CODEWITHSUNDEEP
sqljsonsql-servert-sql
user7331869
user7331869

Reputation:

Expand row with an JSON Array inside a Column to multiple rows

I encountered a SQL-Data-Migration (SQL Server) problem and i hope, you can guide me in the right direction.

Assume, we have the table DataTable (names simplified) with the following columns:

DataID | SomeForeignKey | SpecificDataValues | OtherSpecificDataValues
int    | int            | String             | String
-------+----------------+--------------------+------------------------
0      | 1              | ['1','2']          | ['1', '2']

where SpecificDataValues and OtherSpecificDataValues are JSON arrays (like ['1', '2'])

Now i want to migrate this table (with an SQL Migration script at best) to a new table:

DataValuesTable
DataID | SomeForeignKey | SpecificDataValues | OtherSpecificDataValues
-------+----------------+--------------------+------------------------
0      | 1              | 1                  | 1
1      | 1              | 2                  | 2

So, i basically want to generate a new row in a new table for each value, stored in "SpecificDataValues" and "OtherSpecificDataValues"

I already checked, that there are SQL functions to work with JSON (OPENJSON, JSON_QUERY) but i was not able to produce the desired result using this tools.

I hope, you can show me the right direction.

Upvotes: 3

Views: 4610

Answers (1)

Salman Arshad
Salman Arshad

Reputation: 272006

One solution is to use OPENJSON twice. It will generate Array(2) x Array(2) = 4 rows, the ones you're interested in are the ones where the index positions match:

SELECT DataID, SomeForeignKey, j1.value, j2.value
FROM t
CROSS APPLY OPENJSON(SpecificDataValues) AS j1
CROSS APPLY OPENJSON(OtherSpecificDataValues) AS j2
WHERE j1.[key] = j2.[key]

Another solution is to use OPENJSON and JSON_VALUE together:

SELECT DataID, SomeForeignKey, j1.value, JSON_VALUE(OtherSpecificDataValues, CONCAT('$[', CAST(j1.[key] AS INT), ']'))
FROM t
CROSS APPLY OPENJSON(SpecificDataValues) AS j1

Note that your "JSON" is invalid. Strings must be enclosed inside ".

Demo on db<>fiddle

Upvotes: 6

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