Re: SQL Existence Check

["Nathan M. Andelin" <nandelin@xxxxxxxxxxxxxxxxxxx>]

> From: <David.X.Kahn@gsk.com>
> SETLL will always set the current record pointer. CHAIN will
> set it only when a record is found. If Nathan were to re-run his
> test where the record is never found I think he would find that
> the CHAINs are quicker than the SETLLs.

Apparently correct.  In this 3rd version of the program, none of the 10,000
keys tested exist in the file.  In this case the SETLL loop takes 37% more
time than the CHAIN loop.

     Ftestiol   IF   E           K DISK

      /COPY RDWEB/QRPGLESRC,RDSTRAPI#1

     D myKey           S             12A   Dim(10000)
     D i               S             10I 0
     D Now             S               Z
     D Then            S               Z
     D Secs            S             10I 0
     D myNumber        S             10I 0
     D myString        S            128A   Varying

      *-----------------------------------------------------------------

     C                   For       i = 1 to 10000
     C                   Eval      myKey(i) = strDistinct
     C                   EndFor

      *-----------------------------------------------------------------

     C                   Time                    Then
     C                   For       i = 1 to 10000
     C     myKey(i)      Setll     testiol                                70
     C                   EndFor

      *-----------------------------------------------------------------

     C                   Time                    Now
     C     Now           Subdur    Then          Secs:*MS
     C     Secs          Dsply

      *-----------------------------------------------------------------

     C                   Time                    Then
     C                   For       i = 1 to 10000
     C     myKey(i)      Chain     testior                            70
     C                   EndFor

      *-----------------------------------------------------------------

     C                   Time                    Now
     C     Now           Subdur    Then          Secs:*MS
     C     Secs          Dsply

      *-----------------------------------------------------------------

     C                   Eval      myNumber = tifld1
     C                   Eval      myNumber = tifld2
     C                   Eval      myNumber = tifld3
     C                   Eval      myNumber = tifld4
     C                   Eval      myNumber = tifld5

     C                   Eval      myString = tifld6
     C                   Eval      myString = tifld7
     C                   Eval      myString = tifld8
     C                   Eval      myString = tifld9
     C                   Eval      myString = tiflda
     C                   Eval      myString = tifldb
     C                   Eval      myString = tifldc
     C                   Eval      myString = tifldd
     C                   Eval      myString = tiflde
     C                   Eval      myString = tifldf
     C                   Eval      myString = tifldg

     C                   SetOn                                        LR
     C                   Return

Nathan M. Andelin
www.relational-data.com