 13 '18 at 22:59
@MarcinZukowski: No i don't have column named as parse_json. it is inbuilt in snowflake for parsing JSON.
– OMG
Nov 14 '18 at 5:05
Well, your query is written like you did. Please run "DESC TABLE v" and provide output
– Marcin Zukowski
Nov 14 '18 at 5:06