将年数和月数转换为月数

Question

My table holds a column that holds a value that signifies a period in years and months, in other words:

ID  PERIOD
1   1Y
2   1Y1M
3   11M
4   5Y2M

When doing a select statement how to I convert/calculate this into number of months? It's easy to deal with values that are 'Y' only or 'M' only but not sure how to do for example ID 2 and ID 4 from the example above.

The result from the select statement if I'd select all above would be:

12
13
11
62

Answer 1

You can use

SUBSTRING(period, 1, CHARINDEX('Y', Period) - 1)

To get only the numbers before the Y, and then multiply it by 12.

And for the cases where the Y would not be present could be handled with CASE

So Something like:

   SELECT  CASE WHEN Period LIKE '%Y' THEN
             CAST(SUBSTRING(period, 1, CHARINDEX('Y', Period) - 1)) * 12 
             +
             CAST(REPLACE(SUBSTRING(period, CHARINDEX('Y', Period)),'M',''))
           ELSE
             CAST(REPLACE(period,'M',''))

Answer 2

You can do the following

SELECT *, TRY_CAST(REPLACE(Y, 'Y', '') AS INT) * 12 + 
          TRY_CAST(REPLACE(REPLACE(Period, Y, ''), 'M', '') AS INT)
FROM
(
  VALUES
  (1,   '1Y'),
  (2,   '1Y1M'),
  (3,   '11M'),
  (4,   '5Y2M'),
  (5, 'Whatever')
) T(Id, Period)
CROSS APPLY
(
  VALUES
  (LEFT(Period, CHARINDEX('Y', Period)))
) CI(Y)

Returns:

+----+----------+----+------------------+
| Id |  Period  | Y  | (No column name) |
+----+----------+----+------------------+
|  1 | 1Y       | 1Y |               12 |
|  2 | 1Y1M     | 1Y |               13 |
|  3 | 11M      |    |               11 |
|  4 | 5Y2M     | 5Y |               62 |
|  5 | Whatever |    |                  |
+----+----------+----+------------------+

---

Here is an other way

SELECT Id, 
       Period, 
       (REPLACE(Years, 'Y', '') * 12) + REPLACE(Months, 'M', '') TotalMonths
FROM
(
  VALUES
  (1, '1Y'),
  (2, '1Y1M'),
  (3, '5M'),
  (4, '10Y11M'),
  (5, 'Whatever write Y and M')
) T(Id, Period)
CROSS APPLY
(
  VALUES
  (
    LEFT(Period, CHARINDEX('Y', Period)), REPLACE(REPLACE(Period, 'Y', ''), 'M', '')
  )
) TT(Years, Value)
CROSS APPLY
(
  VALUES
  (
    REPLACE(Period, Years, '')
  )
) TTT(Months)
WHERE TRY_CAST(TT.Value AS INT) IS NOT NULL;

db-fiddle

Answer 3

There aren't so many possibilities. Build a reference table:

select identity(int) as period_id, v.*
into periods p
from (values ('1M', 1),
             ('2M', 2),
             . . . 
             ('1Y', 12),
             ('1Y1M', 13),
             . . .
    ) v(period, months);

This can easily be constructed using a spreadsheet. Or even a recursive CTE.

I am suggesting this for a serious reason: you should not be doing calculations on string representations like this. These values should be treated as foreign key references to a table. And, in fact, they should be using the primary key (the identity column) rather than the string.

You will probably find malformed strings as you go about fixing this. That is a good thing, from a data quality perspective.

Answer 4

Another alternative

select id, (parsename (clean,2) * 12) + (parsename(clean,1)) as months
from t
cross apply (select replace(replace(case when charindex('M', period)=0 then period + '0M' 
                                         when charindex('Y', period)=0 then '0Y' + period
                                         else period end,'M',''),'Y','.') as clean) t2

Outputs

+----+--------+
| id | months |
+----+--------+
|  1 |     12 |
|  2 |     13 |
|  3 |     11 |
|  4 |     62 |
+----+--------+

Answer 5

I think something like this would work

Data

drop table if exists #tTEST;
go
select * INTO #tTEST from (values 
(1, '1Y'),
(2, '1Y1M'),
(3, '11M'),
(4, '5Y2M')) V(ID, [Period]);

Query

select
  t.*, 
  isnull(substring(t.[Period], 1, nullif(y_ndx, 0)-1)*12, 0) +
  isnull(substring(t.[Period], y_ndx+1, nullif(m_ndx, 0)-isnull(y_ndx, 0)-1),0) answer
from #tTEST t
     cross apply (select len(t.[Period]) p_len,
                         CHARINDEX('Y', t.[Period]) y_ndx,
                         CHARINDEX('M', t.[Period]) m_ndx) ndx;

Results

ID  Period  answer
1   1Y      12
2   1Y1M    13
3   11M     11
4   5Y2M    62

Answer 6

Try this below Scalar-Valued Function to get period value

CREATE FUNCTION [dbo].[fn_sum](  
@delimited NVARCHAR(MAX),  
@delimiter NVARCHAR(100)  
) RETURNS INT 
AS  
BEGIN    
DECLARE @value INT
set @delimited=replace(replace(@delimited,'M','M,'),'Y','Y,')
DECLARE @xml XML  
SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>'  

SELECT @value=sum(cast(replace(replace(r.value('.','Nvarchar(MAX)'),'M',''),'Y','') AS INT) 
* CASE WHEN right(r.value('.','Nvarchar(MAX)'),1)='M' THEN 1 ELSE 12 END) 
FROM @xml.nodes('/t') AS records(r)  

RETURN  @value
END 

Check below sample output

DECLARE @tblPeriod AS TABLE(id VARCHAR(100),period VARCHAR(100))
INSERT INTO @tblPeriod(id,period) VALUES(1,'1Y'),(2,'1Y1M'),(3,'11M'),(4,'5Y2M') 

SELECT id,period,dbo.[fn_sum](period,',') AS period_value  FROM @tblPeriod a