I have simple mysql table with 2 fields:
atr_id | atr_val_id
1 | 100
1 | 200
1 | 300
2 | 100
3 | 100
3 | 200
4 | 200
How can I select for example all atr_ids values that have atr_val_id = 100 AND 200 and nothing more ? This will be only atr_id = 3 in that example.
Or for example only 200, this will be atr_id = 4
There could be be simpler ways to do. Here goes one solution that uses mix of group by, group_contact, distinct, sort and having clause to fetch your desired result
All atr_ids values that have atr_val_id = 100 AND 200 and nothing more. This will return only atr_id = 3
select
atr_id
from
tbl_test
group by
atr_id
having
group_concat(distinct atr_val_id order by atr_val_id asc) = '100,200'
and for only 200, this will return atr_id = 4
select
atr_id
from
tbl_test
group by
atr_id
having
group_concat(distinct atr_val_id order by atr_val_id asc) = '200'
Another way is to use the NOT EXISTS as below
select atr_id
from table_name t1
where atr_val_id in(100,200)
AND NOT EXISTS
(
select 1
from table_name t2
where t2.atr_id = t1.atr_id
group by t2.atr_id
having count(t2.atr_val_id) < 2 OR count(t2.atr_val_id) > 2
)
group by t1.atr_id
;
You can change the count part depending on the number of atr_val_id you are looking at. Say for just 200 its going to be
having count(t2.atr_val_id) < 1 OR count(t2.atr_val_id) > 1
And subsequently you need to change the IN clause as well.
Related
I'm trying to select the sum of the values in the isOK column for each Name separated, BUT only if isOK = 1 on Day = 2.
The query for the following example table tablename
Name | Day | isOK
char | int | int
-----------------
Flo | 1 | 1
Seb | 1 | 1
Tim | 1 | 0
Flo | 2 | 1
Seb | 2 | 0
Tim | 2 | 1
should give Flo: 2 and Tim: 1, but not Seb: 1, since his isOK on Day = 2 is 0.
I've tried using SUM(isOK) with IF constructs, but it's just not working. My alternative solution, to select all Name where isOK = 1 first and select the SUM(isOK) for each of the names is slow and seems in need of improvement.
I guess it's not that difficult, but I've been trying for hours now and I just can't combine my two queries into one.
One way to do this is to use a conditional expression together with a having clause like this:
select name, sum(isOk) ok_sum
from your_table
group by name
having sum(case when day = 2 and isOK = 1 then 1 else 0 end) > 0;
With your sample data the result would be:
name ok_sum
Flo 2
Tim 1
As MySQL evaluates boolean expressions as 1 or 0 it should be possible to reduce the condition to this:
having sum(day = 2 and isOK = 1) > 0;
Another way to do it would be to use a correlated subquery that makes sure there exists a row with Day = 2 and isOk = 1 for the Name:
select t1.name, sum(t1.isOk) ok_sum
from your_table t1
where exists (
select 1
from your_table t2
where t2.day = 2 and t2.isOK = 1 and t1.name = t2.name
)
group by t1.name
See this fiddle
TRY this :
SELECT
name, SUM(isok) AS isOk
FROM
table
GROUP BY `name`
HAVING SUM(`day` = 2 AND isok = 1) > 0;
SELECT x.name, SUM(y.isOK) total
FROM my_table x
JOIN my_table y
ON y.name = x.name
WHERE x.day = 2
AND x.isok=1
GROUP
BY x.name;
Is it possible to select the next lower number from a table without using limit.
Eg: If my table had 10, 3, 2 , 1 I'm trying to select * from table where col > 10.
The result I'm expecting is 3. I know I can use limit 1, but can it be done without that?
Try
SELECT MAX(no) no
FROM table1
WHERE no < 10
Output:
| NO |
------
| 3 |
SQLFiddle
Try this query
SELECT
*
FROM
(SELECT
#rid:=#rid+1 as rId,
a.*
FROM
tbl a
JOIN
(SELECT #rid:=0) b
ORDER BY
id DESC)tmp
WHERE rId=2;
SQL FIDDLE:
| RID | ID | TYPE | DETAILS |
------------------------------------
| 2 | 28 | Twitter | #sqlfiddle5 |
Another approach
select a.* from supportContacts a inner join
(select max(id) as id
from supportContacts
where
id in (select id from supportContacts where id not in
(select max(id) from supportContacts)))b
on a.id=b.id
SQL FIDDLE:
| ID | TYPE | DETAILS |
------------------------------
| 28 | Twitter | #sqlfiddle5 |
Alternatively, this query will always get the second highest number based on the inner where clause.
