I pretty much forgot how to do this with SQL actually, the thing is I have an SQL select statement such as:
SELECT COUNT(*) NAME FROM `SomeTable` WHERE `SomeID` = xxx GROUP BY `Field`
Which does return a table with one field and many records containing numbers. What I want is to get the a single value of the maximum number among those records.
it should help you. just use MAX() function
SELECT MAX(COUNT(*)) FROM `SomeTable` WHERE `SomeID` = xxx GROUP BY `Field`
Just add LIMIT:
SELECT COUNT(`NAME`) AS `NUM` FROM `SomeTable` WHERE `SomeID` = xxx GROUP BY `Field` ORDER BY `NUM` DESC LIMIT 1
select max(name) from
(
SELECT COUNT(*) NAME FROM `SomeTable` WHERE `SomeID` = xxx GROUP BY `Field`)a
SELECT COUNT(*) NAME
FROM `SomeTable`
WHERE `SomeID` = xxx
GROUP BY `Field`
order by NAME desc
limit 1
I want is to get the a single value of the maximum number among those
records.
Try this:
SELECT t1.*
FROM SomeTable t1
(
SELECT SomeID, Max(Field) MaxField
FROM SomeTable
GROUP BY SomeID
) t2 ON t1.SomeID = t2.SomeID AND t1.Field = t2.MaxField
WHERE t1.SomeID = xxx
The syntax for the MAX function is:
SELECT MAX(expression )
FROM tables
WHERE predicates;
Example
SELECT MAX(salary) as "Highest salary"
FROM employees;
If you just want a single result of the highest value against a filter, just use Max:
SELECT MAX(MyField) NAME
FROM `SomeTable`
WHERE `SomeID` = xxx;
Alternatively, if you want an aggregate grouped by another column, the syntax is like so:
SELECT Name, MAX(MyField) as MaxOfMyField -- Or COUNT(MyField) if you want the Count
FROM `SomeTable`
WHERE `SomeID` = xxx
GROUP By Name;
Related
This is the result of a UNION of two SELECT
SELECT count(*) FROM
((SELECT session_id_current_user from test.tws_analytics
WHERE (add_date BETWEEN '2022-05-15' AND '2022-05-15') AND ((pathURL='vues/login.php' AND name_current_user='') OR (pathURL='' AND searchURL='?job=forgotten' AND name_current_user=''))
AND session_id_current_user NOT IN
(SELECT session_id_current_user from test.tws_analytics
WHERE (pathURL <> 'vues/login.php' AND searchURL <> '?job=forgotten') AND add_date BETWEEN '2022-05-15' AND '2022-05-15' order by session_id_current_user)
order by session_id_current_user)
UNION
(SELECT name_current_user from test.tws_analytics where add_date BETWEEN '2022-05-15' AND '2022-05-15' AND name_current_user IS NOT NULL AND name_current_user <> ''))
AS tem
The result is 11.
What I want to do is to select this result with other columns like this :
SELECT count(session),count(name), [AND tem.count(*)] FROM ....
This is the general idea, though i didn't know how to implement it.
a simplified general answer would be
select * from (select count(*) numsessions from sessions), (select count(*) numusers from users)
this will give 2 different counts, i didn't include the logics that you provided, but that will need to be done inside the 2 subqueries.
I have a simple group by query:
SELECT timestamp, COUNT(users)
FROM my_table
GROUP BY users
How do I add a sum_each_day column that will sum the users count of each row and will aggregate it forward to the next row and so on
The output should be like this:
timestamp | users | sum_each_day
2015-11-27 1 1
2015-11-28 5 6
2015-11-29 3 9
2015-11-30 7 16
Thanks in advance
You could use a sub-query, like this:
SELECT timestamp,
num_users,
(SELECT COUNT(users)
FROM my_table
WHERE timestamp <= main.timestamp) sum_users
FROM (
SELECT timestamp,
COUNT(users) num_users
FROM my_table
GROUP BY timestamp
) main
If you really need this in mysql it'll cost some performance but i believe a sub query with a count will solve it:
SELECT t1.timestamp, count (), select count () from my_table t2 where t2.timestamp <= t1.timestamp From my_table t1 Group by users
If you display this data through a scripting language like PHP it would be easier to keep a counter and display the aggregate per row.
I would do this using variable:
SET #total := 0;
SELECT timestamp, DayCount, (#total := #total + DayCount) AS Total
FROM
(SELECT timestamp, COUNT(users) AS DayCount
FROM my_table
GROUP BY timestamp) AS t1
Fiddler: I am not using your table structure here, but you can get idea
If I understand correclty, this will work:
set #c=0;
SELECT `timestamp`,sum(`users`),(select #c:=#c+sum(`users`))
FROM `my_table`
group by `timestamp`;
I have one table and i want to check that for one column all value are same.
following is the entry in my table.
two column
rid,value
(1,1)
(1,1)
(2,1)
(2,0)
(2,0)
(3,0)
(3,0)
(3,0)
I want query which gives me rid 1 because all of its value is 1. all record for rid 1 has value 1 and rid 2 and 3 does not has all value as 1 so they should not be selected.
