请查找示例数据：
h_company_id company_nm mainphone1 phone_cnt
20816 800 Flowers 5162377000 3
20816 800 Flowers 5162377131 1
20820 1st Source Corp. 5742353000 3
20821 1st United Bancorp 5613633400 2
20824 3D Systems Inc. 8033273900 4
20824 3D Systems Inc. 8033464010 1
11043 3I Group PLC 2079757115 1
11043 3I Group PLC 2079753731 15
期望的输出：
h_company_id company_nm mainphone1 phone_cnt mainphone2 phone_cnt2
20816 800 Flowers 5162377000 3 5162377131 1
20820 1st Source Corp. 5742353000 3 NULL NULL
20821 1st United Bancorp 5613633400 2 NULL NULL
20824 3D Systems Inc. 8033273900 4 8033464010 1
11043 3I Group PLC 2079757115 1 2079753731 15
（在记事本/ excel上面复制）
嗨，大家好，
我想将列mainphone1和phone_cnt的记录转换为新列mainphone2，phone_cnt2，以便列h_company_id中的数据应该是唯一的意味着应该有只有h_company_id的单个条目。
提前致谢！
答案 0 :(得分：3)
从行转换为列称为PIVOT，有几种不同的方法可以在SQL Server中完成。
聚合/ CASE：您可以将聚合函数与CASE表达式一起使用。这可以通过将row_number()窗口函数应用于表中的数据来实现：
select h_company_id, company_nm,
max(case when seq = 1 then mainphone1 end) mainphone1,
max(case when seq = 1 then phone_cnt end) phone_cnt1,
max(case when seq = 2 then mainphone1 end) mainphone2,
max(case when seq = 2 then phone_cnt end) phone_cnt2
from
(
select h_company_id, company_nm, mainphone1, phone_cnt,
row_number() over(partition by h_company_id order by mainphone1) seq
from yourtable
) d
group by h_company_id, company_nm;
见SQL Fiddle with Demo。 CASE表达式检查序列号是否具有值1或2，然后将数据放在列中。
UNPIVOT / PIVOT：由于您希望PIVOT存在于两列中的数据，因此您需要首先取消mainphone1和phone_cnt列的UNPIVOT以使它们进入同一列，然后应用PIVOT功能。
UNPIVOT代码将类似于以下内容：
select h_company_id, company_nm,
col+cast(seq as varchar(10)) col,
value
from
(
select h_company_id, company_nm,
cast(mainphone1 as varchar(15)) mainphone,
cast(phone_cnt as varchar(15)) phone_cnt,
row_number() over(partition by h_company_id order by mainphone1) seq
from yourtable
) d
unpivot
(
value
for col in (mainphone, phone_cnt)
) unpiv;
见Demo。此查询以下列格式获取数据：
| H_COMPANY_ID | COMPANY_NM | COL | VALUE |
---------------------------------------------------------------
| 11043 | 3I Group PLC | mainphone1 | 2079753731 |
| 11043 | 3I Group PLC | phone_cnt1 | 15 |
| 11043 | 3I Group PLC | mainphone2 | 2079757115 |
| 11043 | 3I Group PLC | phone_cnt2 | 1 |
| 20816 | 800 Flowers | mainphone1 | 5162377000 |
然后将PIVOT函数应用于col：
select h_company_id, company_nm,
mainphone1, phone_cnt1, mainphone2, phone_cnt2
from
(
select h_company_id, company_nm,
col+cast(seq as varchar(10)) col,
value
from
(
select h_company_id, company_nm,
cast(mainphone1 as varchar(15)) mainphone,
cast(phone_cnt as varchar(15)) phone_cnt,
row_number() over(partition by h_company_id order by mainphone1) seq
from yourtable
) d
unpivot
(
value
for col in (mainphone, phone_cnt)
) unpiv
) src
pivot
(
max(value)
for col in (mainphone1, phone_cnt1, mainphone2, phone_cnt2)
) piv;
多个联接：您也可以多次加入您的桌面以获得结果。
;with cte as
(
