将 csv 字段拆分为多列

Question

我有一个包含 200,000 行数据的表。现场与会者已以 csv 格式填充；用逗号分隔符分隔。我想将此字段拆分为最多 7 个不同的列，标记为 Field1、Field2 等。

示例数据：

pkEventBooking  Attendees
166935          p1193,c21867,c21827,c21963,c18069,c19222,
195867          p1193,c21827,c22572,c19222,c22573,c21963,c18069,

新格式

pkEventBooking   Field1   Field2   Field3    Field 4   Field5   Field6  Field7
166935           p1193    c21867   c21827    c21963    c18069   c19222
195867           p1193    c21827   c22572    c19222    c22573   c21963  c18069,

Answer 1

评论太长了。

我倾向于以逗号分隔的格式导出数据并重新导入。 有了这么多数据，您可以执行以下操作：

1. 运行

   select * from example

   或

   select pkEventBooking + ',' + Attendees

2. 将数据复制到Excel中
3. 转到“数据”功能区并选择“文本到列”
4. 另存为文件
5. 导入带有制表符分隔符的文件

顺便说一下，我真的建议您将数据结构为两列：

• pk活动书籍
• 出席者

然后有多个列。

您可以直接在数据库中执行此操作：

insert into EventAttendees(EventBookId, Attendee)
    select t.EventBookId, ss.Attendee
    from table t cross apply
         dbo.SplitString(Attendees) as ss(Attendee);

您可以通过谷歌搜索“SQL Server splitstring”来获取此类函数的定义。

Answer 2

测试数据

DECLARE @TABLE TABLE (pkEventBooking INT,  Attendees NVARCHAR(MAX))
INSERT INTO @TABLE VALUES 
(166935 , 'p1193,c21867,c21827,c21963,c18069,c19222'),
(195867 , 'p1193,c21827,c22572,c19222,c22573,c21963,c18069')

查询

;WITH Split_Names (pkEventBooking, Attendees)
AS
(
 SELECT pkEventBooking,
       CONVERT(XML,'<Attendees><Attendee>'  
   + REPLACE(Attendees,',', '</Attendee><Attendee>') + '</Attendee></Attendees>') AS Attendees
 FROM @Table
)

 SELECT pkEventBooking,      
 Attendees.value('/Attendees[1]/Attendee[1]','varchar(100)') AS Attendees1,    
  Attendees.value('/Attendees[1]/Attendee[2]','varchar(100)') AS Attendees2,
   Attendees.value('/Attendees[1]/Attendee[3]','varchar(100)') AS Attendees3,
    Attendees.value('/Attendees[1]/Attendee[4]','varchar(100)') AS Attendees4,
     Attendees.value('/Attendees[1]/Attendee[5]','varchar(100)') AS Attendees5,
      Attendees.value('/Attendees[1]/Attendee[6]','varchar(100)') AS Attendees6,
       Attendees.value('/Attendees[1]/Attendee[7]','varchar(100)') AS Attendees7
 FROM Split_Names

结果

╔════════╦════════════╦════════════╦════════════╦════════════╦════════════╦════════════╦════════════╗
║ Value  ║ Attendees1 ║ Attendees2 ║ Attendees3 ║ Attendees4 ║ Attendees5 ║ Attendees6 ║ Attendees7 ║
╠════════╬════════════╬════════════╬════════════╬════════════╬════════════╬════════════╬════════════╣
║ 166935 ║ p1193      ║ c21867     ║ c21827     ║ c21963     ║ c18069     ║ c19222     ║ NULL       ║
║ 195867 ║ p1193      ║ c21827     ║ c22572     ║ c19222     ║ c22573     ║ c21963     ║ c18069     ║
╚════════╩════════════╩════════════╩════════════╩════════════╩════════════╩════════════╩════════════╝

Answer 3

另一种选择是使用一些 JSON 与

CROSS APPLY

示例

 Select A.[pkEventBooking]
       ,Pos1 = JSON_VALUE(JS,'$[0]')
       ,Pos2 = JSON_VALUE(JS,'$[1]')
       ,Pos3 = JSON_VALUE(JS,'$[2]')
       ,Pos4 = JSON_VALUE(JS,'$[3]')
       ,Pos5 = JSON_VALUE(JS,'$[4]')
       ,Pos6 = JSON_VALUE(JS,'$[5]')
       ,Pos7 = JSON_VALUE(JS,'$[6]')
 From  YourTable A
Cross Apply (values ('["'+replace(string_escape([Attendees],'json'),',','","')+'"]') ) B(JS)

结果

Answer 4

样本数据：

DECLARE @TABLE TABLE (pkEventBooking INT,  Attendees NVARCHAR(MAX))
INSERT INTO @TABLE VALUES 
(166935 , 'p1193,c21867,c21827,c21963,c18069,c19222'),
(195867 , 'p1193,c21827,c22572,c19222,c22573,c21963,c18069')

查询：

SELECT 
   pkEventBooking,
   MAX(CASE WHEN SplitIndex = 1 THEN Value END) AS Column1,
   MAX(CASE WHEN SplitIndex = 2 THEN Value END) AS Column2,
   MAX(CASE WHEN SplitIndex = 3 THEN Value END) AS Column3,
   MAX(CASE WHEN SplitIndex = 4 THEN Value END) AS Column4,
   MAX(CASE WHEN SplitIndex = 5 THEN Value END) AS Column5,
   MAX(CASE WHEN SplitIndex = 6 THEN Value END) AS Column6,
   MAX(CASE WHEN SplitIndex = 7 THEN Value END) AS Column7
FROM (
   SELECT 
      pkEventBooking,
      Value,
      ROW_NUMBER() OVER (PARTITION BY pkEventBooking ORDER BY (SELECT NULL)) AS SplitIndex
   FROM @TABLE
   CROSS APPLY STRING_SPLIT(Attendees, ',')
) AS T
GROUP BY pkEventBooking;