Exist a better way to do what the following SQL query does? I have the feeling that table1 will be searched twice and may be that can be avoided with some trick and increase the efficient of the query, but I just can't figure out how ;( Here is the query (in MySQL):
SELECT a, SUM(count)
FROM table1
GROUP BY a
HAVING SUM(count) = (SELECT SUM(count) as total FROM table1 GROUP BY a ORDER BY total DESC LIMIT 1)
The goal is return the number(s) with the major accumulate, with its accumulate.
being table1 a two field table like:
a,count
1,10
1,30
1,0
2,1
2,100
2,4
3,10
4,50
4,55
The result with that data sample is:
2,105
4,105
Thanks in advance.
SELECT a, total FROM
(SELECT a AS a, SUM(COUNT) AS total
FROM table1
GROUP BY a) AS xyz
HAVING total = MAX(total)
Hope this will work for you
This sub-query is executed only once, and you don't have to bother with creating any pre-query as other answers may suggest (although doing so this is still correct, just not needed). Database engine will realise, that the sub-query is not using any variable dependent on the other part of the query. You can use EXPLAIN to see how the query is executed.
More on the topic in this answer:
https://stackoverflow.com/a/658954/1821029
I think you could probably do it by moving your HAVING sub-select query into its on prequery. Since it will always include a single row, you won't require any "JOIN", and it does not have to keep recomputing the COUNT(*) every time the HAVING is applied. Do it once, then the rest
SELECT
a,
SUM(count)
FROM
table1,
( SELECT SUM(count) as total
FROM table1
GROUP BY a
ORDER BY total DESC
LIMIT 1 ) PreQuery
GROUP BY
a
HAVING
SUM(count) = PreQuery.Total
This query return one row with two columns:
1- a list of comma separated values of "a" column, which have the biggest "Total"
2- and the biggest Total value
select group_concat(a), Total
from
(select a, sum(count) as Total
from table1
group by a) OnTableQuery
group by Total
order by Total desc
limit 1
Note that it queries table1 just one time. The query was already tested.
Related
Similar to this issue: MySQL 5.7 group by latest record
I'm not sure how to do this properly in 5.7. Also with possibility of 2nd sort column. Working query in 5.6 that I'm trying to replicate in 5.7:
SELECT id FROM test
GROUP BY category
ORDER BY sort1 DESC, sort2 DESC
id is not always the highest, so MAX(id) does not work.
Looking into the link above, the solution for single sort should be:
SELECT t1.*
FROM test t1
INNER JOIN (
SELECT category, max(sort) AS sort FROM test GROUP BY category
) t2 ON t2.category = t1.category AND t2.sort = t1.sort
But how will it work with 2 sorting?
You are using GROUP BY the wrong way.
Think of group by as a way to separate data row into different groups. Each group has multiple rows, based on the value of group by column.
Once you get those groups, selecting table columns (as in: select *) is like picking any row from that group randomly. This is not helpful nor useful.
Usually once we group records (or rows), we need to find meta information about those records. For example: get us the count of records in that group (as in: select count(*)), or the sum of values of a specific column in that group (as in: select sum(price)), or get the min, max or avg values.
So in a nutshell, when you use group by you should use on of the aggregation functions with it, otherwise it's not going to do you any good.
Why don't you have the ORDER BY at your outer query, instead?
SELECT *
FROM (
SELECT 100 AS id, 1 AS category, NULL AS sort
UNION
SELECT 200 AS id, 1 AS category, 2 AS sort
) dt
GROUP BY category
ORDER BY sort DESC;
It seems that what happened to the data when it was grouped, it took the first data while neglecting the ORDER BY DESC. On your first query, it ordered descending first then group by took the first record which is 200. And yes, this shouldn't be the way you should use GROUP BY. It is used in conjunction with aggregate functions.
when you select a column in a group by query that is not one of the columns you are grouping by, (ie, your id) you have no control over the value unless you use another aggregate function. If you want to sort, use MIN or MAX:
SELECT MAX(id), category, FROM `test2`
GROUP BY category; -- always returns 200
SELECT MIN(id), category, FROM `test2`
GROUP BY category; -- always returns 100
How to get the total records details along with the query which is having a limit
My MySQL query is given below
SELECT name FROM employees LIMIT 10
which will give me 10 rows, but the total records will be around 6000. How can we get that also along with the above query
Use a subquery:
SELECT name, (SELECT COUNT(*) FROM employees) AS cnt
FROM employees
LIMIT 10
Use a Sub-query
SELECT nameFROM,
(select count(1) from employees) as total_count
From employees
LIMIT 10
or CROSS JOIN
SELECT nameFROM,total_count
From employees
CROSS JOIN (select count(1) total_count from employees) A
LIMIT 10
Adding ORDER BY to your query will make the result consistent
You can use SQL_CALC_FOUND_ROWS option in your query which will tell MySQL to count total number of rows disregarding LIMIT clause. You still need to execute a second query in order to retrieve row count, but it’s a simple query and not as complex as your query which retrieved the data. Usage is pretty simple. In you main query you need to add SQL_CALC_FOUND_ROWS option just after SELECT and in second query you need to use FOUND_ROWS() function to get total number of rows
SELECT SQL_CALC_FOUND_ROWS name FROM employees WHERE LIMIT 10
SELECT FOUND_ROWS();
(SELECT NAME, 1 AS 'Obs' FROM employees LIMIT 10)
union select 'RecsinTable', count(*) from employees
How would I combine these two SQL queries into a single query for greater speed and efficiency?
