Executing a MySQL count query?

Question

I have the following table:

UID | ID  | Type
1   | 1   | product
1   | 2   | product
1   | 3   | service
1   | 4   | product
1   | 5   | product
2   | 6   | service
1   | 7   | order
2   | 8   | invoice
2   | 9   | product

I want to end up with:

UID | product | service | invoice | order
1   |  4      |  1      |  0      |  1
2   |  1      |  1      |  1      |  0

What will the SQL query look like? Or at least the most suitable?

Answer 1

If you really only need these four types, you can hard-code the values ​​as follows:

select UID,
    count(case when type='product' then 1 else null end) as product,
    count(case when type='service' then 1 else null end) as service,
    count(case when type='invoice' then 1 else null end) as invoice,
    count(case when type='order' then 1 else null end) as order
from MyTable
group by UID
order by UID    

Answer 2

What you want to do is a summary operation that is not directly supported by SQL syntax. However, this is not too complicated and conceptually involves 2 steps:

• "" ​​ . CASE WHEN... ELSE... END (, decode() oracle). CASE WHEN , RDBMSes
• GROUP BY (SUM, MIN, MAX ..), .

:

mysql> select * from foo;
+------+------+---------+
| uid  | id   | type    |
+------+------+---------+
|    1 |    1 | product | 
|    1 |    2 | product | 
|    1 |    3 | service | 
|    1 |    4 | product | 
|    1 |    5 | product | 
|    2 |    6 | service | 
|    1 |    7 | order   | 
|    2 |    8 | invoice | 
|    2 |    9 | product | 
+------+------+---------+

1 , "" :

select uid
     , case when type = 'product' then 1 else 0 end as is_product
     , case when type = 'service' then 1 else 0 end as is_service
     , case when type = 'invoice' then 1 else 0 end as is_invoice
     , case when type = 'order' then 1 else 0 end as is_order
  from foo;

:

+------+------------+------------+------------+----------+
| uid  | is_product | is_service | is_invoice | is_order |
+------+------------+------------+------------+----------+
|    1 |          1 |          0 |          0 |        0 | 
|    1 |          1 |          0 |          0 |        0 | 
|    1 |          0 |          1 |          0 |        0 | 
|    1 |          1 |          0 |          0 |        0 | 
|    1 |          1 |          0 |          0 |        0 | 
|    2 |          0 |          1 |          0 |        0 | 
|    1 |          0 |          0 |          0 |        1 | 
|    2 |          0 |          0 |          1 |        0 | 
|    2 |          1 |          0 |          0 |        0 | 
+------+------------+------------+------------+----------+

is_ *, ( "" ):

select uid
     , sum(is_product) as count_product
     , sum(is_service) as count_service
     , sum(is_invoice) as count_invoice
     , sum(is_order)   as count_order
  from (
         select uid
              , case when type = 'product' then 1 else 0 end as is_product
              , case when type = 'service' then 1 else 0 end as is_service
              , case when type = 'invoice' then 1 else 0 end as is_invoice
              , case when type = 'order' then 1 else 0 end as is_order
           from foo
       ) x
 group by uid;

( , , : MySQL, , , -, , - SQL , !)

:

+------+---------------+---------------+---------------+-------------+
| uid  | count_product | count_service | count_invoice | count_order |
+------+---------------+---------------+---------------+-------------+
|    1 |             4 |             1 |             0 |           1 | 
|    2 |             1 |             1 |             1 |           0 | 
+------+---------------+---------------+---------------+-------------+

.

Answer 3

CASE- MySQL, ( ):

  SELECT t.uid,
         SUM(CASE WHEN t.type = 'product' THEN COUNT(*) END) as PRODUCT,
         SUM(CASE WHEN t.type = 'service' THEN COUNT(*) END) as SERVICE,
         SUM(CASE WHEN t.type = 'invoice' THEN COUNT(*) END) as INVOICE,
         SUM(CASE WHEN t.type = 'order' THEN COUNT(*) END) as ORDER
    FROM TABLE t
GROUP BY t.uid, t.type

Answer 4

-, " ", , MySQL .

, GROUP BY / DISTINCT , . , :

SELECT Type, COUNT(ID), UID
  FROM tablename
 GROUP BY UID, Type

Answer 5

(, , -, ) mytable