SQL - Select discint AND count from JOIN query - mysql

I have below query:
SELECT u.*
(SELECT sum(trs.amount)
FROM transactions trs
WHERE u.id = trs.user AND trs.type = 'Recycle' AND
trs.TIME >= UNIX_TIMESTAMP(CURDATE())
) as amt
FROM (SELECT DISTINCT user_by
FROM xeon_users_rented
) AS xur JOIN
users u
ON xur.user_by = u.username
LIMIT 50
Which selects some data from my database. The above query works fine. However, I would like to also select count(*) from xeon_users_rented where user_by = u.username This is what I have attempted:
SELECT u.*
(SELECT sum(trs.amount)
FROM transactions trs
WHERE u.id = trs.user AND trs.type = 'Recycle' AND
trs.TIME >= UNIX_TIMESTAMP(CURDATE())
) as amt,
(SELECT DISTINCT count(*)
FROM xeon_users_rented
WHERE xur.user_by = u.username
) AS ttl
FROM (SELECT DISTINCT user_by
FROM xeon_users_rented
) AS xur JOIN
users u
ON xur.user_by = u.username
LIMIT 50
However, that gives me the total number of rows in xeon_users_rented as ttl - not the total distinct rows where username = user_by

I think you can do what you want just by tinkering with your subquery a little. That is, change the select distinct to a group by:
SELECT u.*, xur.cnt,
(SELECT sum(trs.amount)
FROM transactions trs
WHERE u.id = trs.user AND trs.type = 'Recycle' AND
trs.TIME >= UNIX_TIMESTAMP(CURDATE())
) as amt
FROM (SELECT user_by, COUNT(*) as cnt
FROM xeon_users_rented
GROUP BY user_by
) xur JOIN
users u
ON xur.user_by = u.username
LIMIT 50;
Some notes:
SELECT DISTINCT is not really necessary, because you can do the same logic using GROUP BY. So, it is more important to understand GROUP BY.
You are using LIMIT with no ORDER BY. That means that you can get a different set of rows each time you run the query. Bad practice.

Related

sum of 2 different queries result

I have to do sum from 2 different tables and show it using MySQL:
total comments from table 1, total comments from table 2: What i have tried so far is,
SELECT u.name as name, u.username as username,
( SELECT SUM(total) FROM (SELECT (COUNT(nc.id)) as total FROM table1 as nc WHERE nc.user_id = u.id) UNION ALL (SELECT COUNT(pc.id) AS total FROM table2 pc WHERE pc.user_id = u.id) as finalTotal ) as total_comments
FROM user as u
GROUP BY u.id
It is giving me this error:
Every derived table must have its own alias
If I understand what you need, you have to modify the query like this :
select u.name,u.username,total as total_comments
from user as u
left join (
select id,sum(total) as total
from(
select nc.id,count(1) as total
from table1 as nc
group by nc.id
union all
select pc.id,count(1) as total
from table2 as pc
group by pc.id
) as t group by id
) comments on comments.id = u.id
Before UNION ALL put alias like you giving for all..
SELECT u.name as name, u.username as username,
( SELECT SUM(total) FROM (SELECT (COUNT(nc.id)) as total FROM table1 as nc WHERE nc.user_id = u.id) as vr UNION ALL (SELECT COUNT(pc.id) AS total FROM table2 pc WHERE pc.user_id = u.id) as finalTotal ) as total_comments
FROM user as u
GROUP BY u.id
SELECT u.name as name, u.username as username,
( SELECT SUM(total) FROM
(SELECT (COUNT(nc.id)) as total FROM table1 as nc WHERE nc.user_id = u.id
UNION ALL SELECT COUNT(pc.id) AS total FROM table2 pc WHERE pc.user_id = u.id) as finalTotal ) as total_comments
FROM user as u
GROUP BY u.id
Remove brackets before and after UNION ALL and check.

