fenway

Assuming you've accounted for the obvious boundary cases, you can just use %newpath%, right?

Besides, who cares?

Maybe you should see what the first parameter represents.

I don't follow -- specify your full path to the cnf file when your startup mysqld.

That doesn't mean they aren't keywords -- and they are poor choices for column names.

Actually, if you simply set the timezone when you connect to the DB, mysql will take care of all of this for you.

That depends which fields you need, and from which tables -- you might need a 4 table join.

We don't encourage posting entire scripts.... if you store the submitter ID with the data, then it should be trivial to filter it on the way out with a WHERE clause. Not sure if you're asking for mysql or php help.

Are you sure you posted in the correct thread?

The first error means your query failed -- use mysql_error() to determine what the problem was. If you'd like our help, show us both the query and the error.

Well, then you need to specify this content id in your ON clause.

ENUMs work too.

What's (db IP)? Is that in your config file?

Why so many queries?

I would recommend proper JOIN syntax... it will serve you well in the long run.

Yes, you could *try* and get them all back with a three-table join, but then I'd imagine that you'll likely have the usual layout questions & problems when iterating through a recordset that represents multiple entities.

Check mysql_error() after the first query and you'll see the problem.

Sounds like you didn't switch your delimeter.

Well, if you're trying to do this all in one query, you'll get stuck. Why not simply issue a simple one for each section, and then have PHP do the layout work?

What do you have now -- table structure? Show us, and we can show you what to change?

We don't need the code -- we need to see the output that isn't "distinct".

Well, if you need to ensure you get back at least one group of three, then you'll have to loop through groups of three repeatedly... I suppose you could "loop" in an SP or in application code. A simple SUM() with LIMIT 3 can help you out.

Sounds like you should be able to CONCAT() them together and simply check...

Those don't total 120... what am I missing?

Why not a LEFT JOIN... IS NULL? NOT EXISTS might work too.