(no subject)

Bob Sneidar bobs at twft.com
Fri May 21 21:17:30 EDT 2010


UPDATE: 

Well I discovered what my problem was. I commented out the method I employed which used sqlYoga for all the queries, and wrote a new method using only RevDB API calls. The sqlYoga method took about 1 to 1.5 seconds per iteration, so for 100 single record queries, let's say roughly 2 minutes. The RevDB method took less than 10 seconds for 100 queries. 

I guess what I am saying is that if you need to do lots of small queries like I am doing, and you don't mind learning a little SQL, the RevDB commands and functions are going to be a TON more efficient. But if you are doing a few queries here and there, the performance hit would be minimal, and sqlYoga can save you enormous amounts of trouble learning SQL and get you going much quicker. 

Bob


On May 21, 2010, at 3:31 PM, Bob Sneidar wrote:

> Hi all. 
> 
> Anyone have any idea how fast the queries to the On-Rev SQL databases are on average? I was using a method where I was looking up a key from one query of 100 records at a time and querying a mySql database at my On-Rev site ONE RECORD AT A TIME! I discovered that the queries were taking about a second, which now that I think about it, is about right, with internet lag and all. 
> 
> So now I have to think about using joins. My problem now becomes, I am looking up values from a table in one database, in a table  tables residing in another database. Can I even create a join like that? Don't the tables have to reside in the same database to do a join like that? <sigh>
> 
> Barring that, I suppose I can compile a list of key values, and do a single query for records with key values in that string, then loop through and add the records that don't exist, but how to do that with a single query! Sheesh! If I were using Foxpro this would be child's play. 
> 
> I'm wondering if the Relational aspect of Trevor's sqlYoga would be able to do this, but again, I think I have to be working with two tables in the same database for that to work. Trevor?
> 
> Bob
> 
> 



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