I have this table in my mysql:
| id | category_id | region_id | score |
+----+-------------+-----------+-------+
| 1 | 1 | 1 | 78 |
| 2 | 1 | 2 | 65 |
| 3 | 1 | 3 | 98 |
| 4 | 1 | 4 | 45 |
| 5 | 1 | 5 | 78 |
| 6 | 1 | 1 | 98 |
| 7 | 1 | 2 | 32 |
| 8 | 1 | 3 | 56 |
| 9 | 1 | 4 | 89 |
| 10 | 1 | 5 | 65 |
+----+-------------+-----------+-------+
I want to get the 5 latest id but order my result table base on region id so I used this code
SELECT *
FROM tb_scores
WHERE category_id = 1
ORDER
BY id DESC
, region_id ASC
LIMIT 5
but the result only sorted the id as desc but not the region_id as ASC..to explain briefly I want this kind of result.
| id | category_id | region_id | score |
+----+-------------+-----------+-------+
| 6 | 1 | 1 | 98 |
| 7 | 1 | 2 | 32 |
| 8 | 1 | 3 | 56 |
| 9 | 1 | 4 | 89 |
| 10 | 1 | 5 | 65 |
+----+-------------+-----------+-------+
try this:
SELECT *
FROM (SELECT *
FROM tb_scores
WHERE category_id = 1
ORDER BY id DESC,region_id ASC LIMIT 5) t
ORDER BY region_id
get your data in subquery and apply order by region_id on the subquery result.
Related
in my table https://www.db-fiddle.com/f/4yPorU6k3SjQ5nmhgi1wGo/0
i use query
SELECT id
FROM test
ORDER BY id <= 7 DESC, id DESC
i want to order everything from 7 to lesser by the bigger then everything else by the lesser
my query give me
| id |
| --- |
| 6 |
| 3 |
| 2 |
| 1 |
| 65 |
| 35 |
| 34 |
| 33 |
| 12 |
| 11 |
| 11 |
| 10 |
but i want to give me
| id |
| --- |
| 6 |
| 3 |
| 2 |
| 1 |
| 10 |
| 11 |
| 11 |
| 12 |
| 33 |
| 34 |
| 35 |
| 65 |
Consider a conditional sort, like so:
select id
from test
order by
case when id <= 7 then id end desc,
id
Demo on DB Fiddle:
| id |
| -: |
| 6 |
| 3 |
| 2 |
| 1 |
| 10 |
| 11 |
| 11 |
| 12 |
| 33 |
| 34 |
| 35 |
| 65 |
I think you want:
SELECT id
FROM test
ORDER BY id <= 7 DESC,
(CASE WHEN id <= 7 THEN id END) DESC,
id ASC
I have a table like this:
mysql> select * from studentscore;
+------------+-----------+-------+
| student_id | cource_id | score |
+------------+-----------+-------+
| 1 | 1 | 80 |
| 1 | 2 | 90 |
| 1 | 3 | 85 |
| 1 | 4 | 78 |
| 2 | 2 | 53 |
| 2 | 3 | 77 |
| 2 | 5 | 80 |
| 3 | 1 | 71 |
| 3 | 2 | 70 |
| 3 | 4 | 80 |
| 3 | 5 | 65 |
| 3 | 6 | 75 |
| 4 | 2 | 90 |
| 4 | 3 | 80 |
| 4 | 4 | 70 |
| 4 | 6 | 95 |
| 5 | 1 | 60 |
| 5 | 2 | 70 |
| 5 | 5 | 80 |
| 5 | 6 | 69 |
| 6 | 1 | 76 |
| 6 | 2 | 88 |
| 6 | 3 | 87 |
| 7 | 4 | 80 |
| 8 | 2 | 71 |
| 8 | 3 | 58 |
| 8 | 5 | 68 |
| 9 | 2 | 88 |
| 10 | 1 | 77 |
| 10 | 2 | 76 |
| 10 | 3 | 80 |
| 10 | 4 | 85 |
| 10 | 5 | 83 |
| 11 | 3 | 80 |
| 12 | 4 | 99 |
| 13 | 5 | 74 |
+------------+-----------+-------+
I want to show student_id and students' average scores that are higher than 80.
The output I want is like this:
+------------+-------------------+
| student_id | Average |
+------------+-------------------+
| 1 | 83.25 |
| 4 | 83.75 |
| 6 | 83.66666666666667 | // and how can I make this result shorter like 83.67?
| 7 | 80 |
| 9 | 88 |
| 10 | 80.2 |
| 11 | 80 |
| 12 | 99 |
+------------+-------------------+
I've tried the following codes
mysql> select student_id, avg(score) as average_score
-> from studentscore
-> group by student_id
-> where avg(score) >= 80;
and it gave me an syntax error.
I know by rules the where clause should go before the group by clause but I can't because the where clause depends on the result from the group by clause, and if I switch their position it will give me another error("Invalid use of group function").
Can some one tell me how to get the table I want?
use "having" instead of "where"
use having instead of where.
