Understanding the number of matched rows in LEFT JOIN

Question

Here is my table structure: SQL Fiddle

CREATE TABLE mytable (
    id int,
    related int
);

INSERT into mytable VALUES(1, NULL);
INSERT into mytable VALUES         


        

Answer 1

In first case, we need to match values 1, 2, 3 with NULL, 1 and 1. Since it is left join, NULL will stay with no match and 1's will be matched with 1 from other table, thus 3 records.

In the second case, we have values 1, 2, 3. 2 and 3 have no match and will result in two rows, but 1 has 2 matches and will result in 2 additional rows, which is 4 rows.

Generally, having:

... LeftTable [LT] left join RightTable [RT] on [LT].[joinCol] = [RT].pjoinCol] ...

will work like this:

take all values from LT.joinCol, try to match with values in RT.joinCol. If some value has n matches in RT.joinCol, then it will result in n rows. If the row has no match, it will still result in one, un-matched record.

In your 1st case, 2 values have 1 match => 1 + 1 = 2 records. One value has no match => 1 record, 2 + 1 = 3.

In your 2nd case, 2 values have no match => thus 2 records, one value has 2 matches => 2 records, 2 + 2 = 4 :)