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我有一个 SQL 表，其中包含一个包含 JSON 数据的列。JSON 值之一如下所示：

{
    "_id": "5a450f038104ca3cb0ff74b5",
    "index": 3,
    "guid": "20d807c5-bddc-44b9-97fe-fd18af1b6066",
    "isActive": false,
    "balance": "$2,832.38",
    "picture": "http://placehold.it/32x32",
    "age": 23,
    "eyeColor": "brown",
    "firstname": "Genevieve",
    "lastname": "Green",
    "gender": "female",
    "company": "PROFLEX",
    "email": "genevievegreen@proflex.com",
    "phone": "+1 (919) 464-2866",
    "address": "107 Clermont Avenue, Rew, California, 4298",
    "about": "Magna pariatur ut enim nulla pariatur ad Lorem amet. Proident nulla exercitation id Lorem commodo minim cillum irure exercitation labore nostrud nostrud sint. Voluptate commodo ea commodo quis Lorem laborum culpa voluptate enim nulla enim duis.\r\n",
    "registered": "2016-02-16T09:51:25 +05:00",
    "latitude": -16.492643,
    "longitude": -71.782118,
    "tags": [
      "in",
      "non",
      "eiusmod",
      "labore",
      "dolor",
      "laboris",
      "ullamco"
    ],
    "friends": [
      {
        "id": 0,
        "name": "Mccoy Berg",
        "interests": [
          "Music",
          "Birding",
          "Chess"
        ]
      },
      {
        "id": 1,
        "name": "Chase Mcfadden",
        "interests": [
          "Software",
          "Chess",
          "History"
        ]
      },
      {
        "id": 2,
        "name": "Michele Dodson",
        "interests": [
          "Football",
          "Birding",
          "Movies"
        ]
      }
    ],
    "greeting": "Hello, Genevieve! You have 2 unread messages.",
    "favoriteFruit": "strawberry"
  }

我可以执行一个检索名字和姓氏以及所有朋友的查询，如下所示：

SELECT
  JSON_VALUE(JsonValue, '$.firstname') as FirstName,
  JSON_VALUE(JsonValue, '$.lastname') as LastName,
  JSON_QUERY(JsonValue, '$.friends') as FriendsList,
From <MyTable>
Where JSON_VALUE(JsonValue,'$.lastname') = 'Green'

所写的查询返回 FriendsList 的 JSON 字符串，如下所示：

[  
   {  
      "id":0,
      "name":"Mccoy Berg",
      "interests":[  
         "Music",
         "Birding",
         "Chess"
      ]
   },
   {  
      "id":1,
      "name":"Chase Mcfadden",
      "interests":[  
         "Software",
         "Chess",
         "History"
      ]
   },
   {  
      "id":2,
      "name":"Michele Dodson",
      "interests":[  
         "Football",
         "Birding",
         "Movies"
      ]
   }
]

我实际上想要的只是一组朋友的名字，如下所示：

["Mccoy Berg", "Chase Mcfadden", ...]

我确信这是可能的，但我对 JSON 的了解有限。

4

1 回答 1

3

在以您期望的格式创建数组时，SQL Server 有点有趣。默认情况下，它总是将 JSON 数组创建为键：值对。STUFF()您可以通过使用和FOR XML使用 JSON来解决此问题。

您首先创建一个仅返回名称的子查询，然后您可以将STUFF这些名称放入FriendsList字段中，如下所示：

SELECT
FirstName    = JSON_VALUE(JsonValue, '$.firstname')
,LastName    = JSON_VALUE(JsonValue, '$.lastname')
,FriendsList = '[' + STUFF(
                        (SELECT ',' + '"' + [name] + '"'
                            FROM OPENJSON(JsonValue,'$.friends')
                            WITH ( [name] NVARCHAR(100) '$.name') 
                        FOR XML PATH (''))
                        , 1, 1, '') 
            + ']'
FROM <MyTable>
于 2017-12-28T16:59:52.707 回答