New answers tagged order-by
I admit this is sleazy but if you're in a pinch try returning the top number of rows in the subquery. Returning the top 100 percent doesn't work but if you want to go through the trouble you can query the number of rows and pass that to TOP as a variable. I tested this on a database set to compatability level 80 so I think it should work with SQL 2000. ...
I agree with zxc that your instructions are not clear. However I think what you are asking for is to get the highest number and group together all the letters that have an instance of that number. Then go to the next highest number. Is correct? I did this in SQL Server but it ought to work in MySQL as well. --populate my temp table. I threw in some D's ...
I did an upgrade to percona mysql 5.6 and it solves a problem. Both type of queries runs equally. Fortunately - equally fast.
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