I want to search TABLE1 and count which number_id has the most 5's in experience column.
TABLE1
+-------------+------------+
| number_id | experience |
+-------------+------------+
| 20 | 5 |
| 20 | 5 |
| 19 | 1 |
| 18 | 2 |
| 15 | 3 |
| 13 | 1 |
| 10 | 5 |
+-------------+------------+
So in this case it would be number_id=20
Then do an inner join on TABLE2 and map the number that matches the number_id in TABLE1.
TABLE2
+-------------+------------+
| id | number |
+-------------+------------+
| 20 | 000000000 |
| 29 | 012345678 |
| 19 | 123456789 |
| 18 | 223456789 |
| 15 | 345678910 |
| 13 | 123457898 |
| 10 | 545678910 |
+-------------+------------+
So the result would be:
000000000 (2 results of 5)
545678910 (1 result of 5)
So far I have:
SELECT number, experience, number_id, COUNT(*) AS SUM FROM TABLE1
INNER JOIN TABLE2 ON TABLE1.number_id = TABLE2.id
WHERE experience = '5' order by SUM LIMIT 10
But it's returning just
545678910
How can I get it to return both results and by order of number of instances of 5 in the experience column?
Thanks
This query will give you the results that you want. The subquery fetches all the number_id that have experience values of 5. The SUM(experience=5) works because MySQL uses a value of 1 for true and 0 for false. The results of the subquery are then joined to table2 to give the number field. Finally the results are ordered by the number of experience=5:
SELECT t2.number, t1.num_fives
FROM (SELECT number_id, SUM(experience = 5) AS num_fives
FROM table1
WHERE experience = 5
GROUP BY number_id) t1
JOIN table2 t2
ON t2.id = t1.number_id
ORDER BY num_fives DESC
Output:
number num_fives
000000000 2
545678910 1
SQLFiddle Demo
Add a group by clause:
SELECT number, experience, number_id, COUNT(*) AS SUM
FROM TABLE1
JOIN TABLE2 ON TABLE1.number_id = TABLE2.id
WHERE experience = '5'
GROUP BY 1, 2, 3 -- <<< Added this clause
ORDER BY SUM
LIMIT 10
Related
Is it possible to use LIMIT based on another column inside a subquery in MySQL? Here is a working query of what I mean.
SELECT id, name,
(SELECT AVG(value) FROM t2 WHERE t1id = t1.id ORDER BY value DESC LIMIT 4) as average
FROM t1
However I'd like to replace the "4" to a field inside t1.
Something like this where table t1 has fields id, name, size:
SELECT id, name,
(SELECT AVG(value) FROM t2 WHERE t1id = t1.id ORDER BY value DESC LIMIT t1.size) as average
FROM t1
I could join t1 and t2, but I'm not sure that works for this. Does it?
Edit:
Here's some sample data to show what I mean:
Table t1
| id | name | Size |
|----|------|------|
| 1 | Bob | 4 |
| 2 | Joe | 3 |
| 3 | Sam | 4 |
Table t2
| t1id | value |
|------|-------|
| 1 | 16 |
| 1 | 14 |
| 1 | 12 |
| 1 | 10 |
| 1 | 8 |
| 2 | 10 |
| 2 | 8 |
| 2 | 6 |
| 2 | 4 |
| 3 | 20 |
| 3 | 15 |
| 3 | 10 |
| 3 | 5 |
| 3 | 2 |
Expected result:
| id | name | avg |
|----|------|------|
| 1 | Bob | 13 |
| 2 | Joe | 8 |
| 3 | Sam | 12.5 |
Notice that the average is the average of only the top t1.size values. For example the average for Bob is 13 and not 12 (based on 4 values and not 5) and the average for Joe is 8 and not 7 (based on 3 values and not 4).
In MySQL, you have little choice other than LEFT JOIN and aggregation:
SELECT t1.id, t1.name, AVG(t2.value) as average
FROM t1 LEFT JOIN
(SELECT t2.*,
ROW_NUMBER() OVER (PARTITION BY t1id ORDER BY VALUE desc) as seqnum
FROM t2
) t2
on t2.t1id = t1.id AND seqnum <= t1.size
GROUP BY t1.id, t1.name;
Here is a db<>fiddle.
No, you cannot use a column reference in a LIMIT clause.
https://dev.mysql.com/doc/refman/8.0/en/select.html has detailed documentation about MySQL's SELECT statement including all its clauses.