Using group by and having can get what you want:
SELECT rid, value
FROM my_table
GROUP BY rid
HAVING COUNT( distinct value) = 1
UPDATE
According to the comment, filter the value will get the result:
SELECT *
FROM
(
SELECT rid, value
FROM my_table
GROUP BY rid
HAVING COUNT( distinct value) = 1
) AS T1
WHERE value = 1
If the values would only be 1 or 0, then you could do this trick:
SELECT rid, value
FROM my_table
GROUP BY rid
HAVING COUNT( * ) = SUM(value)
You can do like this:
CREATE TABLE my_table (
id varchar(255),
col_value varchar(255)
);
INSERT INTO my_table
VALUES
('1','1'),
('1','1'),
('2','1'),
('2','1'),
('2','1'),
('2','4'),
('3','1'),
('3','1');
Query for selection:
SELECT src.* FROM
(
SELECT DISTINCT t1.* FROM my_table AS t1
) AS src
WHERE src.id NOT IN(
SELECT test.id
FROM
(
SELECT DISTINCT t1.* FROM my_table AS t1
) AS test
GROUP BY test.id
HAVING COUNT(*) > 1
)
fiddle here.
Let's say I was looking for the second most highest record.
Sample Table:
CREATE TABLE `my_table` (
`id` int(2) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`value` int(10),
PRIMARY KEY (`id`)
);
INSERT INTO `my_table` (`id`, `name`, `value`) VALUES (NULL, 'foo', '200'), (NULL, 'bar', '100'), (NULL, 'baz', '0'), (NULL, 'quux', '300');
The second highest value is foo. How many ways can you get this result?
The obvious example is:
SELECT name FROM my_table ORDER BY value DESC LIMIT 1 OFFSET 1;
Can you think of other examples?
I was trying this one, but LIMIT & IN/ALL/ANY/SOME subquery is not supported.
SELECT name FROM my_table WHERE value IN (
SELECT MIN(value) FROM my_table ORDER BY value DESC LIMIT 1
) LIMIT 1;
Eduardo's solution in standard SQL
select *
from (
select id,
name,
value,
row_number() over (order by value) as rn
from my_table t
) t
where rn = 1 -- can pick any row using this
This works on any modern DBMS except MySQL. This solution is usually faster than solutions using sub-selects. It also can easily return the 2nd, 3rd, ... row (again this is achievable with Eduardo's solution as well).
It can also be adjusted to count by groups (adding a partition by) so the "greatest-n-per-group" problem can be solved with the same pattern.
Here is a SQLFiddle to play around with: http://sqlfiddle.com/#!12/286d0/1
This only works for exactly the second highest:
SELECT * FROM my_table two
WHERE EXISTS (
SELECT * FROM my_table one
WHERE one.value > two.value
AND NOT EXISTS (
SELECT * FROM my_table zero
WHERE zero.value > one.value
)
)
LIMIT 1
;
This one emulates a window function rank() for platforms that don't have them. It can also be adapted for ranks <> 2 by altering one constant:
SELECT one.*
-- , 1+COALESCE(agg.rnk,0) AS rnk
FROM my_table one
LEFT JOIN (
SELECT one.id , COUNT(*) AS rnk
FROM my_table one
JOIN my_table cnt ON cnt.value > one.value
GROUP BY one.id
) agg ON agg.id = one.id
WHERE agg.rnk=1 -- the aggregate starts counting at zero
;
Both solutions need functional self-joins (I don't know if mysql allows them, IIRC it only disallows them if the table is the target for updates or deletes)
The below one does not need window functions, but uses a recursive query to enumerate the rankings:
WITH RECURSIVE agg AS (
SELECT one.id
, one.value
, 1 AS rnk
FROM my_table one
WHERE NOT EXISTS (
SELECT * FROM my_table zero
WHERE zero.value > one.value
)
UNION ALL
SELECT two.id
, two.value
, agg.rnk+1 AS rnk
FROM my_table two
JOIN agg ON two.value < agg.value
WHERE NOT EXISTS (
SELECT * FROM my_table nx
WHERE nx.value > two.value
AND nx.value < agg.value
)
)
SELECT * FROM agg
WHERE rnk = 2
;
(the recursive query will not work in mysql, obviously)
You can use inline initialization like this:
select * from (
select id,
name,
value,
#curRank := #curRank + 1 AS rank
from my_table t, (SELECT #curRank := 0) r
order by value desc
) tb
where tb.rank = 2
SELECT name
FROM my_table
WHERE value < (SELECT max(value) FROM my_table)
ORDER BY value DESC
LIMIT 1
SELECT name
FROM my_table
WHERE value = (
SELECT min(r.value)
FROM (
SELECT name, value
FROM my_table
ORDER BY value DESC
LIMIT 2
) r
)
LIMIT 1
When I tried:
select *
from some_table
group by table_id
order by timestamp desc;
I am gettings rows which have least timestamp values for that particular table_id(which I use for grouping)
How can I get the rows which have highest timestamp values for that particular table_id.
I also tried:
select *
from some_table
group by table_id
having max(timestamp)
order by timestamp desc;
which gives the same result as in the 1st case.
Thank You
select *
from your_table t
inner join
(
select table_id, max(created_timestamp) as mts
from your_table
group by table_id
) x on x.table_id = t.table_id
and x.mts = t.created_timestamp
In MySQL you can do
select *, max(created_timestamp) as mts
from your_table
group by table_id
but that will not make sure you get the corresponding data to your max(created_timestamp) but only to your table_id
SELECT * FROM (SELECT * FROM some_table ORDER BY timestamp) t1 GROUP BY t1.id