将不同的cte json对象组合在一起

tnkciper  于 2021-08-01  发布在  Java
关注(0)|答案(1)|浏览(312)

我很难理解将两个cte json对象组合在一起的概念。来自这两者的数据由一个字段woid进行匹配,该字段在这个json中是“workorderid”。我不知道该怎么办。
需要的结果:

{
    "header": {
        "InstanceName": "string"
    },
{
    "workOrderList": [
        {
            "workOrderID": "string",
            "serviceAddressID": "string",
            "routeID": "string",
            "workOrderSubTypes": {
                "workOrderSubTypeList": [
                    {
                        "workOrderSubTypeID": "string",
                        "instructions": "string",
                        "pricePerService": "string",
                        "serviceQuantity": 0
                    }
                ]
            }
        }
    ]
}

这是迄今为止创建的,并创建了两个部分。我似乎找不到正确的方法将它们组合成一个json对象,以便在下一步中使用。我使用cte创建json的每个部分,从最内部的部分开始向外。我被困在“结合工作顺序和子类型”的点上。也许这应该在我已经创建的步骤中完成;我不确定?

-- Create work order subtype
With fSubType1 (sSubType1) as (select json_object(     
'workOrderSubTypeID' : SvcSeq,
'instructions' : trim(Instr),
'pricePerService' : InvAmt,
'serviceQuantity' : Qty)              
  FROM PADWOD)   

-- Create work order subtype array 
 , fSubType2 (sSubType2) as (values json_array  (                     
           select sSubType1  from fSubType1 format json))  

-- Create work order subtype List
 , fSubType3 (sSubType3) as ( select json_object  ('workOrderSubTypeList'  : 
        sSubType2 Format json) from fSubType2)   

-- Create work order 
, fWorkOrder1 (sWorkOrder1) as (select json_object(     
'workOrderID' : WOID,
'serviceAddressID' : trim(Acct#),
'routeID' : RouteID) 
FROM PADWOH)

--**Combine the workorder and subtypes - not sure how to do?**
, fWorkOrder2 (sWorkOrder2) as (select  json_object('workOrderList' : sworkorder1 format json), 
json_object('workOrderSubTypes' : ssubtype3)

-- Create the work order array
, fWorkOrder3 (sWorkOrder3) as (values json_array  ( 
select sWorkOrder1 from fWorkOrder1 format json))   

-- Create the header info
, fheaderData (sheaderData) as (select  json_object( 
'InstanceName' : trim(Cntry) )
from padxmlhdr
where cntry = 'US')

-- Final json
select json_object('header' : sheaderData format json, 
'workOrderData' value sWorkOrder1 format json)
from fheaderData, fWorkOrder1;

以下是上述文件中使用的数据:

PADWOH
workOrderID                           serviceAddressID        routeID   
85789003-b2af-1a3a-898a-0004ac1acb95  BRAU0244-0001           161819    
562ef003-49af-1a3d-898a-0004ac1acb95  21CM0005-0001           161812    
37a11004-f8f8-1a46-95c9-0004ac1acb95  AMER1412-0001           191114    

PADWOD
workOrderID                           subTypeID  pricePerService  serviceQuantity  instructions 
85789003-b2af-1a3a-898a-0004ac1acb95    00001             96.36             0      Test Inst02  
85789003-b2af-1a3a-898a-0004ac1acb95    00127             10.03             0      Test Inst03  
37a11004-f8f8-1a46-95c9-0004ac1acb95    00051               .00             0      Test Inst01   

PADXMLHDR
COUNTRY      
INSTANCE NAME
     US

这是预期的json(我想,我不确定子类型记录会是什么样子——或者在没有数据显示时应该是什么样子)。我认为它不会创造任何东西,但我现在不知道。)

{
    "header": {
        "InstanceName": "US"
    },
    "workOrderList": [
        {
            "workOrderID": "85789003-b2af-1a3a-898a-0004ac1acb95",
            "serviceAddressID": "BRAU0244-0001",
            "routeID": "161819",
            "workOrderSubTypes": {
                "workOrderSubTypeList": [
                    {
                        "workOrderSubTypeID": "00001",
                        "instructions": "Test Inst02",
                        "pricePerService": "96.36",
                        "serviceQuantity": "0"
                    },
                    {
                        "workOrderSubTypeID": "00127",
                        "instructions": "Test Inst03",
                        "pricePerService": "10.03",
                        "serviceQuantity": "0"
                    }
                ]
            }
        },
        {
            "workOrderID": "562ef003-49af-1a3d-898a-0004ac1acb95",
            "serviceAddressID": "21CM0005-0001",
            "routeID": "161812",
            "workOrderSubTypes": {
                "workOrderSubTypeList": [
                    {}
                ]
            }
        },
        {
            "workOrderID": "37a11004-f8f8-1a46-95c9-0004ac1acb95",
            "serviceAddressID": "AMER1412-0001",
            "routeID": "191114",
            "workOrderSubTypes": {
                "workOrderSubTypeList": [
                    {
                        "workOrderSubTypeID": "00051",
                        "instructions": "Test Inst01",
                        "pricePerService": ".00",
                        "serviceQuantity": "0"
                    }
                ]
            }
        }
    ]
}
0h4hbjxa

0h4hbjxa1#

只是一个演示它应该如何工作。

WITH 
-- Your sample data
  PADWOH (workOrderID, serviceAddressID, routeID) AS 
(
VALUES
  ('85789003-b2af-1a3a-898a-0004ac1acb95', 'BRAU0244-0001', 161819)   
, ('562ef003-49af-1a3d-898a-0004ac1acb95', '21CM0005-0001', 161812)   
, ('37a11004-f8f8-1a46-95c9-0004ac1acb95', 'AMER1412-0001', 191114)
)
, PADWOD (workOrderID, subTypeID, pricePerService, serviceQuantity, instructions) AS 
(
VALUES
  ('85789003-b2af-1a3a-898a-0004ac1acb95', '00001', 96.36, 0, 'Test Inst02')  
, ('85789003-b2af-1a3a-898a-0004ac1acb95', '00127', 10.03, 0, 'Test Inst03')
, ('37a11004-f8f8-1a46-95c9-0004ac1acb95', '00051',   .00, 0, 'Test Inst01')
)
-- End of your sample data

, PADWOD2 AS 
(
SELECT 
  workOrderID
, JSON_OBJECT
  (
    'workOrdersubTypeID' VALUE subTypeID
  , 'instructions'       VALUE instructions
  , 'pricePerService'    VALUE pricePerService 
  , 'serviceQuantity'    VALUE serviceQuantity
  ) subtypes
FROM PADWOD
)
, PADWOH2 AS
(
SELECT 
  workOrderID
, JSON_OBJECT
(
  'workOrderID'       VALUE workOrderID
, 'serviceAddressID'  VALUE serviceAddressID
, 'routeID'           VALUE routeID
, 'workOrderSubTypes' VALUE JSON_OBJECT 
  (
    'workOrderSubTypeList' VALUE JSON_ARRAY 
    (
      (
      SELECT subtypes 
      FROM PADWOD2 D
      WHERE D.workOrderID = H.workOrderID
      FETCH FIRST 10000 ROW ONLY
      ) FORMAT JSON
    ) FORMAT JSON
  ) FORMAT JSON
) WORKORDER 
FROM PADWOH H
FETCH FIRST 1 ROW ONLY
)

VALUES JSON_OBJECT
(
  'header'        VALUE JSON_OBJECT('InstanceName' VALUE 'US') FORMAT JSON
, 'workOrderList' VALUE JSON_ARRAY((SELECT WORKORDER FROM PADWOH2) FORMAT JSON) FORMAT JSON
);

该查询按原样工作,但此时json_数组标量函数似乎已损坏(在11.5.4.0和11.1.4.5中),因此需要使用这些函数 FETCH FIRST 子句(它必须在没有它们的情况下工作)只是为了得到一些结果(否则会出现各种sqlcode=-901错误)。
您可以在ibm支持下打开一个案例,使其按需工作。。。

更新

预计这些bug将在即将发布的两个db2版本的fixpack中得到解决。

相关问题