How can I make this SQL more Efficient











up vote
0
down vote

favorite












I have written this piece of SQL, I know there are ways to make it run faster, with the right practices.



SELECT DISTINCT
ACCOUNTNUM,
FIRSTNAME AS NAME,
LASTNAME AS SURNAME,
(PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
EMAIL,
(SELECT TOP 1 CREATEDDATE
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM
ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
(SELECT COUNT(TRANSACTIONID)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
(SELECT SUM(PAYMENTAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
(SELECT SUM(DISCAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
FROM
CUSTOMER
LEFT OUTER JOIN
RBOTRANSACTIONTABLE ON CUSTACCOUNT = ACCOUNTNUM


I know if I am using some sort of Joins, I don't have necessarily have to keep saying FROM RBOTRANSACTIONTABLE every time (the code after Email column). The code above works great for my requirement, but I know there are missing gaps in my knowledge, I just don't know what.



I am looking for in-depth answers as to why the above solution is not recommended, and why your solution is.










share|improve this question




















  • 1




    It's helpful to tag database questions with both the appropriate software (MySQL, Oracle, DB2, ...) and version, e.g. sql-server-2014. Differences in syntax and features often affect the answers. Note that tsql narrows the choices, but does not specify the database. Assuming that you are using SQL Server: See paste the plan for a way to include an execution plan in your question.
    – HABO
    Nov 22 at 1:12






  • 1




    You used DISTINCT to cover up your logic error. There is NO reason to join Customer and Transaction when you do the equivalent as subqueries. And you used 4 separate subqueries to the same table for your calculations, each of which is a simple aggregate. That should have been a clue that you could calculate all 4 values with a single pass through the transaction table - as Tim demonstrates. lastly - [top 1 x order by x desc] is the same as max(). The latter is far more readable, more understandable, and less prone to error.
    – SMor
    Nov 22 at 14:18















up vote
0
down vote

favorite












I have written this piece of SQL, I know there are ways to make it run faster, with the right practices.



SELECT DISTINCT
ACCOUNTNUM,
FIRSTNAME AS NAME,
LASTNAME AS SURNAME,
(PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
EMAIL,
(SELECT TOP 1 CREATEDDATE
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM
ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
(SELECT COUNT(TRANSACTIONID)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
(SELECT SUM(PAYMENTAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
(SELECT SUM(DISCAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
FROM
CUSTOMER
LEFT OUTER JOIN
RBOTRANSACTIONTABLE ON CUSTACCOUNT = ACCOUNTNUM


I know if I am using some sort of Joins, I don't have necessarily have to keep saying FROM RBOTRANSACTIONTABLE every time (the code after Email column). The code above works great for my requirement, but I know there are missing gaps in my knowledge, I just don't know what.



I am looking for in-depth answers as to why the above solution is not recommended, and why your solution is.










share|improve this question




















  • 1




    It's helpful to tag database questions with both the appropriate software (MySQL, Oracle, DB2, ...) and version, e.g. sql-server-2014. Differences in syntax and features often affect the answers. Note that tsql narrows the choices, but does not specify the database. Assuming that you are using SQL Server: See paste the plan for a way to include an execution plan in your question.
    – HABO
    Nov 22 at 1:12






  • 1




    You used DISTINCT to cover up your logic error. There is NO reason to join Customer and Transaction when you do the equivalent as subqueries. And you used 4 separate subqueries to the same table for your calculations, each of which is a simple aggregate. That should have been a clue that you could calculate all 4 values with a single pass through the transaction table - as Tim demonstrates. lastly - [top 1 x order by x desc] is the same as max(). The latter is far more readable, more understandable, and less prone to error.
    – SMor
    Nov 22 at 14:18













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have written this piece of SQL, I know there are ways to make it run faster, with the right practices.



SELECT DISTINCT
ACCOUNTNUM,
FIRSTNAME AS NAME,
LASTNAME AS SURNAME,
(PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
EMAIL,
(SELECT TOP 1 CREATEDDATE
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM
ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
(SELECT COUNT(TRANSACTIONID)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
(SELECT SUM(PAYMENTAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
(SELECT SUM(DISCAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
FROM
CUSTOMER
LEFT OUTER JOIN
RBOTRANSACTIONTABLE ON CUSTACCOUNT = ACCOUNTNUM


I know if I am using some sort of Joins, I don't have necessarily have to keep saying FROM RBOTRANSACTIONTABLE every time (the code after Email column). The code above works great for my requirement, but I know there are missing gaps in my knowledge, I just don't know what.



I am looking for in-depth answers as to why the above solution is not recommended, and why your solution is.










share|improve this question















I have written this piece of SQL, I know there are ways to make it run faster, with the right practices.



