I have the following structure:
+----------+--------+---------------------+
| id| gr_id| name | value |
+----------+--------+---------------------+
| 1 | 11 | name | Burro |
| 2 | 11 | submit | 2019/05/10 |
| 3 | 11 | date | 2019/05/17 |
| 4 | 12 | name | Ajax |
| 5 | 12 | submit | 2019/05/10 |
| 6 | 12 | date | 2019/05/18 |
+----------+--------+---------------------+
I have to order it by the date(if the name is date), from highest to lowest date, also it has to keep the groups (gr_id) without mixing the elments.
The desired result would look like this:
+----------+--------+---------------------+
| id| gr_id| name | value |
+----------+--------+---------------------+
| 4 | 12 | name | Ajax |
| 5 | 12 | submit | 2019/05/10 |
| 6 | 12 | date | 2019/05/18 |
| 1 | 11 | name | Burro |
| 2 | 11 | submit | 2019/05/10 |
| 3 | 11 | date | 2019/05/17 |
+----------+--------+---------------------+
How can i implement this?
You'll have to associate the group ordering criteria with all the elements of the group. You can do it through a subquery, or a join.
Subquery version:
SELECT t.*
FROM (SELECT gr_id, value as `date` FROM t WHERE `name` = 'date') AS grpOrder
INNER JOIN t ON grpOrder.gr_id = t.gr_id
ORDER BY grpOrder.`date`
, CASE `name`
WHEN 'name' THEN 1
WHEN 'submit' THEN 2
WHEN 'date' THEN 3
ELSE 4
END
Join version:
SELECT t1.*
FROM t AS t1
INNER JOIN AS t2 ON t1.gr_id = t2.gr_id AND t2.`name` = 'date'
ORDER BY t2.value
, CASE t1.`name`
WHEN 'name' THEN 1
WHEN 'submit' THEN 2
WHEN 'date' THEN 3
ELSE 4
END
Related
I am having trouble with an SQL query. I have two tables.
My first table:
+------------+-------------+---------------+
| id_mission | Some column | Other column |
+------------+-------------+---------------+
| 1 | ... | ... |
| 2 | ... | ... |
+------------+-------------+---------------+
My second table:
+------------+-------------+---------+
| id_mission | id_category | points |
+------------+-------------+---------+
| 1 | 1 | 3 |
| 1 | 2 | 4 |
| 1 | 3 | 4 |
| 1 | 4 | 8 |
| 2 | 1 | -4 |
| 2 | 2 | 3 |
| 2 | 3 | 1 |
| 2 | 4 | -7 |
+------------+-------------+---------+
And I would like to have this kind of result with my SELECT request
+------------+-------------+--------------+---------------+----------------+
| id_mission | Some column | Other column | id_category 1 | id_category X |
+------------+-------------+--------------+---------------+----------------+
| 1 | ... | ... | ... | ... |
| 2 | ... | ... | ... | ... |
+------------+-------------+--------------+---------------+----------------+
I have tried this with the first two column but it doesn't work, I also tried GROUP_CONCAT, it works but it's not the result I want.
SELECT m.id_mission ,mc.id_category 1,mc1.id_category 2
from mission m
left join mission_category mc on m.id_mission = mc.id_mission
left join mission_category mc1 on m.id_mission = mc1.id_mission
Can someone help me?
You can use conditional aggregation. Assuming that you want to pivot the points value per category:
select
t1.*,
max(case when t2.id_category = 1 then points end) category_1,
max(case when t2.id_category = 2 then points end) category_2,
max(case when t2.id_category = 3 then points end) category_3
from t1
inner join t2 on t2.id_mission = t1.id_mission
group by t1.id_mission
This assumes that id_mission is the primary key of t1 (else, you need to enumerate the columns you want in both the select and group by clauses).
I have following schema:
+--+------+-----+----+
|id|device|token|cash|
+--+------+-----+----+
column device is unique and token is not unique and null by default.
What i want to achieve is to set all duplicate token values to default (null) leaving only one with highest cash. If duplicates have same cash leave first one.
I have heard about cursor, but it seems that it can be done with usual query.
I have tried following SELECT only to see if im right about my thought how to achieve this, but it seems im wrong.