SELECT *
FROM
(
SELECT t.col,
(
SELECT COUNT(distinct t2.col)
FROM tableName t2
WHERE t2.col >= t.col
) as rank
FROM tablename t
WHERE col <= 10
) xx
WHERE rank = 2 -- <<== means second highest
SQLFiddle Demo
SQLFiddle Demo (supports duplicate values)
If you want to get next lower number from table
you can get it with this query:
SELECT distinct col FROM table1 a
WHERE 2 = (SELECT count(DISTINCT(b.col)) FROM table1 b WHERE a.col >= b.col);
later again if you want to get third lower number you can just pass 3 in place of 2 in where clause
again if you want to get second higher number, just change the condition of where clause in inner query with
a.col <= b.col
I need help in finding the rows that correspond to the most recent date, the next most recent and the one after that, where some condition ABC is "Y" and group it by a column name XYZ ASC but XYZ can appear multiple times. So, say XYZ is 50, then for the rows in the three years, the XYZ will be 50. I have the following code that executes but returns only two rows out of thousands which is impossible. I tried executing just the date condition but it returned dates that were less than or equal to MAX(DATE)-3 as well. Don't know where I am going wrong.
select * from money.cash where DATE =(
select
MAX(DATE)
from
money.cash
where
DATE > (select MAX(DATE)-3 from money.cash)
)
GROUP BY XYZ ASC
having ABC = "Y";
The structure of the table is as follows (only a schematic, not the real thing).
Comp_ID DATE XYZ ABC $$$$ ....
1 2012-1-1 10 Y SOME-AMOUNT
2 2011-1-1 10 Y
3 2006-1-1 10 Y
4 2011-1-1 20 Y
5 2002-1-1 20 Y
6 2000-1-1 20 Y
7 1998-1-1 20 Y
The desired o/p would be the first three rows for XYZ=10 in ascending order and the most recent 3 dates for XYZ=20.
LAST AND IMPORTANT-This table's values keeps changing as new data comes in. So, the o/p(which will be in a new table) must reflect the dynamics in the 1st/original/above TABLE.
MySQL doesn't have functionallity that is friendly to greatest-n-per-group queries.
One option would be...
- Find the MAX(Date) per group (XYZ)
- Then use that result to find the MAX(Date) of all records before that date
- Then do it again for all records before that date
It's really innefficient, but MySQL hasn't got the functionality required to do this efficiently. Sorry...
CREATE TABLE yourTable
(
comp_id INT,
myDate DATE,
xyz INT,
abc VARCHAR(1)
)
;
INSERT INTO yourTable SELECT 1, '2012-01-01', 10, 'Y';
INSERT INTO yourTable SELECT 2, '2011-01-01', 10, 'Y';
INSERT INTO yourTable SELECT 3, '2006-01-01', 10, 'Y';
INSERT INTO yourTable SELECT 4, '2011-01-01', 20, 'Y';
INSERT INTO yourTable SELECT 5, '2002-01-01', 20, 'Y';
INSERT INTO yourTable SELECT 6, '2000-01-01', 20, 'Y';
INSERT INTO yourTable SELECT 7, '1998-01-01', 20, 'Y';
SELECT
yourTable.*
FROM
(
SELECT
lookup.XYZ,
COALESCE(MAX(yourTable.myDate), lookup.MaxDate) AS MaxDate
FROM
(
SELECT
lookup.XYZ,
COALESCE(MAX(yourTable.myDate), lookup.MaxDate) AS MaxDate
FROM
(
SELECT
yourTable.XYZ,
MAX(yourTable.myDate) AS MaxDate
FROM
yourTable
WHERE
yourTable.ABC = 'Y'
GROUP BY
yourTable.XYZ
)
AS lookup
LEFT JOIN
yourTable
ON yourTable.XYZ = lookup.XYZ
AND yourTable.myDate < lookup.MaxDate
AND yourTable.ABC = 'Y'
GROUP BY
lookup.XYZ,
lookup.MaxDate
)
AS lookup
LEFT JOIN
yourTable
ON yourTable.XYZ = lookup.XYZ
AND yourTable.myDate < lookup.MaxDate
AND yourTable.ABC = 'Y'
GROUP BY
lookup.XYZ,
lookup.MaxDate
)
AS lookup
INNER JOIN
yourTable
ON yourTable.XYZ = lookup.XYZ
AND yourTable.myDate >= lookup.MaxDate
WHERE
yourTable.ABC = 'Y'
ORDER BY
yourTable.comp_id
;
DROP TABLE yourTable;
There are other options, but they're all a bit hacky. Search SO for greatest-n-per-group mysql.
