Question

我们的数据库中有一列包含+个字符串的字符串。无论如何选择每一行的总和？也就是说，我们有5212667+5212662;我想选择10425329。

我试过了CONVERT(INT, ColumnAddList)，但这并没有奏效。

Answer 1

如果您使用SQL Server，则可以拆分并计算总和：

CREATE TABLE tab(ID INT IDENTITY(1,1), col VARCHAR(1000));

INSERT INTO tab(col) VALUES('5212667+5212662'),('1+2+3'),('2'), (NULL), ('1+-1');

SELECT *
FROM tab
CROSS APPLY (
    SELECT [result] = SUM( Split.a.value('.', 'BIGINT'))
    FROM (SELECT [X] = CAST ('<M>'+REPLACE(col, '+', '</M><M>') + '</M>' AS XML)) AS A 
    CROSS APPLY X.nodes ('/M') AS Split(a)
) AS s;

的 LiveDemo

输出：

╔════╦═════════════════╦══════════╗ ║ ID ║ col ║ result ║ ╠════╬═════════════════╬══════════╣ ║ 1 ║ 5212667+5212662 ║ 10425329 ║ ║ 2 ║ 1+2+3 ║ 6 ║ ║ 3 ║ 2 ║ 2 ║ ║ 4 ║ NULL ║ NULL ║ ║ 5 ║ 1+-1 ║ 0 ║ ╚════╩═════════════════╩══════════╝

正确的方法是规范化表格架构。

Answer 2

CREATE FUNCTION dbo.fnSplit(
@sInputList VARCHAR(8000) -- List of delimited items
 , @sDelimiter VARCHAR(8000) = ',' -- delimiter that separates items
 ) RETURNS @List TABLE (item VARCHAR(8000))

BEGIN
 DECLARE @sItem VARCHAR(8000)
 WHILE CHARINDEX(@sDelimiter,@sInputList,0) <> 0
  BEGIN
    SELECT
     @sItem=RTRIM(LTRIM(SUBSTRING(@sInputList,1,CHARINDEX
      (@sDelimiter,@sInputList,0)-1))),  @sInputList=RTRIM(LTRIM(SUBSTRING(@sInputList,CHARINDEX
      (@sDelimiter,@sInputList,0)+LEN(@sDelimiter),LEN(@sInputList))))

  IF LEN(@sItem) > 0
    INSERT INTO @List SELECT @sItem
  END

 IF LEN(@sInputList) > 0
     INSERT INTO @List SELECT @sInputList -- Put the last item in
     RETURN
  END
GO

declare @table table (Val varchar(2000))
insert into @table values('1+1+1+')
insert into @table values('2+2+2+')
insert into @table values('3+3+3+')
insert into @table values('4+4+4+')
select Val, (select sum(convert(decimal(10,2),item)) from dbo.fnSplit(val,'+') ) as summ from @table

希望这有帮助!!!

快乐编码!!!

SQL：将+分隔列表转换为整数

2 个答案: