SQL Server OPENJSON读取嵌套的json
问题描述:
我有一些json,我想在SQL Server 2016中解析。有一个Projects-> Structures-> Properties的层次结构。我想编写解析整个层次的查询,但我不希望指定索引号的任何元素,即我不想做这样的事:SQL Server OPENJSON读取嵌套的json
openjson (@json, '$[0]')
或
openjson (@json, '$.structures[0]')
我有这个想法,我可以读取顶级项目对象的值以及表示它下面的结构的json字符串,然后可以单独解析这些结构。问题是,下面的代码无法正常工作:
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和名称得到恢复没有问题,但由于某种原因,我不能在“结构”举行的嵌套JSON。取而代之的是JSON内容只是它返回NULL:
有谁知道这是可能的,如果是的话,我究竟做错了什么?
答
如果引用JSON对象或数组,你需要指定为JSON子句:
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) AS JSON
) as Projects
参见常见问题解答: https://msdn.microsoft.com/en-us/library/mt631706.aspx#Anchor_6
如果你想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 (......)
答
典型!我在发布问题后才找到答案。你需要指定的列时,使用“作为JSON”关键字返回:
select IdProject, Name, structures
from openjson (@json)
with
(
IdProject uniqueidentifier,
Name nvarchar(100),
structures nvarchar(max) as json
) as Projects
答
使用CROSS APPLY:
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