Question

我有以下数据；

DECLARE @json NVARCHAR(2048) = N'{
    "searchId": -1,
    "statuses": [
        33,
        85,
        90
        ]
}';

并使用以下命令执行返回；

SELECT searchID
    ,x.statuses
FROM OpenJson(@json) WITH (
        searchID VARCHAR(200) '$.searchId'
        ,statuses NVARCHAR(MAX) '$.statuses' AS JSON
        )
CROSS APPLY OPENJSON(statuses, '$') WITH (statuses int '$') AS x

返回；

searchID    statuses
-1          33
-1          85
-1          90

它可以工作，但是id像返回一样是一行，因此“状态”列将显示为

searchID    statuses
-1          33,85,90

我主要使用Stuff（）和ForXML进行了一些实验，但是我认为必然会有更清洁的方式来处理json吗？

Answer 1

您可以仅将状态保留为JSON，然后将演示文稿保留在前端。

DECLARE @json NVARCHAR(2048) = N'{
    "searchId": -1,
    "statuses": [33,85,90]
}';

SELECT searchID
    ,x.statuses
FROM OpenJson(@json) WITH (
        searchID VARCHAR(200) '$.searchId'
        ,statuses NVARCHAR(MAX) '$.statuses' AS JSON
        ) x

输出

searchID              statuses
--------------------- -----------
-1                    [33,85,90]

或者只是从JSON数组中删除“ [”和“]”：

DECLARE @json NVARCHAR(2048) = N'{
    "searchId": -1,
    "statuses": [33,85,90]
}';

SELECT searchID
    ,substring(x.statuses,2,LEN(x.statuses)-2) statuses
FROM OpenJson(@json) WITH (
        searchID VARCHAR(200) '$.searchId'
        ,statuses NVARCHAR(MAX) '$.statuses' AS JSON
        ) x

输出

searchID              statuses
--------------------- -----------
-1                    33,85,90

Answer 2

STRING_AGG返回到varchar。

DECLARE @json NVARCHAR(2048) = N'{
    "searchId": -1,
    "statuses": [
        33,
        85,
        90
        ]
}';
--
SELECT searchID, string_agg(x.statuses, ',') statuses
FROM OpenJson(@json) WITH (
        searchID VARCHAR(200) '$.searchId'
        ,statuses NVARCHAR(MAX) '$.statuses' AS JSON
        )
CROSS APPLY OPENJSON(statuses, '$') WITH (statuses int '$') AS x
GROUP BY searchID;

SQL OPENJSON到单列

2 个答案: