1

这是我声明的 json：

DECLARE @json VARCHAR(MAX) = N'
[
 {
  "mTruckId": -35839339,
  "mPositionId": 68841545,
  "mPositionDateGmt": "laboris ipsum ullamco",
  "mLatitude": -36598160.205007434,
  "mLongitude": 54707169.834195435,
  "mGpsValid": false,
  "mHeading": 114,
  "mSpeed": -888256.4982997179,
  "mAdditionalInformation": {
   "mVin": "voluptate veniam",
   "mOdometer": 25567959.615529776,
   "mEngineHours": -87509827.08880372,
   "mTemperatureSensors": [
    {
     "mUnit": "C",
     "mLabel": "aute in",
     "mValue": -74579140.64111689
    },
    {
     "mUnit": "C",
     "mLabel": "ullamco labore dolore",
     "mValue": -91870052.84894001
    }
   ]
  }
 },
 {
  "mTruckId": 80761376,
  "mPositionId": 88380593,
  "mPositionDateGmt": "sed pariatur ut sint",
  "mLatitude": 62504812.42302373,
  "mLongitude": 14622406.17103973,
  "mGpsValid": false,
  "mHeading": 302,
  "mSpeed": 39030054.634676635,
  "mAdditionalInformation": {
   "mVin": "aute",
   "mOdometer": 74400412.05641022,
   "mEngineHours": 88453976.08453897,
   "mTemperatureSensors": [
    {
     "mUnit": "F",
     "mLabel": "reprehenderit consectetur id ipsum",
     "mValue": 22634605.53841141
    },
    {
     "mUnit": "C",
     "mLabel": "magna consectetur esse",
     "mValue": 72633803.44269562
    }
   ]
  }
 }
]'

这是我从 json 中提取温度传感器数据的代码。我认为它会起作用，因为这个 json 中获取温度传感器数据的层次结构是 root -> mAdditionalInformation -> mTemperatureSensors。

SELECT  Unit,
        Label,
        Value
FROM OPENJSON(@json)
WITH(
    Unit    VARCHAR(15) '$.mAdditionalInformation.mTemperatureSensors.mUnit',
    Label   VARCHAR(50) '$.mAdditionalInformation.mTemperatureSensors.mLabel',
    Value   FLOAT       '$.mAdditionalInformation.mTemperatureSensors.mValue'
)

它返回 2 行都为空，为什么要这样做？我希望它提取 mTemperatureSensors 数据中的每个元素。

Unit    Label   Value
NULL    NULL    NULL
NULL    NULL    NULL
4

2 回答 2

0 
select s.*
from openjson(@json)
with
(
    mTemperatureSensors nvarchar(max) '$.mAdditionalInformation.mTemperatureSensors' as json
) as t
cross/*outer*/ apply openjson(t.mTemperatureSensors)
with
(
mLabel nvarchar(20)
) as s
于 2020-09-18T22:02:41.637 回答
0

一种选择是在将@json的内容插入table( ) 之后OPENJSON逐步应用：CROSS APPLYtab

SELECT  Unit, Label, Value
  FROM tab
 CROSS APPLY OPENJSON(JsonData)
             WITH (
                    TempSens NVARCHAR(MAX) '$.mAdditionalInformation.mTemperatureSensors' AS JSON ) Q1 
 CROSS APPLY OPENJSON (Q1.TempSens) 
             WITH (
                    Unit   NVARCHAR(MAX) '$.mUnit',
                    Label  NVARCHAR(MAX) '$.mLabel',
                    Value  FLOAT         '$.mValue'
                  ) Q2

Demo

或在您的情况下声明一个标量变量@json：

SELECT Unit, Label, Value
  FROM OPENJSON(@json)
             WITH (
                    TempSens NVARCHAR(MAX) '$.mAdditionalInformation.mTemperatureSensors' AS JSON ) Q1
 CROSS APPLY OPENJSON (Q1.TempSens) 
             WITH (
                    Unit   NVARCHAR(MAX) '$.mUnit',
                    Label  NVARCHAR(MAX) '$.mLabel',
                    Value  FLOAT         '$.mValue'
                  ) Q2

Demo

于 2020-09-18T22:18:46.513 回答