SQL查询中的LIMIT返回指定的最大金额

Question

For some reason this is always returning 31 row instead of the expected 2 rows. Not really certain what is going on?

SELECT 
    ins_by          AS userid,
    wds_user_id     AS customer, 
    wds_address_id  AS billing,
    (
        SELECT 
            wds_address_id  AS shipping
        FROM 
            wds_user_address B
        WHERE 
            ins_by != "GUEST" 
        AND 
            wds_address_type_id = 2 
        AND 
            B.ins_by = A.ins_by
    ) AS shipping
FROM 
    wds_user_address A 
WHERE 
    ins_by != "GUEST" 
AND 
    wds_address_type_id = 1
ORDER BY id ASC 
LIMIT 30, 32;

Expecting

+---------------------+----------+---------+----------+
| userid              | customer | billing | shipping |
+---------------------+----------+---------+----------+
| FOO@GMAIL.COM       |      121 |     185 |      186 |
| BAR@GMAIL.COM       |      123 |     189 |      190 |
+---------------------+----------+---------+----------+

Answer 1

Your statement is telling MySQL to return 32 rows.

LIMIT 30,32

That says "skip the first 30 rows, and then return the next 32 rows".

If you want to skip the first 30 rows and return only the next 2 rows, your LIMIT clause needs to be modified.

Answer 2

You need to just replace the limit with 30,2. Like below query

SELECT 
    ins_by          AS userid,
    wds_user_id     AS customer, 
    wds_address_id  AS billing,
    (
        SELECT 
            wds_address_id  AS shipping
        FROM 
            wds_user_address B
        WHERE 
            ins_by != "GUEST" 
        AND 
            wds_address_type_id = 2 
        AND 
            B.ins_by = A.ins_by
    ) AS shipping
FROM 
    wds_user_address A 
WHERE 
    ins_by != "GUEST" 
AND 
    wds_address_type_id = 1
ORDER BY id ASC 
LIMIT 30, 2;

Answer 3

LIMIt 30, 32

Equals to

Move the pointer to the 30th row, and get the next 32 rows.

Change it to

LIMIT 30, 2

If you want it to return 2 rows.