Can I do this in mySQL? Or needs to be PHP?

First let me say that I'm unsure if this is the right forum for this question. My apologies if it is not. One of the few things that I dislike about mySQL is that it cannot do sub-selects. I know you can get around this somewhat by using JOINS, but I am unfamiliar with anything but a basic JOIN. There is a query that I need to run and I'm wonding if someone can tell me if it is possible using mySQL or if I'm going to have to work around it with PHP code. Here are the relevant tables and fields: (main table) table1 -> thing_name (subordinate tables) table2 -> thing_name -> thing_type table3 -> thing_name -> thing_rarity The field "thing_name" is the key in each field and means the same in each. However, there will be records in table1 that have a relational record in table2 but not in table3. There can also be instances wehre there'll be a record in table1 where there is a relational record in table3 but not in table 2. What I need to do is select records in all 3 tables to get the relevant data. I'm not sure if I've made myself overly clear, so here is the query that I've been working with (which isn't working right): SELECT user_haves.card_name FROM user_haves LEFT JOIN cards_crypt ON user_haves.card_name = cards_crypt.name AND cards_crypt.clan IN ( "Brujah", "Caitiff", "Gangrel", "Giovanni" ) AND cards_crypt.release_set IN ( "A", "JV", "V" ) LEFT JOIN cards_library ON user_haves.card_name = cards_library.name AND cards_library.type IN ( "Action" ) AND cards_library.rarity IN ( "Rare" ) WHERE user_haves.user = "joebobbriggs" ORDER BY cards_crypt.capacity, cards_crypt.clan, cards_library.type, cards_library.requires, cards_library.rarity The above query runs, but it doesn't return the correct data. As I said, I'm pretty unfamiliar with JOINS so please don't laugh if the above is blatently incorrect. :p I know I can do what I need to do in PHP by just selecting all the records and filtering the data, but I'm hoping I can do this in mySQL so not only would it be faster, but so I can learn something new. :P Thanks for any help anyone can give. Chris