SELECT DISTINCT
ACCOUNTNUM,
FIRSTNAME AS NAME,
LASTNAME AS SURNAME,
(PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
EMAIL,
(SELECT TOP 1 CREATEDDATE
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM
ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
(SELECT COUNT(TRANSACTIONID)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
(SELECT SUM(PAYMENTAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
(SELECT SUM(DISCAMOUNT)
FROM RBOTRANSACTIONTABLE
WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
FROM
CUSTOMER
LEFT OUTER JOIN
RBOTRANSACTIONTABLE ON CUSTACCOUNT = ACCOUNTNUM


I know if I am using some sort of Joins, I don't have necessarily have to keep saying FROM RBOTRANSACTIONTABLE every time (the code after Email column). The code above works great for my requirement, but I know there are missing gaps in my knowledge, I just don't know what.



I am looking for in-depth answers as to why the above solution is not recommended, and why your solution is.







sql-server tsql relational-database






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 at 5:14









marc_s

566k12610941246




566k12610941246










asked Nov 22 at 0:08









Kush

97111




97111








  • 1




    It's helpful to tag database questions with both the appropriate software (MySQL, Oracle, DB2, ...) and version, e.g. sql-server-2014. Differences in syntax and features often affect the answers. Note that tsql narrows the choices, but does not specify the database. Assuming that you are using SQL Server: See paste the plan for a way to include an execution plan in your question.
    – HABO
    Nov 22 at 1:12






  • 1




    You used DISTINCT to cover up your logic error. There is NO reason to join Customer and Transaction when you do the equivalent as subqueries. And you used 4 separate subqueries to the same table for your calculations, each of which is a simple aggregate. That should have been a clue that you could calculate all 4 values with a single pass through the transaction table - as Tim demonstrates. lastly - [top 1 x order by x desc] is the same as max(). The latter is far more readable, more understandable, and less prone to error.
    – SMor
    Nov 22 at 14:18














  • 1




    It's helpful to tag database questions with both the appropriate software (MySQL, Oracle, DB2, ...) and version, e.g. sql-server-2014. Differences in syntax and features often affect the answers. Note that tsql narrows the choices, but does not specify the database. Assuming that you are using SQL Server: See paste the plan for a way to include an execution plan in your question.
    – HABO
    Nov 22 at 1:12






  • 1




    You used DISTINCT to cover up your logic error. There is NO reason to join Customer and Transaction when you do the equivalent as subqueries. And you used 4 separate subqueries to the same table for your calculations, each of which is a simple aggregate. That should have been a clue that you could calculate all 4 values with a single pass through the transaction table - as Tim demonstrates. lastly - [top 1 x order by x desc] is the same as max(). The latter is far more readable, more understandable, and less prone to error.
    – SMor
    Nov 22 at 14:18








1




1




It's helpful to tag database questions with both the appropriate software (MySQL, Oracle, DB2, ...) and version, e.g. sql-server-2014. Differences in syntax and features often affect the answers. Note that tsql narrows the choices, but does not specify the database. Assuming that you are using SQL Server: See paste the plan for a way to include an execution plan in your question.
– HABO
Nov 22 at 1:12




It's helpful to tag database questions with both the appropriate software (MySQL, Oracle, DB2, ...) and version, e.g. sql-server-2014. Differences in syntax and features often affect the answers. Note that tsql narrows the choices, but does not specify the database. Assuming that you are using SQL Server: See paste the plan for a way to include an execution plan in your question.
– HABO
Nov 22 at 1:12




1




1




You used DISTINCT to cover up your logic error. There is NO reason to join Customer and Transaction when you do the equivalent as subqueries. And you used 4 separate subqueries to the same table for your calculations, each of which is a simple aggregate. That should have been a clue that you could calculate all 4 values with a single pass through the transaction table - as Tim demonstrates. lastly - [top 1 x order by x desc] is the same as max(). The latter is far more readable, more understandable, and less prone to error.
– SMor
Nov 22 at 14:18




You used DISTINCT to cover up your logic error. There is NO reason to join Customer and Transaction when you do the equivalent as subqueries. And you used 4 separate subqueries to the same table for your calculations, each of which is a simple aggregate. That should have been a clue that you could calculate all 4 values with a single pass through the transaction table - as Tim demonstrates. lastly - [top 1 x order by x desc] is the same as max(). The latter is far more readable, more understandable, and less prone to error.
– SMor
Nov 22 at 14:18












2 Answers
2






active

oldest

votes

















up vote
3
down vote













You may rewrite your query as a join to a subquery which finds aggregates:



SELECT
c.ACCOUNTNUM,
c.FIRSTNAME AS NAME,
c.LASTNAME AS SURNAME,
(c.PHONE + ' ' + c.CELLULARPHONE) AS PHONENUM,
c.EMAIL,
a.LASTVISIT,
COALESCE(a.TOTALVISITS, 0) AS TOTALVISITS,
COALESCE(a.TOTALSALES, 0) AS TOTALSALES,
COALESCE(a.DISCOUNT, 0) AS DISCOUNT
FROM CUSTOMER c
LEFT JOIN
(
SELECT
CUSTACCOUNT,
MAX(CREATEDDATE) AS LASTVISIT,
COUNT(TRANSACTIONID) AS TOTALVISITS,
SUM(PAYMENTAMOUNT) AS TOTALSALES,
SUM(DISCAMOUNT) AS DISCOUNT
FROM RBOTRANSACTIONTABLE
GROUP BY CUSTACCOUNT
) a
ON c.ACCOUNTNUM = a.CUSTACCOUNT;


You should always use proper aliases when referring to columns in the SELECT clause.






share|improve this answer




























    up vote
    0
    down vote













    Tim's is definitely a good way to rewrite the query. But you can also make your version more efficient by getting rid of the count(distinct) and outer join:



    SELECT ACCOUNTNUM,
    FIRSTNAME AS NAME,
    LASTNAME AS SURNAME,
    (PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
    EMAIL,
    (SELECT TOP 1 CREATEDDATE FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
    (SELECT COUNT(TRANSACTIONID) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
    (SELECT SUM(PAYMENTAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
    (SELECT SUM(DISCAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
    FROM CUSTOMER c;


    With the right indexes, this could even have better performance than the group by/join version.






    share|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote













      You may rewrite your query as a join to a subquery which finds aggregates:



      SELECT
      c.ACCOUNTNUM,
      c.FIRSTNAME AS NAME,
      c.LASTNAME AS SURNAME,
      (c.PHONE + ' ' + c.CELLULARPHONE) AS PHONENUM,
      c.EMAIL,
      a.LASTVISIT,
      COALESCE(a.TOTALVISITS, 0) AS TOTALVISITS,
      COALESCE(a.TOTALSALES, 0) AS TOTALSALES,
      COALESCE(a.DISCOUNT, 0) AS DISCOUNT
      FROM CUSTOMER c
      LEFT JOIN
      (
      SELECT
      CUSTACCOUNT,
      MAX(CREATEDDATE) AS LASTVISIT,
      COUNT(TRANSACTIONID) AS TOTALVISITS,
      SUM(PAYMENTAMOUNT) AS TOTALSALES,
      SUM(DISCAMOUNT) AS DISCOUNT
      FROM RBOTRANSACTIONTABLE
      GROUP BY CUSTACCOUNT
      ) a
      ON c.ACCOUNTNUM = a.CUSTACCOUNT;


      You should always use proper aliases when referring to columns in the SELECT clause.






      share|improve this answer

























        up vote
        3
        down vote













        You may rewrite your query as a join to a subquery which finds aggregates:



        SELECT
        c.ACCOUNTNUM,
        c.FIRSTNAME AS NAME,
        c.LASTNAME AS SURNAME,
        (c.PHONE + ' ' + c.CELLULARPHONE) AS PHONENUM,
        c.EMAIL,
        a.LASTVISIT,
        COALESCE(a.TOTALVISITS, 0) AS TOTALVISITS,
        COALESCE(a.TOTALSALES, 0) AS TOTALSALES,
        COALESCE(a.DISCOUNT, 0) AS DISCOUNT
        FROM CUSTOMER c
        LEFT JOIN
        (
        SELECT
        CUSTACCOUNT,
        MAX(CREATEDDATE) AS LASTVISIT,
        COUNT(TRANSACTIONID) AS TOTALVISITS,
        SUM(PAYMENTAMOUNT) AS TOTALSALES,
        SUM(DISCAMOUNT) AS DISCOUNT
        FROM RBOTRANSACTIONTABLE
        GROUP BY CUSTACCOUNT
        ) a
        ON c.ACCOUNTNUM = a.CUSTACCOUNT;


        You should always use proper aliases when referring to columns in the SELECT clause.






        share|improve this answer























          up vote
          3
          down vote










          up vote
          3
          down vote









          You may rewrite your query as a join to a subquery which finds aggregates:



          SELECT
          c.ACCOUNTNUM,
          c.FIRSTNAME AS NAME,
          c.LASTNAME AS SURNAME,
          (c.PHONE + ' ' + c.CELLULARPHONE) AS PHONENUM,
          c.EMAIL,
          a.LASTVISIT,
          COALESCE(a.TOTALVISITS, 0) AS TOTALVISITS,
          COALESCE(a.TOTALSALES, 0) AS TOTALSALES,
          COALESCE(a.DISCOUNT, 0) AS DISCOUNT
          FROM CUSTOMER c
          LEFT JOIN
          (
          SELECT
          CUSTACCOUNT,
          MAX(CREATEDDATE) AS LASTVISIT,
          COUNT(TRANSACTIONID) AS TOTALVISITS,
          SUM(PAYMENTAMOUNT) AS TOTALSALES,
          SUM(DISCAMOUNT) AS DISCOUNT
          FROM RBOTRANSACTIONTABLE
          GROUP BY CUSTACCOUNT
          ) a
          ON c.ACCOUNTNUM = a.CUSTACCOUNT;


