Updating a second record set in VBS

Posted on 2005-03-19
Medium Priority
Last Modified: 2010-04-06
I have captured the following numbers to a string called clients numbers as the record set was read

response.write(strClientNumbers) will give the following result
the number of characters has been captured in the following
response.write(strCount) would give a value of 21

I am looking for some sort of loop to do the following to use these client number to update a second record set at the end of the ASP page.

sql="UPDATE client SET Newsletters='Newsletters Collected' WHERE [Client Number]=101"
sql="UPDATE client SET Newsletters='Newsletters Collected' WHERE [Client Number]=175"
sql="UPDATE client SET Newsletters='Newsletters Collected' WHERE [Client Number]=217"


unless there is a way to update another record set when the record set that generated these client number is still open

Question by:kempvet
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LVL 53

Accepted Solution

Ryan Chong earned 2000 total points
ID: 13582400
try use the "IN" clause in your sql statement, so that you no need to have multiple sql statement to update your records, try like:

sql="UPDATE client SET Newsletters='Newsletters Collected' WHERE [Client Number] In (" & strClientNumbers & ") "
conn.execute sql


Expert Comment

ID: 13589784
strClientNumbers is a string?

I am not sure on it ... if it was in array or string

for i=0 to 21
sql="UPDATE client SET Newsletters='Newsletters Collected' WHERE [Client Number]=" & strClientNumbers(i)
'here somthing to execute your sql
next i

I know it is not a big help, but small idea too.

Thanks !
Bj Mac Donel
LVL 53

Expert Comment

by:Ryan Chong
ID: 13590482
It can be done by a loop just want donel mentioned above, but this will sacrifice the performance, since the field(s) to be updated is same.

so, by using a single sql that can performance the same thing will be the better choice.

Just my $0.02, cheers

Author Comment

ID: 13590836
will try a few of these over the next couple of days - just a little pushed for time at the moment but I am sure the answer is in one of these replies, thanks

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