试图找出如何编写SQL查询以返回mysql数据库的两个表的非交叉，遇到问题

Question

Sorry for total noobishness, but I'm trying to write SQL for the following problem involving a mysql database and I'm at a loss.

Here's the problem. Given the following two tables:

**Table1**
PID WEBSITE     COMP_SITE   KEYCODE     SITE_TYPE   KEYWORDS
14  google.com  bing.com    w3874yf8    alpha       happy+campers
26  radio.com   alice.com   98ygdsfg    beta        slip+slop
13  simon.com   fat.org     98dfyg77    delta       sunrise

**Table2**
PID WEBSITE     KEYWORDS
14  google.com  happy+campers
14  yahoo.com   fat+albert
14  aol.com     chump+change

I want a query to return EVERY row in Table1 for which there are NO rows in Table2 matching the same (ID,WEBSITE,KEYWORDS) chunk of the row. In the above example this query would return rows 2 and 3 of Table1 .

Note that no columns in either table containing individual values unique to the table, though the ID/WEBSITE/KEYWORDS chunk of a row will be unique.

I feel I'm making this far more complicated than it need to be. :/

Answer 1

select * from Table1 as t1
left join Table2 as t2 
    on  t1.PID =t2.PID
        and t1.WEBSITE  =t2.WEBSITE     
        and t1.KEYWORDS = t2.KEYWORDS
where t2.PID is null

Answer 2

There are two common ways to approach this. The first is a left outer join . Use this construct to match all rows, keeping rows in the first table, and then filter out the ones with a match:

select t1.*
from table1 t1 left outer join
     table2 t2
     on t1.PID = t2.PID and
        t1.WEBSITE = t2.WEBSITE and
        t1.KEYWORDS = t2.KEYWORDS
where t2.PID is null;

The second method is not exists :

select t1.*
from table1 t1
where not exists (select 1
                  from table2 t2
                  where t1.PID = t2.PID and
                        t1.WEBSITE = t2.WEBSITE and
                        t1.KEYWORDS = t2.KEYWORDS
                 );

Both these methods can take advantage of an index on table2 :

create index table2_PID_WEBSITE_KEYWORDS on table2(PID, WEBSITE, KEYWORDS);