54

我有一些我想在 SQL Server 2016 中解析的 json。有一个 Projects->Structures->Properties 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]')

或者

openjson (@json, '$.structures[0]')

我有这样的想法，即我可以读取顶级项目对象的值以及表示其下方结构的 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，我无法将嵌套的 json 保存在“结构”中。它只是返回 NULL 而不是 json 内容：

在此处输入图像描述

有谁知道这是否可行，如果可以，我做错了什么？

4

3 回答 3

100

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
于 2016-12-13T07:03:45.170 回答
69

如果引用 JSON 对象或数组，则需要指定 AS JSON 子句：

select IdProject, Name, structures
from   openjson (@json)
with
(
    IdProject uniqueidentifier,
    Name nvarchar(100),
    structures nvarchar(max) AS JSON
) as Projects

请参阅常见问题解答：https ://docs.microsoft.com/en-us/sql/relational-databases/json/solve-common-issues-with-json-in-sql-server?view=sql-server-ver15#return -a-nested-json-sub-object-from-json-text-with-openjson

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 (......)
于 2016-05-13T20:00:02.657 回答
7

典型的！我在发布问题后找到了答案。在指定要返回的列时，您需要使用“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
于 2016-05-13T19:55:56.840 回答