在 SQL Server 2016 中使用 JSON 进行字符串聚合

Question

I would like to format a json string '[{"_":7},{"_":13},{"_":17}]' as '[7,13,17]' Tried with REPLACE Method in TSQL. I have to use REPLACE method three times to get the desire result.

SELECT REPLACE(REPLACE(REPLACE('[{"_":7},{"_":13},{"_":17}]','},{"_":',', '),'{"_":',''),'}','')

is there a better way to do that? I am using SQL Server 2016.

After some comments for this post, This my actual issue.

I have some customer data. Customer Table

CustomerId | Name
    1         ABC
    2         XYZ
    3         EFG

each customer has some area of interest. Customer Area of Interest

CustomerAreaInterestId | FK_CustomerId | FK_AreaOfInterestId
      1                       1            2
      2                       1            3
      3                       1            5
      4                       2            1
      5                       2            2
      6                       3            3
      7                       3            4

Area of interest table

   AreaOfInterestId | Description
       1                Interest1
       2                Interest2
       3                Interest3
       4                Interest4
       5                Interest5

In the final result set, I have to include area of interest id's as an array of value

[
{
    "CustomerName": "ABC",
    "AreaofInterest": "[2,3,5]"
},
{
    "CustomerName": "XYZ",
    "AreaofInterest": "[1,2]"
},
{
    "CustomerName": "EFG",
    "AreaofInterest": "[3,4]"
}
]

The result consists with some other data's as well. I have omitted for the code brevity.

Answer 1

Short Version

Cast the numeric field to text before trying to aggregate it

---

From the comments, it looks like the real question is how to use JSON to aggregate strings in SQL Server 2016, as shown in this answer .

SELECT 
 JSON_VALUE(
   REPLACE(
     (SELECT _ = someField FROM someTable FOR JSON PATH)
    ,'"},{"_":"',', '),'$[0]._'
) 

or, rewritten for clarity :

SELECT 
 JSON_VALUE(  REPLACE(
                    (SELECT _ = someField 
                     FROM someTable 
                     FOR JSON PATH)
              ,'"},{"_":"',', ')
 ,'$[0]._') 

That query works only with string fields. One needs to understand what it does before it can be adopted to other types.

• The inner query generates a JSON string from a field's values, eg '[{"_":"value1"},{"_":"value2"}]' .
• REPLACE replaces the quotes and separators between objects, changing that array of objects to '[{"_":"value1,value2"}]' . That's a single object in an array, whose single attribute is a comma-separated string.
• JSON_VALUE(...,,'$[0]._') extracts the _ attribute of that single array item.

That trick can't be used with numeric values because they don't have quotes. The solution is to cast them to text first:

SELECT 
 JSON_VALUE(  REPLACE(
                    (SELECT _ = CAST(someNumber as nvarchar(20))
                     FROM someTable 
                     FOR JSON PATH)
              ,'"},{"_":"',', ')
 ,'$[0]._') 

Eg :

declare @t table (id int)
insert into @t 
values
(7),
(13),
(17)


SELECT 
   JSON_VALUE(   REPLACE(
                        (SELECT _ = cast(ID as nvarchar(20)) 
                         FROM @t 
                         FOR JSON PATH)
    ,'"},{"_":"',', '),'$[0]._') 

The only change from the original query is the cast clause.

This produces :

7, 13, 17

This conversion is localized so care must be taken with decimals and dates, to avoid producing unexpected results, eg 38,5, 40,1 instead of 38.5, 40.1 .

PS: That's no different than the XML technique, except STUFF is used there to cut off the leading separator. That technique also needs casting numbers to text, eg :

SELECT STUFF(
    (  SELECT N', ' + cast(ID as nvarchar(20)) 
       FROM @t FOR XML PATH(''),TYPE)
    .value('text()[1]','nvarchar(max)'),
    1,2,N'')

Answer 2

Yes you could do it with only 2 replace :

SELECT REPLACE(REPLACE('[{"_":7},{"_":13},{"_":17}]','{"_":',''),'}','')

DEMO HERE

Except if you really need a space after coma which is not what you asked to be honest.

Answer 3

If you want to use only JSON functions (not string-based approach), the next example may help:

DECLARE @json nvarchar(max) = N'[{"_":7},{"_":13},{"_":17}]'
DECLARE @output nvarchar(max) = N'[]'

SELECT @output = JSON_MODIFY(@output, 'append $', j.item)
FROM OPENJSON(@json) WITH (item int '$."_"') j

SELECT @output AS [Result]

Result:

Result
[7,13,17]

Of course, the approach based on string aggregation is also a possible solution:

DECLARE @json nvarchar(max) = N'[{"_":7},{"_":13},{"_":17}]'

SELECT CONCAT(
   N'[',
   STUFF(
      (
      SELECT CONCAT(N',', j.item)
      FROM OPENJSON(@json) WITH (item int '$."_"') j
      FOR XML PATH('')
      ), 1, 1, N''
   ),
   N']'
)