Bracket notation for SQL Server json_value? - json

Question

This works:
select json_value('{ "a": "b" }', '$.a')
This doesn't work:
select json_value('{ "a": "b" }', '$["a"]')
and neither does this:
select json_value('{ "a": "b" }', '$[''a'']')
In JSON, these are the same:
foo = { "a": "b" }
console.log(foo.a)
console.log(foo["a"])
What am I missing? I get an error trying to use bracket notation in SQL Server:
JSON path is not properly formatted. Unexpected character '"' is found at position 2

Answer 1

No sooner do I ask, than I stumble on an answer. I couldn't find this in any documentation anywhere, but select json_value('{ "a": "b" }', '$."a"') works. Bracket notation is not supported, but otherwise invalid keys can be escaped with quotation marks, e.g. select json_value('{ "I-m so invalid][": "b" }', '$."I-m so invalid]["') when in JavaScript that would be foo["I-m so invalid]["]

Answer 2

MsSql reserves this for array index. SQL parses all JSON as a string literal, instead of as an object(JSON or ARRAY) with any hidden key.
Some of what SQL can do will vary with version. Here's a crash course on the annoying (but also really powerful, and fast once in place) requirements. I'm posting more than you need because a lot of the documentation for JSON through MsSql is lacking, and doesn't do justice to how strong it is with JSON.
MsDoc here: https://learn.microsoft.com/en-us/sql/relational-databases/json/json-data-sql-server?view=sql-server-ver15
In this example, we are working with a JSON "object" to separate the data into columns. Note how calling a position inside of an array is weird.
declare #data nvarchar(max) = N'[{"a":"b","c":[{"some":"random","array":"value"},{"another":"random","array":"value"}]},{"e":"f","c":[{"some":"random","array":"value"},{"another":"random","array":"value"}]}]'
--make sure SQL is happy. It will not accept partial snippets
select ISJSON(#data)
--let's look at the data in tabular form
select
json1.*
, json2.*
from openjson(#data)
with (
a varchar --note there is no "path" specified here, as "a" is a key in the first layer of the object
, c nvarchar(max) as JSON --must use "nvarchar(max)" and "as JSON" or SQL freaks out
, c0 nvarchar(max) N'$.c[0]' as JSON
) as json1
cross apply openjson(json1.c) as json2
You can also pull out the individual values, if needed
select oj.value from openjson(#data) as oj where oj.[key] = 1;
select
oj.value
, JSON_VALUE(oj.value,N'$.e')
, JSON_VALUE(oj.value,N'$.c[0].some')
, JSON_VALUE(#data,N'$[1].c[0].some') --Similar to your first example, but uses index position instead of key value. Works because SQL views the "[]" brackets as an array while trying to parse.
from openjson(#data) as oj
where oj.[key] = 1