从存储为字符串的数字中删除所有尾随小数点

Question

I have a couple of strings ( nvarchar data type), one is a whole number and one has decimal points trailing. My goal is to remove decimals and have all values as a whole number.

I tried the code below but it gives me an error for the value with no decimals. Is there a way to accomplish this without a case expression. I'll be using this new column in a join.

SELECT [SOW]
  --,LEFT([SOW], CHARINDEX('.', [SOW])-1) as 'TestColumn'
FROM [dbo].[t_Schedule_kdm]
WHERE sow  in ('15229.11','11092')

Output:
11092
15229.11

My desired Output:
11092
15229

Answer 1

Just append a dot character so that you'll always find an index:

LEFT(SOW, CHARINDEX('.', SOW + '.') - 1)

It's not clear whether you need to cast the result of that expression to an integer value.

Answer 2

Convert first to the most precision number you could ever have eg decimal(9,2) then convert to an int . You can't convert directly from a decimal string to an int.

SELECT [Value]
  , CONVERT(int,CONVERT(decimal(9,2),[Value]))
FROM (
    VALUES ('15229.11'),('11092')
) x ([Value]);