SELECT price FROM table LIMIT 30
SELECT AVG(price) as avg FROM table
It will not be faster or more efficient, but I can see needs for it (such as trying to compare a value to the average), something like this:
SELECT t.value, avgSubQ.theAvg, t.value / avgSubQ.theAvg AS relativeToAvg
FROM table AS t
INNER JOIN (SELECT AVG(value) AS theAvg FROM table) AS avgSubQ
LIMIT 30;
as avg will generate 1 record, you could just use a cross join.
SELECT price, B.mavg
FROM table
CROSS JOIN (SELECT AVG(price) as mavg FROM table) B
LIMIT 30
Try this query
SELECT price,avg(price) as avg from table LIMIT 30
Here every row will duplicate the average price, but you'll get what you want in one query.
In my SQL query I am selecting data with GROUP BY and ORDER BY clauses. The table has the same numbers across multiple rows with different times in each row. So I think I want to apply a GROUP BY clause.
However in the results return the oldest time with the number, but I need the most recent time.
SELECT * FROM TABLE GROUP BY (numbers) ORDER BY time DESC
The query appears as if it should first apply GROUP BY and then ORDER BY... but the results do not appear to work this way.
Is there any way to fix this?
SELECT *
FROM table t
WHERE time = (
SELECT max(time)
FROM table
WHERE t.numbers = numbers
)
work-around is to re-write the query as:
SELECT * FROM (SELECT * FROM table ORDER BY time DESC) AS t GROUP BY numbers;
SELECT * FROM table
WHERE time IN (
SELECT MAX(time)
FROM table
GROUP BY numbers
)
According to the manual you can add desc to the group by list:
Example:
group by item1, item2 desc, item3
with or without rollup.
I've tried this and it works in Ubuntu version 5.5.58. The reference page is:
https://dev.mysql.com/doc/refman/5.7/en/group-by-modifiers.html
SELECT * FROM TABLE GROUP BY numbers DESC;
This will give you last record from group.
Thanks
I'am trying to understand what causes the following, maybe you could help me:
I have a query like:
select field1,fieldDate from table1
union all
select field1,fieldDate from table2
order by fieldDate desc
and the another one like this:
select field1,field2,fieldDate from table1
union all
select field1,field2,fieldDate from table2
order by fieldDate desc
So basically they are the same with the exception that in the second I retrieve an extra field.
Now, both results come with a diferent ordering, but just for the cases that the dates are exacly the same. For example there are 2 rows (row1,row2) with date 2009-11-25 09:41:55. For query 1 row1 comes before row2 and for query 2 row2 comes before row1.
Does somebody knows why this happens?
Thanks,
Regards
The ordering based on any fields that you don't explicitly order by is undefined, and the optimizer can change the ordering if it thinks that results in a better execution plan. Given two rows with the exact same value in the order by field you can not depend on them being in any particularly order in relation to each other unless you explicitly order by another field with different values.
Can you do this
select * from ( select
field1,field2,fieldDate, 0 as ordercol from table1
union all select
field1,field2,fieldDate, 1 as ordercol from table2) t1
order by fieldDate desc, ordercol asc
Straight from the MySQl manual, to user order by on a union you have to parenthesis the individual tables.
(select field1,fieldDate from table1)
union all
(select field1,fieldDate from table2)
order by fieldDate desc
This is not SQL standards compliant! The code you entered should order the union of both tables but to my surprise MySQL has the above syntax.
The order in which rows with the same fieldDate are returned can differ for each query execution. Usually this order will be the same but you should not count on it. If you want any extra ordering state more order by fields.
EDIT: This answer is wrong: the order by works on the entire union. I'll leave it here to save others the trouble :)
Your order by only works on the second part of the union. You can use a subquery to make the order by work on the entire union:
select field1,field2,fieldDate
from (
select field1,field2,fieldDate
from table1
union all
select field1,field2,fieldDate
from table2
) SubQueryName
order by fieldDate desc