It says:
The LIMIT clause can be used to constrain the number of rows returned by the SELECT statement. LIMIT takes one or two numeric arguments, which must both be nonnegative integer constants, with these exceptions:
Within prepared statements, LIMIT parameters can be specified using ? placeholder markers.
Within stored programs, LIMIT parameters can be specified using integer-valued routine parameters or local variables.
Expressions, including subqueries, are not mentioned as legal argument in the LIMIT clause.
A simple solution would be to do your task in two queries: the first to get the size and then use that value as a constant value in the second query that includes the LIMIT.
Not every task needs to be done in a single SQL statement.
Today I have posted a question and got a good answer: Stuck in building mysql query.
I though it helped me, but I've discovered that it returns wrong data. So I'm reposting the question here, with an answer I received, as well I will explain the problem why it is not working for me.
Example of data:
id | item_id | user_id | bid_price
----------------------------------
1 | 1 | 11 | 1
2 | 1 | 12 | 2
3 | 1 | 13 | 3
4 | 1 | 14 | 1
5 | 1 | 15 | 4
6 | 2 | 16 | 2
7 | 2 | 17 | 1
8 | 3 | 18 | 2
9 | 3 | 19 | 3
10 | 3 | 18 | 2
Expected result:
id | item_id | user_id | bid_price
----------------------------------
1 | 1 | 11 | 1
7 | 2 | 17 | 1
8 | 3 | 18 | 2
Offered solution:
select m.id, m.item_id, m.user_id, m.bid_price
from my_table m
inner join (
select item_id, min(id) min_id, min(bid_price) min_price
from my_table
where item_id IN (1,2,3)
group by item_id
) t on t.item_id = m.item_id
and t.min_price= m.bid_price
and t.min_id = m.id
The problem:
In the sub query the minimum ID is selected entire the group by (item_id) statement and doesn't reflects according to minimum bid_price.
In other words, the minimum id is selected not depending on the price field at all. So, in the result I will get minimum price and minimum id of the group, but this will not be the same row! The id can be related to the row with another bet_price value.
How this query can be adjusted? Thank you in advance!
SELECT min(m.id) AS id, m.item_id, m.user_id, m.bid_price
FROM my_table m
INNER JOIN (
SELECT item_id, min(bid_price) AS min_price
FROM my_table
GROUP BY item_id
) t ON t.item_id = m.item_id
AND t.min_price= m.bid_price
GROUP BY item_id
Output
id item_id user_id bid_price
1 1 11 1
7 2 17 1
8 3 18 2
Live Demo
http://sqlfiddle.com/#!9/a52dc6/13
SELECT DISTINCT
t1.item_id,
t1.bid_price
FROM tab1 t1
WHERE NOT exists(SELECT 1
FROM tab1 t2
WHERE t2.item_id = t1.item_id
AND t2.bid_price < t1.bid_price)
AND t1.item_id IN (1, 2, 3);
http://sqlfiddle.com/#!9/615e0a/5
I have the following table:
+----+-----------+-----------+
| id | teacherId | studentId |
+----+-----------+-----------+
| 1 | 1 | 4 |
| 2 | 1 | 2 |
| 3 | 1 | 1 |
| 4 | 1 | 3 |
| 5 | 2 | 2 |
| 6 | 2 | 1 |
| 7 | 2 | 3 |
| 8 | 3 | 9 |
| 9 | 3 | 6 |
| 10 | 1 | 6 |
+----+-----------+-----------+
I need a query to find two teacherId's with maximum number of common studentId's.
In this case teachers with teacherIds 1,2 have common students with studentIds 2, 1, 3, which is greater than 1,3 having common students 6.
Thanks in Advance!
[Edit]: After several hours I've had the following solution:
SELECT * FROM (
SELECT r1tid, r2tid, COUNT(r2tid) AS cnt
FROM (
SELECT r1.teacherId AS r1tid, r2.teacherId AS r2tid
FROM table r1
INNER JOIN table r2 ON r1.studentId=r2.studentId AND r1.teacherId!=r2.teacherId
ORDER BY r1tid
) t
GROUP BY r1tid, r2tid
ORDER BY cnt DESC
) t GROUP BY cnt ORDER BY cnt DESC LIMIT 1;
I was sure that there must exist more short and elegant solution, but I could not find it.