SELECT
*
FROM
db.table
WHERE
db.table.token NOT IN (SELECT
*
FROM
(
SELECT DISTINCT
MAX(db.table.balance)
FROM
db.table
GROUP BY db.table.balance) temp
)
For example:
This table after query
+-----+---------+--------+-------+
| id | device | token | cash|
+-----+---------+--------+-------+
| 1 | dev_1 | tkn_1 | 3 |
| 2 | dev_2 | tkn_1 | 10 |
| 3 | dev_3 | tkn_2 | 10 |
| 4 | dev_4 | tkn_2 | 14 |
| 5 | dev_5 | tkn_3 | 10 |
| 6 | dev_6 | null | 10 |
| 7 | dev_7 | null | 10 |
| 8 | dev_8 | tkn_4 | 11 |
| 8 | dev_8 | tkn_4 | 11 |
| 8 | dev_8 | tkn_5 | 11 |
+-----+---------+--------+-------+
should be:
+-----+---------+--------+-------+
| id | device | token | cash|
+-----+---------+--------+-------+
| 1 | dev_1 | null | 3 |
| 2 | dev_2 | tkn_1 | 10 |
| 3 | dev_3 | null | 10 |
| 4 | dev_4 | tkn_2 | 14 |
| 5 | dev_5 | tkn_3 | 10 |
| 6 | dev_6 | null | 10 |
| 7 | dev_7 | null | 10 |
| 8 | dev_8 | tkn_4 | 11 |
| 8 | dev_8 | null | 11 |
| 8 | dev_8 | tkn_5 | 15 |
+-----+---------+--------+-------+
Thanks in advance :)
Try using an EXISTS subquery:
UPDATE yourTable t1
SET token = NULL
WHERE EXISTS (SELECT 1 FROM (SELECT * FROM yourTable) t2
WHERE t2.token = t1.token AND
t2.cash > t1.cash);
Demo
Note that this answer assumes that there would never be a tie for two token records having the same highest cash amount.
To set exactly one row in the even of duplicates on the maximum cash, use the id:
update t join
(select tt.*,
(select t3.id
from t t3
where t3.token = tt.token
order by t3.cash desc, id desc
) as max_cash_id
from t tt
) tt
on t.id = tt.id and t.id < tt.max_cash_id
set token = null;
I have two tables.
Tab1:
+------------+
| id | title |
+------------+
| 1 | B |
| 2 | C |
| 3 | A |
| 4 | A |
| 5 | A |
| 6 | A |
| ... |
+------------+
Tab2:
+-------------------------------------------+
| id | item_id | item_key | item_value |
+-------------------------------------------+
| 1 | 1 | value | $4 |
| 2 | 1 | url | http://h.com/ |
| 3 | 2 | value | $5 |
| 4 | 3 | url | http://i.com/ |
| 5 | 3 | value | $1 |
| 6 | 3 | url | http://y.com/ |
| 7 | 4 | value | $2 |
| 8 | 4 | url | http://z.com/ |
| 9 | 5 | value | $1 |
| 10 | 5 | url | http://123.com/ |
| ... |
+-------------------------------------------+
item_id is a foreign key from tab1.
How do I make it so I get a table of ids from Tab1 in order according to criteria from both tables. The criteria are the following:
Order ASC by title. If title is the same,
Order DESC by value. If both title and value is the same,
Prioritize items who's 'url' key contains '123.com'.
The resulting table with the ordered results would be:
+------------+
| id | title |
+------------+
| 4 | A |
| 5 | A |
| 3 | A |
| 6 | A |
| 1 | B |
| 2 | C |
| ... |
+------------+
The results should include items that don't have the one, both, or none of the fields from Tab2 set.
As far as I understand, a simple join will do it. You'll have to join Tab2 twice, since you want to order by values from both rows.
SELECT Tab1.id, Tab1.title
FROM Tab1
JOIN Tab2 t2_val ON t2_val.item_id = Tab1.id AND t2_val.item_key='value'
JOIN Tab2 t2_url ON t2_url.item_id = Tab1.id AND t2_url.item_key='url'
ORDER BY title,
t2_val.item_value DESC,
t2_url.item_value LIKE '%123.com%' DESC
An SQLfiddle to test with.