My results using your example data:
Comp_ID | DATE | XYZ | ABC
------------------------------
1 | 2012-1-1 | 10 | Y
2 | 2011-1-1 | 10 | Y
3 | 2006-1-1 | 10 | Y
4 | 2011-1-1 | 20 | Y
5 | 2002-1-1 | 20 | Y
6 | 2000-1-1 | 20 | Y
Here's another way, hopefully more efficient than Dems' answer.
Test it with an index on (abc, xyz, date):
SELECT m.xyz, m.date --- for all columns: SELECT m.*
FROM
( SELECT DISTINCT xyz
FROM money.cash
WHERE abc = 'Y'
) AS dm
JOIN
money.cash AS m
ON m.abc = 'Y'
AND m.xyz = dm.xyz
AND m.date >= COALESCE(
( SELECT im.date
FROM money.cash AS im
WHERE im.abc = 'Y'
AND im.xyz = dm.xyz
ORDER BY im.date DESC
LIMIT 1
OFFSET 2 --- to get 3 latest rows per xyz
), DATE('1000-01-01') ) ;
If you have more than rows with same (abc, xyz, date), the query may return more than 3 rows per xyz (all tied in 3rd place will all be shown).
I am writing a query to grab the items that a specific user_id was the first to use. Here is some sample data -
item_id used_user_id date_used
1 1 2012-08-25
1 2 2012-08-26
1 3 2012-08-27
2 2 2012-08-27
3 1 2012-08-27
4 1 2012-08-21
4 3 2012-08-24
5 3 2012-08-23
query
select item_id as inner_item_id, ( select used_user_id
from test
where test.item_id = inner_item_id
order by date_used asc
limit 1 ) as first_to_use_it
from test
where used_user_id = 1
group by item_id
It returns the correct values
inner_item_id first_to_use_it
1 1
3 1
4 1
but the query is VERY slow on a giant table. Is there a certain index that I can use or a better query that I can write?
i can't get exactly what you mean because in your inner query you have sorted it by their used_user_id and and on your outer query you have filtered it also by their userid. Why not do this directly?
SELECT DISTINCT item_id AS inner_item_id,
used_user_id AS first_to_use_it
FROM test
WHERE used_user_id = 1
UPDATE 1
SELECT b.item_id,
b.used_user_id AS first_to_use_it
FROM
(
SELECT item_ID, MIN(date_used) minDate
FROM tableName
GROUP BY item_ID
) a
INNER JOIN tableName b
ON a.item_ID = b.item_ID AND
a.minDate = b.date_used
WHERE b.used_user_id = 1
x_Id | y_Id | z_Id
----- |----- |-----
1 | 1 | 1
2 | 1 | 1
3 | 1 | 1
4 | 1 | 1
5 | 1 | 1
1 | 2 | 3
I am relatively new at programming and I cant figure out this MySql query. I need to select x_Id only where ((y_Id = 1 AND z_Id = 1) AND (y_Id = 2 AND z_Id = 3)).
Therefore, using these numbers as an example the only thing that should be selected is (x_Id =) 1.
**All of these columns are in the same table
The closest I have come is by using this query:
SELECT
*
FROM
`relationships`
WHERE
y_id = 1 AND
z_id = 1
UNION
SELECT
*
FROM
`relationships`
WHERE
z_id = 3 AND
y_id = 2
However, this returns all the x_ids and x_id = 1 again as a duplicate.
**I am using sqlPro and MySql 5
no need to Union.
Updated after seeing comments:
select
*
from relationships T1
INNER JOIN relationshipsT2 on t1.x_Id = t2.x_Id where
((T1.y_Id = 1 AND T1.z_Id = 1) AND (T2.y_Id = 2 AND T2.zz_Id= 3))
also you can only return x_Id instead of *
If you are only interested in the x_id value you can use the query above, but just add DISTINCT and project only the x_id value.
Example:
SELECT
DISTINCT x_id
FROM
`relationships`
WHERE
y_id = 1 AND z_id = 1
UNION
SELECT
DISTINCT x_id
FROM
`relationships`
WHERE
z_id = 3 AND y_id = 2
There are few other way how to do it, which are even easier such as use OR in the WHERE clause.
Updated after seeing comments:
Using an aggregate SUM() you can total up the number of conditions met per value of x_id. If the total is > 1, both conditions are met somewhere in the table.
SELECT DISTINCT x_id FROM (
SELECT
x_id,
SUMCASE WHEN (y_id = 1 AND z_id = 1) OR (z_id = 3 AND y_id = 2) THEN 1 ELSE 0 END) AS hasboth
FROM relationships
GROUP BY x_id
HAVING hasboth > 1
) subq