1

我有以下数据：

CREATE TABLE mytable
 (
  ID int,
  Product nvarchar(50),
  Usage nvarchar(255),
 );

INSERT INTO mytable VALUES (99346,'Universal light','Art and Culture');
INSERT INTO mytable VALUES (99346,'Universal light','Health and Care');
INSERT INTO mytable VALUES (99346,'Universal light','Hotel and Wellness');
INSERT INTO mytable VALUES (99346,'Universal light','Education and Science');

我创建了以下代码来获取我的 JSON 输出：

SELECT DISTINCT T1.ID
      ,T1.Product
      ,(SELECT T2.Usage
        FROM mytable T2
        WHERE T1.ID=T2.ID
        FOR JSON PATH) as 'Usage'
FROM mytable T1
FOR JSON PATH

它输出以下结果：

[
    {
        "ID": 99346,
        "Product": "Universal light",
        "Usage": [
            {
                "Usage": "Art and Culture"
            },
            {
                "Usage": "Health and Care"
            },
            {
                "Usage": "Hotel and Wellness"
            },
            {
                "Usage": "Education and Science"
            }
        ]
    }
]

我希望得到如下结果，但不知道如何更改语法：

[
    {
        "ID": 99346,
        "Product": "Universal light",
        "Usage": [ "Art and Culture" , "Health and Care" , "Hotel and Wellness" , "Education and Science"
         ]
    }
]

对此非常感谢任何帮助。

编辑

如果我使用这个初始数据，在第 3 行的末尾我有一个额外的 ' ' 解决方案不起作用，没有错误或警告：

INSERT INTO mytable VALUES (99346,'Universal light','Art and Culture');
INSERT INTO mytable VALUES (99346,'Universal light','Education and Science');
INSERT INTO mytable VALUES (99346,'Universal light','Health and Care ');
INSERT INTO mytable VALUES (99346,'Universal light','Hotel and Wellness');
INSERT INTO mytable VALUES (99346,'Universal light','Offices and Communication');
INSERT INTO mytable VALUES (99346,'Universal light','Presentation and Retail');

我尝试使用 TRIM，如下所示：

SELECT Distinct ID
      ,Product
      ,json_query(QUOTENAME(STRING_AGG('"' + STRING_ESCAPE(TRIM(Usage), 'json') + '"', char(44)))) AS 'Usage'
FROM mytable
GROUP BY ID
      ,Product
FOR JSON PATH;

不幸的是，它不起作用，整个数组 'Usage' 以某种方式被忽略了，查看结果：

[
    {
        "ID": 99346,
        "Product": "Universal light"
    }
]
4

1 回答 1

1

您可以STRING_AGG像这样使用构建数组：

SELECT  DISTINCT ID
      ,Product
      ,json_query(QUOTENAME(STRING_AGG('"' + STRING_ESCAPE(Usage, 'json') + '"', char(44)))) AS 'Usage'
FROM mytable T1
GROUP BY ID
      ,Product
FOR JSON PATH;

如果您未使用 SQL Sever 2017 或更高版本，则可以使用 XML PATH 连接值。

SELECT DISTINCT T1.ID
      ,T1.Product
      ,
       (
         '[' +
          STUFF
          (
                (
                SELECT ',' + '"' + T2.Usage + '"' 
                FROM mytable T2
                WHERE T1.ID=T2.ID
                FOR XML PATH, TYPE
                ).value('.', 'NVARCHAR(MAX)')
                ,1
                ,1
                ,''
            )
          + ']'
       ) as 'Usage'
FROM mytable T1
FOR JSON PATH

供您编辑使用：

SELECT Distinct ID
      ,Product
      ,json_query('[' + (STRING_AGG('"' + STRING_ESCAPE(TRIM(Usage), 'json') + '"', char(44))) + ']') AS 'Usage'
FROM mytable
GROUP BY ID
      ,Product
FOR JSON PATH;

问题是QUOTENAME输入限制为 128 个字符，并NULL在添加更多记录时返回。

于 2020-04-29T09:15:03.220 回答