          You should always use proper aliases when referring to columns in the SELECT clause.






          share|improve this answer












          You may rewrite your query as a join to a subquery which finds aggregates:



          SELECT
          c.ACCOUNTNUM,
          c.FIRSTNAME AS NAME,
          c.LASTNAME AS SURNAME,
          (c.PHONE + ' ' + c.CELLULARPHONE) AS PHONENUM,
          c.EMAIL,
          a.LASTVISIT,
          COALESCE(a.TOTALVISITS, 0) AS TOTALVISITS,
          COALESCE(a.TOTALSALES, 0) AS TOTALSALES,
          COALESCE(a.DISCOUNT, 0) AS DISCOUNT
          FROM CUSTOMER c
          LEFT JOIN
          (
          SELECT
          CUSTACCOUNT,
          MAX(CREATEDDATE) AS LASTVISIT,
          COUNT(TRANSACTIONID) AS TOTALVISITS,
          SUM(PAYMENTAMOUNT) AS TOTALSALES,
          SUM(DISCAMOUNT) AS DISCOUNT
          FROM RBOTRANSACTIONTABLE
          GROUP BY CUSTACCOUNT
          ) a
          ON c.ACCOUNTNUM = a.CUSTACCOUNT;


          You should always use proper aliases when referring to columns in the SELECT clause.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 22 at 0:20









          Tim Biegeleisen

          210k1381129




          210k1381129
























              up vote
              0
              down vote













              Tim's is definitely a good way to rewrite the query. But you can also make your version more efficient by getting rid of the count(distinct) and outer join:



              SELECT ACCOUNTNUM,
              FIRSTNAME AS NAME,
              LASTNAME AS SURNAME,
              (PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
              EMAIL,
              (SELECT TOP 1 CREATEDDATE FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
              (SELECT COUNT(TRANSACTIONID) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
              (SELECT SUM(PAYMENTAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
              (SELECT SUM(DISCAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
              FROM CUSTOMER c;


              With the right indexes, this could even have better performance than the group by/join version.






              share|improve this answer

























                up vote
                0
                down vote













                Tim's is definitely a good way to rewrite the query. But you can also make your version more efficient by getting rid of the count(distinct) and outer join:



                SELECT ACCOUNTNUM,
                FIRSTNAME AS NAME,
                LASTNAME AS SURNAME,
                (PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
                EMAIL,
                (SELECT TOP 1 CREATEDDATE FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
                (SELECT COUNT(TRANSACTIONID) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
                (SELECT SUM(PAYMENTAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
                (SELECT SUM(DISCAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
                FROM CUSTOMER c;


                With the right indexes, this could even have better performance than the group by/join version.






                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Tim's is definitely a good way to rewrite the query. But you can also make your version more efficient by getting rid of the count(distinct) and outer join:



                  SELECT ACCOUNTNUM,
                  FIRSTNAME AS NAME,
                  LASTNAME AS SURNAME,
                  (PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
                  EMAIL,
                  (SELECT TOP 1 CREATEDDATE FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
                  (SELECT COUNT(TRANSACTIONID) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
                  (SELECT SUM(PAYMENTAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
                  (SELECT SUM(DISCAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
                  FROM CUSTOMER c;


                  With the right indexes, this could even have better performance than the group by/join version.






                  share|improve this answer












                  Tim's is definitely a good way to rewrite the query. But you can also make your version more efficient by getting rid of the count(distinct) and outer join:



                  SELECT ACCOUNTNUM,
                  FIRSTNAME AS NAME,
                  LASTNAME AS SURNAME,
                  (PHONE + ' ' + CELLULARPHONE) AS PHONENUM,
                  EMAIL,
                  (SELECT TOP 1 CREATEDDATE FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM ORDER BY CREATEDDATE DESC) AS LASTVISIT, -- LAST VISIT,
                  (SELECT COUNT(TRANSACTIONID) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALVISITS, -- TOTAL VISITS,
                  (SELECT SUM(PAYMENTAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS TOTALSALES, -- TOTAL SALES,
                  (SELECT SUM(DISCAMOUNT) FROM RBOTRANSACTIONTABLE WHERE CUSTACCOUNT = ACCOUNTNUM) AS DISCOUNT
                  FROM CUSTOMER c;


                  With the right indexes, this could even have better performance than the group by/join version.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 22 at 0:30









                  Gordon Linoff

                  747k34285390




                  747k34285390






























                       

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