Why sql subtraction two colon












-2















code:



SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 



                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,





                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2,





                               examplecolon1 -(2 * examplecolon2) from


.................



example:



subtraction error :



examplecolon1 -(2 * examplecolon2)



why solved the problem?






sql







share|improve this question

























  • please provide sample data and expected output

    – fa06
    Nov 20 '18 at 12:37






  • 1





    You cannot re-use column aliases in the same SELECT where they are defined. You need to use a subquery, CTE, or similar mechanism to define the logic.

    – Gordon Linoff
    Nov 20 '18 at 12:37
















-2















code:



SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 



                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,





                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2,





                               examplecolon1 -(2 * examplecolon2) from


.................



example:



subtraction error :



examplecolon1 -(2 * examplecolon2)



why solved the problem?






sql







share|improve this question

























  • please provide sample data and expected output

    – fa06
    Nov 20 '18 at 12:37






  • 1





    You cannot re-use column aliases in the same SELECT where they are defined. You need to use a subquery, CTE, or similar mechanism to define the logic.

    – Gordon Linoff
    Nov 20 '18 at 12:37














-2












-2








-2








code:



SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 



                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,





                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2,





                               examplecolon1 -(2 * examplecolon2) from


.................



example:



subtraction error :



examplecolon1 -(2 * examplecolon2)



why solved the problem?






sql







share|improve this question
















code:



SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 



                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,





                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2,





                               examplecolon1 -(2 * examplecolon2) from


.................



example:



subtraction error :



examplecolon1 -(2 * examplecolon2)



why solved the problem?






sql




sql






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 '18 at 12:38









fa06

15k2917




15k2917










asked Nov 20 '18 at 12:36









michaelmichael

13




13













  • please provide sample data and expected output

    – fa06
    Nov 20 '18 at 12:37






  • 1





    You cannot re-use column aliases in the same SELECT where they are defined. You need to use a subquery, CTE, or similar mechanism to define the logic.

    – Gordon Linoff
    Nov 20 '18 at 12:37



















  • please provide sample data and expected output

    – fa06
    Nov 20 '18 at 12:37






  • 1





    You cannot re-use column aliases in the same SELECT where they are defined. You need to use a subquery, CTE, or similar mechanism to define the logic.

    – Gordon Linoff
    Nov 20 '18 at 12:37

















please provide sample data and expected output

– fa06
Nov 20 '18 at 12:37





please provide sample data and expected output

– fa06
Nov 20 '18 at 12:37




1




1





You cannot re-use column aliases in the same SELECT where they are defined. You need to use a subquery, CTE, or similar mechanism to define the logic.

– Gordon Linoff
Nov 20 '18 at 12:37





You cannot re-use column aliases in the same SELECT where they are defined. You need to use a subquery, CTE, or similar mechanism to define the logic.

– Gordon Linoff
Nov 20 '18 at 12:37












1 Answer
1






active

oldest

votes


















0














You need to use a subquery or a CTE (if your backend supports) to do that. ie:



select newcolon1, newcolon2,

   examplecolon1, examplecolon2,

   examplecolon1 -(2 * examplecolon2) as myNewColumn

from

(SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 

                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,

                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2

from myTable1 ms

[leftrightinner] join  myTable2 mss on ... ) sq;





share|improve this answer
























  • AKA derived table.

    – jarlh
    Nov 20 '18 at 13:20











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














You need to use a subquery or a CTE (if your backend supports) to do that. ie:



select newcolon1, newcolon2,

   examplecolon1, examplecolon2,

   examplecolon1 -(2 * examplecolon2) as myNewColumn

from

(SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 

                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,

                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2

from myTable1 ms

[leftrightinner] join  myTable2 mss on ... ) sq;





share|improve this answer
























  • AKA derived table.

    – jarlh
    Nov 20 '18 at 13:20
















0














You need to use a subquery or a CTE (if your backend supports) to do that. ie:



select newcolon1, newcolon2,

   examplecolon1, examplecolon2,

   examplecolon1 -(2 * examplecolon2) as myNewColumn

from

(SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 

                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,

                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2

from myTable1 ms

[leftrightinner] join  myTable2 mss on ... ) sq;





share|improve this answer
























  • AKA derived table.

    – jarlh
    Nov 20 '18 at 13:20














0












0








0







You need to use a subquery or a CTE (if your backend supports) to do that. ie:



select newcolon1, newcolon2,

   examplecolon1, examplecolon2,

   examplecolon1 -(2 * examplecolon2) as myNewColumn

from

(SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 

                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,

                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2

from myTable1 ms

[leftrightinner] join  myTable2 mss on ... ) sq;





share|improve this answer













You need to use a subquery or a CTE (if your backend supports) to do that. ie:



select newcolon1, newcolon2,

   examplecolon1, examplecolon2,

   examplecolon1 -(2 * examplecolon2) as myNewColumn

from

(SELECT DISTINCT mss.colon1 AS newcolon1, ms.colon2 AS newcolon2, 

                               (SELECT        COUNT(mmm.numbertop) AS Expr1

                               FROM            listtable AS mmm

                               WHERE        (ms.number = mmm.number) ) AS examplecolon1,

                               (SELECT        COUNT(stls.numbertop) AS Expr1

                               FROM           listtable2 AS stls

                               WHERE        ms.number = stls.number ) AS examplecolon2

from myTable1 ms

[leftrightinner] join  myTable2 mss on ... ) sq;






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 20 '18 at 12:42









Cetin BasozCetin Basoz

11k11527




11k11527













  • AKA derived table.

    – jarlh
    Nov 20 '18 at 13:20



















  • AKA derived table.

    – jarlh
    Nov 20 '18 at 13:20

















AKA derived table.

– jarlh
Nov 20 '18 at 13:20





AKA derived table.

– jarlh
Nov 20 '18 at 13:20




















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