Here's the difference:
with where you can write a predicate that will be applied to each row
with having you can write a predicate that will applied to each group
and in your case, the 2nd is the only solution that can work.
select student_id, avg(score) as average_score
from studentscore
group by student_id
having avg(score) >= 80;
where applies a filter to your data before grouping has taken place, whereas having applies a filter post-grouping. round(,2) will format as you also ask:
select student_id, round(avg(score), 2) as average_score
from studentscore
group by student_id
having average_score >= 80;
I using MySql and I have...
orders
+--------+---------+-------------+
| ID | PRO_ID | CUS_ID |
+--------+---------+-------------+
| 1 | 1 | 2 |
| 2 | 1 | 79 |
| 3 | 1 | 6 |
| 4 | 2 | 41 |
| 5 | 2 | 28 |
| 6 | 2 | 33 |
| 7 | 3 | 2 |
| 8 | 4 | 9 |
| 9 | 4 | 23 |
| 10 | 5 | 43 |
| 11 | 5 | 45 |
| 12 | 5 | 78 |
| 13 | 5 | 67 |
| 14 | 6 | 12 |
| 15 | 6 | 95 |
How to get 5 top PRO_ID and their count. Example
+---------+-------------+
| PRO_ID | COUNT |
+---------+-------------+
| 5 | 4 |
| 1 | 3 |
| 2 | 3 |
| 4 | 2 |
| 6 | 2 |
my orders table have 2 millions rows, i need a query faster. Thanks for any help!
Apply indexing for that particular field and go on with this query....i hope it will give you better performance
select PRO_ID,count(PRO_ID) as prod_count
from orders
group by PRO_ID
order by prod_count desc
limit 5
You can do like:
select PRO_ID, COUNT(*) as total
from test
group by PRO_ID
ORDER BY total desc, PRO_ID
limit 5
here is fiddle
select PRO_ID,count(PRO_ID) as count
from orders
group by PRO_ID
order by count desc
limit 5
If the query seems to slow then try adding index on your table
i want to find solution for my query problem. I need to find the SUM of all priceProduct*quantity and separated with each of productcategory. I have already made a query, but it takes longer time to executed it. this is my query,
SELECT
pb.ProductCategoryID,
pb.ProductCategoryDescription,
(SELECT
SUM((SELECT pd.HPP FROM `price details` pd WHERE pd.ProductID = pdt.ProductID ORDER BY pd.PriceDetailID DESC LIMIT 1)*
(SELECT StockProductBallance FROM `stock product` sp WHERE sp.ProductID = pdt.ProductID ORDER BY sp.StockProductID DESC LIMIT 1))
FROM product pdt
WHERE pdt.ProductCategoryID = pb.ProductCategoryID
) AS Total
FROM `product category` pb
GROUP BY pb.ProductCategoryID
this my example table
table product:
+------+-------+
| id_p | id_pc |
+------+-------+
| 1 | 3 |
| 2 | 4 |
| 3 | 3 |
| 4 | 4 |
+------+-------+
table productcategory:
+-------+---------+
| id_pc | pc_name |
+-------+---------+
| 3 | new_pc |
| 4 | old_pc |
+-------+---------+
table price details:
+---------------+------+-----+
| PriceDetailID | id_p | hpp |
+---------------+------+-----+
| 1 | 1 | 100 |
| 2 | 1 | 110 |
| 3 | 1 | 120 |
| 4 | 2 | 200 |
| 5 | 2 | 210 |
| 6 | 2 | 220 |
+---------------+------+-----+
table stockProduct:
+-----------------+------+---------------+
| id_stockProduct | id_p | stockballance |
+-----------------+------+---------------+
| 1 | 1 | 10 |
| 2 | 1 | 11 |
| 3 | 1 | 12 |
| 4 | 2 | 20 |
| 5 | 2 | 21 |
| 6 | 2 | 22 |
+-----------------+------+---------------+
Really need your help guys, for better query..
i have a table called rc_language_type_table with:
id language
1 english
2 Xhosa
3 afrikaans
etc
then i have a table rc_language_type_assoc_table with:
profile_id | language_type_id |
+------------+------------------+
| 3 | 1 |
| 13 | 1 |
| 15 | 1 |
| 16 | 1 |
where i have profiles and each profile is connected to a language id in a 1 to many
so then i did:
select *,count(*) from rc_language_type_assoc_table group by language_type_id;
+------------+------------------+----------+
| profile_id | language_type_id | count(*) |
+------------+------------------+----------+
| 3 | 1 | 96 |
| 3 | 2 | 19 |
| 3 | 3 | 18 |
| 64 | 4 | 51 |
| 94 | 5 | 10 |
| 37 | 6 | 26 |
| 3 | 7 | 21 |
| 3 | 8 | 4 |
| 3 | 9 | 6 |
| 88 | 10 | 4 |
| 3 | 11 | 3 |
+------------+------------------+----------+
what i want now is: instead having the language_type_id i want to display the actual language...how would i do this please???
i tried:
select *, count(*)
from rc_language_type_assoc_table, rc_language_type_table
group by language_type_id
where rc_language_type_assoc_table.language_type_id = rc_language_type_table.id;
but i get a syntax error...
please help??
thank you
GROUP BY should be "after" the WHERE statement and not before
select *, count(*)
from rc_language_type_assoc_table, rc_language_type_table
where rc_language_type_assoc_table.language_type_id = rc_language_type_table.id
group by language_type_id ;