如何在sql中使用GROUPBY子句替换字符串？

在下表中，我需要根据组“Yearmonth”将列“Formula”中的字符串替换为匹配的输入“VALUE”列

IDNUM  formula                      INPUTNAME      VALUE    YEARMONTH
---------------------------------------------------------------------
1      imports(398)+imports(399)    imports(398)    17.000  2003:1
2      imports(398)+imports(399)    imports(398)    56.000  2003:2
3      imports(398)+imports(399)    imports(399)    15.000  2003:1
4      imports(398)+imports(399)    imports(399)    126.000 2003:2

例如：从上表中，我需要输出为

Idnum Formula        Yearmonth
 1.    17.00 +15.00  2003:1
 2.    56.00 +126.00 2003:2

我尝试了下面的查询，但未能实现。如何做到这一点

SELECT
    REPLACE(FORMULA, INPUTName, AttributeValue) AS realvalues, 
    yearmonth 
FROM table1 
GROUP BY yearmonth

---

将

用于XML路径

用于连接：

SELECT
    IDNUM = MIN(IDNUM),
    FORMULA = 
        (SELECT STUFF(
            (SELECT ' +' + CONVERT(VARCHAR(10), Value)
            FROM Table1
            WHERE YEARMONTH = t1.YEARMONTH
            FOR XML PATH(''))
        ,1, 2, '')),
    YEARMONTH
FROM Table1 t1
GROUP BY YEARMONTH

---

不幸的是，这种类型的替换需要递归：

with t as (
      select t.*,
             row_number() over (partition by yearmonth order by idnum) as seqnum,
             count(*) over (partition by yearmonth) as cnt
      from table t
     ),
     cte as (
      select t.seqnum, t.yearmonth, t.cnt,
             replace(formula, inputname, value) as formula
      from t
      where seqnum = 1
      union all
      select cte.seqnum, cte.yearmonth, cte.cnt,
             replace(formula, t.inputername, t.value)
      from cte join
           t
           on cte.yearmonth = t.yearmonth and t.seqnum = cte.seqnum + 1
    )
select row_number() over (order by (select null)) as id,
       formula
from cte
where seqnum = cnt;

如果您需要列idnum，请尝试以下操作

declare @t table (idnum int identity(1,1),formula varchar(100),Yearmonth date)
insert into @t
    SELECT t1.yearmonth,
   formula=REPLACE( (SELECT value AS [data()]
       FROM table t2
      WHERE t2.YEARMONTH= t1.YEARMONTH
      ORDER BY value
        FOR XML PATH('')
        ), ' ', '+')
  FROM table t1
  GROUP BY YEARMONTH ;

---

这个很好用。。。非常感谢……我有一个问题，这适用于表中的任何类型的字符串还是仅适用于这种类型的字符串？？似乎这个查询适用于公式中有“+”的字符串。。。。有没有办法只替换字符串名而不考虑数学符号？谢谢你的建议。。。但是我相信querey需要一些修改。。我有点难以获得输出…嗨，戈登。。。。我尝试了代码执行，但只得到了null值…你能帮我实现这一点吗one@JamesRodix . . . 你能安装SQL小提琴吗？嗨。。。谢谢你的建议。。。Felix Pamittan的建议都是一样的。。。请听我对他的建议的评论。。那可能吗？？

declare @t table (idnum int identity(1,1),formula varchar(100),Yearmonth date)
insert into @t
    SELECT t1.yearmonth,
   formula=REPLACE( (SELECT value AS [data()]
       FROM table t2
      WHERE t2.YEARMONTH= t1.YEARMONTH
      ORDER BY value
        FOR XML PATH('')
        ), ' ', '+')
  FROM table t1
  GROUP BY YEARMONTH ;