fenway

If the subsequent tables refer to a LEFT JOIN'ed table, then yes.

Troubleshoot what?

Your VALUES are mis-quoted. You really should be using a DB wrapper.

This sounds like a follow-up to you other question.

If you use LEFT JOIN, then any conditions on the LEFT JOIN-ed table fields needs to be the ON clause, not the where clause.

What do you have so far? Hard to follow.

That's exactly what I'm saying.

That's likely a unicode game.

Then that's how long it takes -- you're not exactly benchmarking here.

Echo $query.

Oh, that's different -- if you're simply storing it, and not querying it or updating it, then sure, nothing wrong with that! TEXT holds up to 65K -- LONGTEXT, 4GB. But if this table is going to have a lot of rows, you might want to consider breaking this out into it's own table.

Other than too many indexes, there's little you can do.

You shouldn't be storing that kind of formatted data in a row-based database.

It says "NULL" because you have no where clause -- so it has to read all the rows from one of the tables.

Not MySQL.

Don't separate date & time fields like that -- otherwise, you'll have to merge then back together for your where clause.

Sorry, I missed that somehow -- then you want " <> ANY ">

Well, you imported it somehow -- so there is a raw text file somehwere.

You are here to help. Anyway, you didnt ask anything, i dont see queston mark on end, so.. My question was :"That depends on how much of your data you want back if it crashes."

You'd use a self-join to find the record you want -- then join the result back to whatever other tables(s) you need.

MySQL tells you why it fails.

Unless you show us the sql file, we can't help you.

MySQL has a LOAD FROM INFILE.

I assume you mean "greater than or equal to", based on a count of 5 for day 1, not 4. select a.day, sum( if( b.daycount >= a.day, 1, 0 ) ) AS cnt from tableA AS a cross join tableB AS b

You know, there are plenty of browser utilities that will allow you to override style sheets as you see fit.