Can we use call function within a subroutine to call another subroutine??

[Anonymous]

//1st subroutine

Sub createanddrop (vTableName)

Store $(vTableName) into [$(vTableName).qvd];

Drop Table [$(vTableName)];

End Sub

// 2nd subroutine

sub pop

FOR  i=1 to NoOfTables();

LET vTable=TableName($(i) -1);

CALL createanddrop($(vTable));

NEXT i

END SUB

//Main call to subroutine

Call pop;

This is my code....I want to use call within a subroutine, so that I can pass parameter to another subroutine.

Thanks

[Anonymous]

Hi Jonathan,

You can look at my code...I am using CALL createanddrop('$(vTable)');......Still not working..

Could You plz check out the flow of execution in this code....Bcoz dis is the first time I am using nested subroutine...I am not sure, whether It is correct.

First call is made to pop subroutine...then again inside pop another  subroutine createanddrop is being called....So plz verify from ur side....It is correct or not...

[Anonymous]

Thank you Luis for ur reply...Everything  working fine, but my 2nd table is not getting drop..

Emp1:

LOAD * INLINE [

    EmpName, Salary

    Ram, 4587

    Shyam, 1500

    John, 5000

    James, 2500

    Jim, 1000

];

Emp2:

LOAD * INLINE [

    EmpId, Office

    1, UAE

    2, Ind

    3, Uk

    5, Us

    4, Pune

];

Emp3:

LOAD * INLINE [

    EId, EOffice

    1, UAE

    2, Ind

    3, Uk

    5, Us

    4, Pune

];

[jonathandienst]

Fabrice is correct. Always work from the bottom when dropping tables in a loop:

sub pop

LET vTableCount = NoOfTables();

FOR  i=0 to vTableCount - 1;

     LET vTable=TableName(0);

     CALL createanddrop('$(vTable)');

NEXT i

END SUB

HTH

Jonathan

[Anonymous]

Please read this..............

I really appreciate efforts of everyone......After dealing with the code I made some changes and now I am using same function in single subroutine....This is my code...........////Single subroutine//////////

sub pop

LET vTableCount = NoOfTables();

FOR  i=0 to vTableCount - 1;

LET vTable=TableName(0);

Store $(vTable) into [$(vTable).qvd];

Drop Table [$(vTable)];

NEXT i

END SUB

call pop

Now I am having two different code with same functionality.....So my question is Which one is Optimized and which one will you prefer???

//1st subroutine

Sub createanddrop (vTableName)

Store $(vTableName) into [$(vTableName).qvd];

Drop Table [$(vTableName)];

End Sub

// 2nd subroutine

sub pop

LET vTableCount = NoOfTables();

FOR  i=0 to vTableCount - 1;

LET vTable=TableName(0);

CALL createanddrop('$(vTable)');

NEXT i

END SUB

call pop

[jonathandienst]

Hi

I doubt you will see any performance difference as the time consuming operations are the same, but always test to confirm the assumption. If you can't see a difference in testing, then I suggest that you use the simpler script.

HTH

Jonathan

[Anonymous]

hi Jonathan,

I just reloaded same table using both functions one by one......I found something unexpected result

In 1st case, i.e, nested call , It is taking 31 sec

In 2nd case, i.e, Single subroutine, It is taking 37 sec..

It is completely opposite of my expectation...I was thinking nested call will take more time but I was wrong...Dont know how