A little complicated, because when you do the join you will get multiple rows. Here is an approach that aggregates tab2 before doing the join:
select t1.*
from Tab1 t1 left outer join
(select id,
max(case when item_key = 'value' then item_value end) as value,
max(case when item_key = 'url' then item_value end) as url
from Tab2 t2
group by id
) t2
on t1.id = t2.id
order by t1.title, t2.value desc,
(t2.url like '%123.com%') desc;
translations
+---------+----------------+----------+---------+
| id_user | id_translation | referrer | id_word |
+---------+----------------+----------+---------+
| 1 | 3 | NULL | 4 |
| 1 | 17 | NULL | 3 |
| 2 | 17 | NULL | 5 |
| 2 | 17 | NULL | 1 |
| 2 | 17 | NULL | 7 |
words
+----+------+
| id | word |
+----+------+
| 4 | out |
+----+------+
users_translations
+---------+----------------+----------+---------+
| id_user | id_translation | referrer | id_word |
+---------+----------------+----------+---------+
| 1 | 17 | 1 | 4 |
| 2 | 17 | 2 | 4 |
| 3 | 18 | NULL | 4 |
I need to select all translations for current word and id_translation, but if in the row referrer = 1 (current user), then I don't need another results (translations from another users for current word), if there is no referrer = 1, show all.
SELECT DISTINCT `t`.*, `ut`.`id_user` AS tuser
FROM translations AS t
LEFT JOIN users_translations AS ut ON `t`.`id` = `ut`.`id_translation`
INNER JOIN words ON `words`.`id` = `ut`.`id_word` OR `words`.`id` = `t`.`id_word`
WHERE (`word` = 'help')
ORDER BY `t`.`translation` ASC
+----+-------------+---------+---------+-------+
| id | translation | id_word | id_user | tuser |
+----+-------------+---------+---------+-------+
| 17 | допомагати | 4 | 1 | 2 |
| 17 | допомагати | 4 | 1 | 1 |
First row doesn't need, because we have tuser = 1. When there is no tuser = 1, all results should be returned.
I don't understand how to build select statement and I will be very appreciative that somebody shows me how to make it work.
First thing that comes to mind
--add this to your where clause
id_user <=
CASE WHEN EXISTS(SELECT * FROM translations WHERE id_user = 1 AND id_word = words.id_word)
THEN 1
ELSE (SELECT MAX(Id) FROM translations)
END
I've got a table that looks like:
Table 1 ->
+----+--------+--------+
| id | name | author |
+----+--------+--------+
| 1 | First | Me |
| 2 | Second | You |
+----+--------+--------+
Table 2 ->
+-----+------------+-----------+------------+
| mid | table1_id | key | value |
+-----+------------+-----------+------------+
| 1 | 1 | desc | hello |
| 2 | 1 | begin_day | monday |
| 3 | 1 | end_day | tuesday |
| 4 | 2 | desc | goodbye |
| 5 | 2 | begin_day | wednesday |
| 6 | 2 | end_day | friday |
+-----+------------+-----------+------------+
The relationship here is that the id in table 1 corresponds to the table1_id in table 2.
The output that I'm trying to get is...
+----+---------+---------+-------------+-----------+-----------+
| id | name | author | desc | begin_day | end_day |
+----+---------+---------+-------------+-----------+-----------+
| 1 | First | Me | hello | monday | tuesday |
| 1 | Second | You | goodbye | wednesday | friday |
+----+---------+---------+-------------+-----------+-----------+
I've tried several different join statements -- all a variation of the below. I'm not that well versed in MySQL queries, however.
SELECT * FROM table_1 LEFT JOIN table_2 on table_1.id = table_2.table1_id
Which produces...
+----+----------+----------+----------+------------+-----------+
| id | mid | name | author | key | value |
+----+----------+----------+----------+------------+-----------+
| 1 | 1 | First | Me | desc | hello |
| 1 | 2 | First | Me | begin_day | monday |
| 1 | 3 | First | Me | end_day | tuesday |
| 2 | 4 | Second | You | desc | goodbye |
| 2 | 5 | Second | You | begin_day | wednesday|
| 2 | 6 | Second | You | end_day | friday |
Obviously, iterating over this join statement produces 6 results, 1 for each row in table 2 that matches the id in table 1. How can I avoid this with a proper query statement?
Thank you in advance.
You can use a case statement if you know all of the columns you will be getting, as follows:
Select distinct table_1.*,
case when table_2.key='desc' then value end as desc,
case when table_2.key='begin_day' then value end as begin_day,
case when table_2.key='end_day' then value end as end_day
FROM table_1 LEFT JOIN table_2 on table_1.id = table_2.table1_id
Hope this helps!
SELECT
table_1.*,
MAX(IF(key='desc', value, NULL)) AS 'desc',
MAX(IF(key='begin_day', value, NULL)) AS begin_day,
MAX(IF(key='end_day', value, NULL)) AS end_day
FROM table_1
LEFT JOIN table_2 ON (id = table1_id)
GROUP BY id;