Two millions data sheets, questions about joint queries

Two millions data sheets, the problem of joint query
Two data tables, a, B.
A, B table has two fields: Exp_code, CODE;
Now to find out a, B table in the Exp_code field equals ' R8_002_xz, and Exp_code not the same record, with the following statement, if two tables Exp_code record tens of thousands of cases, basically the crash, not to mention to millions. Ask how to improve.

SELECT * from a where exp_code= ' r8_002_xz ' and CODE not in (select CODE from B)


------Solution--------------------
SELECT * from a where exp_code= ' r8_002_xz ' and CODE not in (select CODE from B)

=======================================================

SELECT * from a LEFT join B on A.code=b.code
where a.exp_code= ' R8_002_xz ' and b.code is null

Look at your code on the index, such as no, build index, Exp_code also built


------Solution--------------------
Upstairs positive solution, but for millions record database is still slow, can be limited by limit
------Solution--------------------
SELECT * from a where exp_code= ' r8_002_xz ' and code not exist (select code from B where exp_code!= ' R8_002_xz ')
The syntax above may be a bit problematic, but the logical relationship should be right.
Requires:version> 4.1 to support subqueries.
------Solution--------------------
Is this what you asked for?
SELECT * from a LEFT join B on A.code=b.code and A.exp_code=b.exp_code
where a.exp_code= ' r8_002 ' and b.code is null


------Solution--------------------
Nested select is less efficient, try to use inner join, left join instead
------Solution--------------------
It's wrong. Change to the following:
SELECT * from a where exp_code= ' r8_002_xz ' and Code not in (select code from B where exp_code!= ' r8_002_xz ');
Or
SELECT * from a where exp_code= ' r8_002_xz ' and NOT EXISTS (select code from B where exp_code!= ' R8_002_xz ' and a.code=b. code);



------Solution--------------------
SELECT * from a LEFT join B on A.code=b.code and A.exp_code=b.exp_code and B.code is null

All you need to do is to find the records in a that are not in B, and then press B to append the structure!

Hard, "Farmer" peer!

------Solution--------------------
Hahaha .... Met with ...