select h_company_id, company_nm, mainphone1, phone_cnt,
row_number() over(partition by h_company_id order by mainphone1) seq
from yourtable
)
select c1.h_company_id,
c1.company_nm,
c1.mainphone1,
c1.phone_cnt phone_cnt1,
c2.mainphone1 mainphone2,
c2.phone_cnt phone_cnt2
from cte c1
left join cte c2
on c1.h_company_id = c2.h_company_id
and c2.seq = 2
where c1.seq = 1;
动态SQL：最后，如果您想要转换未知数量的值，则需要实现动态SQL以获得结果：
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT ',' + QUOTENAME(col+cast(seq as varchar(10)))
from
(
select row_number() over(partition by h_company_id order by mainphone1) seq
from yourtable
) d
cross apply
(
select 'mainphone', 1 union all
select 'phone_cnt', 2
) c (col, so)
group by seq, so, col
order by seq, so
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT h_company_id, company_nm,' + @cols + '
from
(
select h_company_id, company_nm,
col+cast(seq as varchar(10)) col,
value
from
(
select h_company_id, company_nm,
cast(mainphone1 as varchar(15)) mainphone,
cast(phone_cnt as varchar(15)) phone_cnt,
row_number() over(partition by h_company_id order by mainphone1) seq
from yourtable
) d
unpivot
(
value
for col in (mainphone, phone_cnt)
) unpiv
) x
pivot
(
max(value)
for col in (' + @cols + ')
) p '
execute(@query)
见SQL Fiddle with Demo。所有结果都是：
| H_COMPANY_ID | COMPANY_NM | MAINPHONE1 | PHONE_CNT1 | MAINPHONE2 | PHONE_CNT2 |
-----------------------------------------------------------------------------------------
| 20820 | 1st Source Corp. | 5742353000 | 3 | (null) | (null) |
| 20821 | 1st United Bancorp | 5613633400 | 2 | (null) | (null) |
| 20824 | 3D Systems Inc. | 8033273900 | 4 | 8033464010 | 1 |
| 11043 | 3I Group PLC | 2079753731 | 15 | 2079757115 | 1 |
| 20816 | 800 Flowers | 5162377000 | 3 | 5162377131 | 1 |
答案 1 :(得分：0)
以下内容可行（假设您的表名为company）：
SELECT
c1.h_company_id,
c1.company_nm,
c1.mainphone1,
c1.phone_cnt,
c2.mainphone1 AS mainphone2,
c2.phone_cnt AS phone_cnt2
FROM
company AS c1
LEFT JOIN
company AS c2 ON c2.h_company_id = c1.h_company_id
但是，为了尊重良好做法，将数据分成两个表是不是更好？
company表，包含2列：h_company_id（PK）和company_nm phone表格，包含4列：phone_id（PK），h_company_id（FK），mainphone和phone_cnt 它允许您为每个公司提供任意数量的电话号码（包括没有）。
答案 2 :(得分：0)
尝试此查询。这将有助于你
SELECT t.H_COMPANY_ID,t.COMPANY_NM, a.mainphone1,a.PHONE_CNT,b.mainphone1 mainphone2,b.PHONE_CNT PHONE_CNT2 FROM table_name t
INNER JOIN
(
SELECT h_company_id,phone_cnt,mainphone1 FROM table_name
WHERE mainphone1
IN(
SELECT max(mainphone1) mainphone1 FROM table_name GROUP BY h_company_id
)
)a ON t.H_COMPANY_ID = a.h_company_id
INNER JOIN
(
SELECT h_company_id,phone_cnt,mainphone1 FROM table_name
WHERE mainphone1
IN(
SELECT min(mainphone1) mainphone1 from table_name GROUP BY h_company_id
)
)b ON t.H_COMPANY_ID = b.H_COMPANY_ID
GROUP BY t.H_COMPANY_ID,a.mainphone1,t.COMPANY_NM,a.PHONE_CNT,b.mainphone1,b.PHONE_CNT