You would do this with a self-join. Assuming no duplicates in the table:
select t.teacherid, t2.teacherid, count(*) as NumStudentsInCommon
from table t join
table t2
on t.studentid = t2.studentid and
t.teacherid < t2.teacherid
group by t.teacherid, t2.teacherid
order by NumStudentsInCommon desc
limit 1;
If you had duplicates, you would just replace count(*) with count(distinct studentid), but count(distinct) requires a bit more work.
select t.teacherId, t2.teacherId, sum(t.studentId) as NumStudentsInCommon
from table1 t join
table1 t2
on t.studentId = t2.studentId and
t.teacherId < t2.teacherId
group by t.teacherId, t2.teacherId
order by NumStudentsInCommon desc
I have a MySQL table like this.
| season_id | round_1 | names | score_round_1
| 5 | 10 | John1 | 5
| 5 | 10 | John2 | 3
| 5 | 11 | John3 | 2
| 5 | 11 | John4 | 5
I want to select the records with highest score_round_1 in each round_1(10,11) group .
In this case the first and last rows would be selected.
I tried using the GROUP BY round_1 but that only returns the first row from the two.
Any advice?
Zolka
This is simple
select max(score_round_1),
name
from score
group by round_1
SELECT *
FROM table p1
WHERE score_round_1 = (
SELECT MAX( p2.score_round_1 )
FROM table p2
WHERE p1.round_1 = p2.round_1 ) ANDround_1 !=0
Use aggregate function MAX
SELECT names, MAX(score_round_1) GROUP BY round_1
I have prices in two different tables and want to subtract them (current price-last day price) and ORDER them in DESC form. I was wondering if it can be done using a single MySQL command.
Table Structure
Table 1
id | Item Name | Date | Price
1 | alpha | 2011-10-05 | 10
2 | beta | 2011-10-05 | 12
3 | gamma | 2011-10-05 | 14
Table 2
id | Item Name | Date | Price
1 | alpha | 2011-10-04 | 8
2 | beta | 2011-10-04 | 10
3 | gamma | 2011-10-04 | 12
4 | alpha | 2011-10-03 | 4
5 | beta | 2011-10-03 | 6
6 | gamma | 2011-10-03 | 8
SELECT
table1.id, table1.`Item Name`,
table1.`Date` AS CurrDate, table1.Price AS CurrPrice,
table2.`Date` AS PrevDate, table2.Price AS PrevPrice,
table1.Price - table2.Price AS Difference
FROM table1
LEFT JOIN table2 ON table1.id = table2.id AND table1.`Date` - INTERVAL 1 DAY = table2.`Date`
ORDER BY Difference DESC
There is nothing special about this query except the way I've used the LEFT JOIN. I believe if yesterday's rates for a record are not available, the the last three columns would contain NULL. Output:
id | Item Name | CurrDate | CurrPrice | PrevDate | PrevPrice | Difference
2 | beta | 2011-10-05 | 12 | 2011-10-04 | 10 | 2
3 | gamma | 2011-10-05 | 14 | 2011-10-04 | 12 | 2
1 | alpha | 2011-10-05 | 10 | 2011-10-04 | 8 | 2
SELECT
a.price as price1
, IFNULL(b.price,'(no data)') as price2
, (a.price - IFNULL(b.price,0)) as difference
FROM table1 a
LEFT JOIN table2 b ON (a.`item name` = b.`item name`)
GROUP BY a.`item name`
HAVING IFNULL(b.`date`,'') = MAX(IFNULL(b.`date`,'')
Here's how it works.
It selects data from 2 tables: All data from table1 and matching data from table2.
If it cannot find matching data from table2 it will substitute null values in place of the missing rows. (left join)
Then it groups (group by) rows together based on table1.item name.
This combines multiple rows per item.
The having clause fixes this by only selecting the newest date rows from table2.
A small correction is build into the select and having clauses to deal with the case when there is no data in table2 to match table1.
Your query should be:
SELECT
s.closing as price1
, IFNULL(sh.closing,'(no data)') as price2
, (s.closing - IFNULL(sh.closing,0)) as difference
FROM stocks s
LEFT JOIN stockhistory sh ON (s.symbol = sh.symbol)
GROUP BY s.symbol
HAVING IFNULL(sh.edate,'') = MAX(IFNULL(sh.edate,'')
LIMIT 30 OFFSET 0;