使用CASE根据字符串列中的特定文本创建新列

Question

我找到了几个类似的线程,但没有一个正在工作.我正在尝试为另一列满足某个条件时创建一个新列.

这是我正在使用的代码:

SELECT DISTINCT R.[Column1] AS Person, 
SUM(CASE WHEN  R.[Event] = 'Event1' THEN 1 ELSE NULL END) AS Event1,
    CASE (WHEN L.[Column2] LIKE '%String1%' THEN 'String1'
        ELSE WHEN L.[Column2] LIKE '%String2%' THEN 'String2'
        ELSE WHEN L.[Column2] LIKE '%String3%' THEN 'String3'
        ELSE NULL END) AS NewColumn
  FROM [Database1].[dbo].[Table1] R
  LEFT JOIN
     [Database1].[dbo].[Table2] L
        ON R.[UniqueIdentifier] = L.[UniqueIdentifier]
    WHERE L.[Column2] LIKE '%String1%'
        OR L.[Column2] LIKE '%String2%'
        OR L.[Column2] LIKE '%String3%'
GROUP BY  R.[Column1], L.[Column2]
ORDER BY R.[Event1] DESC

如果我从第2列中取出CASE语句,那么查询工作正常.我想要的结果是三列:Person,String,Event ...用Person和String上的聚合计数事件.

错误是:消息156,级别15,状态1,行3关键字'CASE'附近的语法不正确.

Answer 1

您遇到了一些语法问题：

第一个问题是：CASE (WHEN

第二个问题是：ELSE WHEN

现在应该可以正常运行：

SELECT DISTINCT
       R.[Column1] AS Person,
       SUM(CASE
               WHEN R.[Event] = 'Event1'
               THEN 1
               ELSE NULL
           END) AS Event1,
       (CASE
            WHEN L.[Column2] LIKE '%String1%'
            THEN 'String1'
            WHEN L.[Column2] LIKE '%String2%'
            THEN 'String2'
            WHEN L.[Column2] LIKE '%String3%'
            THEN 'String3'
            ELSE NULL
        END) AS NewColumn
FROM [Database1].[dbo].[Table1] R
     LEFT JOIN [Database1].[dbo].[Table2] L ON R.[UniqueIdentifier] = L.[UniqueIdentifier]
WHERE L.[Column2] LIKE '%String1%'
      OR L.[Column2] LIKE '%String2%'
      OR L.[Column2] LIKE '%String3%'
GROUP BY R.[Column1],
         L.[Column2]
ORDER BY R.[Event1] DESC;