SQL 服务器 OPENJSON 读取嵌套 json
SQL Server OPENJSON read nested json
我有一些 json 想在 SQL Server 2016 中解析。有一个项目->结构->属性的层次结构。我想编写一个解析整个层次结构的查询,但我不想通过索引号指定任何元素,即我不想做这样的事情:
openjson (@json, '$[0]')
或
openjson (@json, '$.structures[0]')
我有这样的想法,我可以读取顶级项目对象的值以及代表其下方结构的 json 字符串,然后可以单独对其进行解析。问题是以下代码不起作用:
declare @json nvarchar(max)
set @json = '
[
{
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Project",
"structures":[
{
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Structure",
"BaseStructure":"Base Structure",
"DatabaseSchema":"dbo",
"properties":[
{
"IdProperty":"618DC40B-4D04-4BF8-B1E6-12E13DDE86F4",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 2",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
},
{
"IdProperty":"FFF433EC-0BB5-41CD-8A71-B5F09B97C5FC",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 1",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
}
]
}
]
}
]';
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max)
) as Projects
IdProject 和 Name 返回没有问题,但由于某种原因我无法获得 'structures' 中的嵌套 json。而不是 json 内容,它只是 returns NULL:
有谁知道这是否可行,如果可行,我做错了什么?
典型!我在发布问题后立即找到了答案。将列指定为 return:
时需要使用 'as json' 关键字
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) as json
) as Projects
如果您引用 JSON 对象或数组,您需要指定 AS JSON 子句:
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) AS JSON
) as Projects
如果你想在返回的结构数组上应用 OPENJSON,你可以使用类似下面的代码:
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) AS JSON
) as Projects
CROSS APPLY OPENJSON (structures) WITH (......)
使用交叉应用:
declare @json nvarchar(max)
set @json = '
[
{
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Project",
"structures":[
{
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Structure",
"BaseStructure":"Base Structure",
"DatabaseSchema":"dbo",
"properties":[
{
"IdProperty":"618DC40B-4D04-4BF8-B1E6-12E13DDE86F4",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 2",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
},
{
"IdProperty":"FFF433EC-0BB5-41CD-8A71-B5F09B97C5FC",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 1",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
}
]
}
]
}
]';
select
Projects.IdProject, Projects.Name as NameProject,
Structures.IdStructure, Structures.Name as NameStructure, Structures.BaseStructure, Structures.DatabaseSchema,
Properties.*
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) as json
)
as Projects
cross apply openjson (Projects.structures)
with
(
IdStructure uniqueidentifier,
Name nvarchar(100),
BaseStructure nvarchar(100),
DatabaseSchema sysname,
properties nvarchar(max) as json
) as Structures
cross apply openjson (Structures.properties)
with
(
IdProperty uniqueidentifier,
NamePreoperty nvarchar(100) '$.Name',
DataType int,
[Precision] int,
[Scale] int,
IsNullable bit,
ObjectName nvarchar(100),
DefaultType int,
DefaultValue nvarchar(100)
)
as Properties
我有一些 json 想在 SQL Server 2016 中解析。有一个项目->结构->属性的层次结构。我想编写一个解析整个层次结构的查询,但我不想通过索引号指定任何元素,即我不想做这样的事情:
openjson (@json, '$[0]')
或
openjson (@json, '$.structures[0]')
我有这样的想法,我可以读取顶级项目对象的值以及代表其下方结构的 json 字符串,然后可以单独对其进行解析。问题是以下代码不起作用:
declare @json nvarchar(max)
set @json = '
[
{
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Project",
"structures":[
{
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Structure",
"BaseStructure":"Base Structure",
"DatabaseSchema":"dbo",
"properties":[
{
"IdProperty":"618DC40B-4D04-4BF8-B1E6-12E13DDE86F4",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 2",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
},
{
"IdProperty":"FFF433EC-0BB5-41CD-8A71-B5F09B97C5FC",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 1",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
}
]
}
]
}
]';
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max)
) as Projects
IdProject 和 Name 返回没有问题,但由于某种原因我无法获得 'structures' 中的嵌套 json。而不是 json 内容,它只是 returns NULL:
有谁知道这是否可行,如果可行,我做错了什么?
典型!我在发布问题后立即找到了答案。将列指定为 return:
时需要使用 'as json' 关键字select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) as json
) as Projects
如果您引用 JSON 对象或数组,您需要指定 AS JSON 子句:
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) AS JSON
) as Projects
如果你想在返回的结构数组上应用 OPENJSON,你可以使用类似下面的代码:
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) AS JSON
) as Projects
CROSS APPLY OPENJSON (structures) WITH (......)
使用交叉应用:
declare @json nvarchar(max)
set @json = '
[
{
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Project",
"structures":[
{
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"IdProject":"97A76363-095D-4FAB-940E-9ED2722DBC47",
"Name":"Test Structure",
"BaseStructure":"Base Structure",
"DatabaseSchema":"dbo",
"properties":[
{
"IdProperty":"618DC40B-4D04-4BF8-B1E6-12E13DDE86F4",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 2",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
},
{
"IdProperty":"FFF433EC-0BB5-41CD-8A71-B5F09B97C5FC",
"IdStructure":"CB0466F9-662F-412B-956A-7D164B5D358F",
"Name":"Test Property 1",
"DataType":1,
"Precision":0,
"Scale":0,
"IsNullable":false,
"ObjectName":"Test Object",
"DefaultType":1,
"DefaultValue":""
}
]
}
]
}
]';
select
Projects.IdProject, Projects.Name as NameProject,
Structures.IdStructure, Structures.Name as NameStructure, Structures.BaseStructure, Structures.DatabaseSchema,
Properties.*
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) as json
)
as Projects
cross apply openjson (Projects.structures)
with
(
IdStructure uniqueidentifier,
Name nvarchar(100),
BaseStructure nvarchar(100),
DatabaseSchema sysname,
properties nvarchar(max) as json
) as Structures
cross apply openjson (Structures.properties)
with
(
IdProperty uniqueidentifier,
NamePreoperty nvarchar(100) '$.Name',
DataType int,
[Precision] int,
[Scale] int,
IsNullable bit,
ObjectName nvarchar(100),
DefaultType int,
DefaultValue nvarchar(100)
)
as Properties