我一直在尝试以我想要的格式从我的数据库中获取JSON对象,因此我运行了以下sql查询:
SELECT PROJECTS.key_code AS CODE, PROJECTS.name AS Name,
PROJECTS.date AS Date, PROJECTS.descr AS Description
FROM PROJECTS LEFT JOIN ACCESS
ON PROJECTS.key_code = ACCESS.key_code
WHERE ACCESS.Ukey_code = '5d8hd5' FOR JSON PATH, WITHOUT_ARRAY_WRAPPER;
以及查询结果如下:
{
"Code": "h5P93G",
"Name": "Project1 test name",
"Date": "2017-09-03",
"Description": "This is a test description 1"
},
"Code": "KYJ482",
"Name": "Project2 test name",
"Date": "2018-10-25",
"Description": "This is a test description 2"
}
但实际上我想要的是不同的。 JSON对象应如下所示:
{
"h5P93G": {
"Name": "Project1 test name",
"Date": "2017-09-03",
"Description": "This is a test description 1"
},
"KYJ482": {
"Name": "Project2 test name",
"Date": "2018-10-25",
"Description": "This is a test description 2"
},
}
那么,我怎么能得到这个JSON对象?
据我所知,你不能用select ... for json
创建带有变量键名的json。
但是,如果你不介意使用变量并且你正在使用Sql Server 2017(否则你不能使用json-modify
的动态密钥),你可以这样做:
declare @a nvarchar(max) = '{}'
select
@a = json_modify(
@a,
concat('$.', p.key_code),
json_query((select p.name, p.[date], p.descr for json path, without_array_wrapper))
)
from projects as p
select @a
如果您使用的是早期版本的Sql Server,您可以使用您可以找到的任何聚合方法将其聚合(为了简单起见,我使用了string_agg
):
select
concat('{', string_agg(
concat('"',p.key_code,'":',p.data),
','
), '}')
from (
select
p.key_code,
(select p.name, p.[date], p.descr for json path, without_array_wrapper) as data
from projects as p
) as p
您可能还会考虑使用string_escape
来防止错误,以防您的密钥包含特殊字符:
select
...
concat('"',string_escape(p.key_code,'json'),'":',p.data),
','
...