MySQL SELECT subquery

I have a calendar and user_result table and I need to join these two queries.
calendar query
SELECT `week`, `date`, `time`, COUNT(*) as count
FROM `calendar`
WHERE `week` = 1
GROUP BY `date`
ORDER BY `date` DESC
and the result is
{"week":"1","date":"2014-08-21","time":"15:30:00","count":"4"}, {"week":"1","date":"2014-08-20","time":"17:30:00","count":"12"}
user_result query
SELECT `date`, SUM(`point`) as score
FROM `user_result`
WHERE `user_id` = 1
AND `date` = '2014-08-20'
and the result is just score 3
My goal is to always show calendar even if the user isn't present in the user_result table, but if he is, SUM his points for that day where calendar.date = user_result.date. Result should be:
{"week":"1","date":"2014-08-21","time":"15:30:00","count":"4","score":"3"}, {"week":"1","date":"2014-08-20","time":"17:30:00","count":"12","score":"0"}
I have tried this query below, but the result is just one row and unexpected count
SELECT c.`week`, c.`date`, c.`time`, COUNT(*) as count, SUM(p.`point`) as score
FROM `calendar` c
INNER JOIN `user_result` p ON c.`date` = p.`date`
WHERE c.`week` = 1
AND p.`user_id` = 1
GROUP BY c.`date`
ORDER BY c.`date` DESC
{"week":"1","date":"2014-08-20","time":"17:30:00","count":"4","score":"9"}
SQL Fiddle
ow sorry, i was edited, and i was try at your sqlfiddle, if you want to show all date from calendar you can use LEFT JOIN, but if you want to show just the same date between calendar and result you can use INNER JOIN, note: in this case INNER JOIN just show 1 result, and LEFT JOIN show 2 results
SELECT c.`week`, p.user_id, c.`date`, c.`time`, COUNT(*) as count, p.score
FROM `calendar` c
LEFT JOIN
(
SELECT `date`, SUM(`point`) score, user_id
FROM `result`
group by `date`
) p ON c.`date` = p.`date`
WHERE c.`week` = 1
GROUP BY c.`date`
ORDER BY c.`date` DESC
I put a pre-aggreate query / group by date as a select for the one person you were interested in... then did a left-join to it. Also, your column names of week, date and time (IMO) are poor choice column names as they can appear to be too close to reserved keywords in MySQL. They are not, but could be confusing..
SELECT
c.week,
c.date,
c.time,
coalesce( OnePerson.PointEntries, 0 ) as count,
coalesce( OnePerson.totPoints, 0 ) as score
FROM
calendar c
LEFT JOIN ( select
r.week,
r.date,
COUNT(*) as PointEntries,
SUM( r.point ) as totPoints
from
result r
where
r.week = 1
AND r.user_id = 1
group by
r.week,
r.date ) OnePerson
ON c.week = OnePerson.week
AND c.date = OnePerson.date
WHERE
c.week = 1
GROUP BY
c.date
ORDER BY
c.date DESC
Posted code to SQLFiddle

mysql: group functions on subqueries with limits

I have a group of users who perform tasks on which they are scored. I'm trying to create a report showing the average of each user's last 50 tasks.
user table: userid, username, usertype
task table: taskid, score, tasktype, userid
If I do:
SELECT u.userid, u.username, (SELECT AVG(score)
FROM task t
WHERE t.userid = u.userid AND t.tasktype = 'task1'
ORDER BY t.taskid DESC LIMIT 50) AS avgscore
FROM user u
WHERE u.usertype = 'utype';
that doesn't work because it does the limit 50 after it does the average of everything.
What I need is this:
SELECT u.userid, u.username, AVG(SELECT t.score
FROM task t
WHERE t.userid = u.userid AND t.tasktype = 'task1'
ORDER BY t.taskid DESC LIMIT 50) AS avgscore
FROM user u
WHERE u.usertype = 'utype';
but that is not valid syntax
I've tried sub-sub queries, but can't get it that way either, as I always get a problem with the limit, or a join, or unknown fields when I reference u.userid in the sub-subquery.
Is there a way to do this?
Use a subquery within the subquery:
SELECT u.userid, u.username,
(SELECT AVG(score)
FROM (select t.*
from task t
WHERE t.userid = u.userid AND t.tasktype = 'task1'
ORDER BY t.taskid DESC
LIMIT 50
) t
) AS avgscore
FROM user u
WHERE u.usertype = 'utype';
EDIT:
I didn't realize that MySQL would not recognize the u.userid. It should according to ANSI rules for the scoping of table aliases.
You can take a different approach which is find the 50th taskid value, and then take everything above that:
select ut.userid, ut.username, avg(t.score)
from (SELECT u.userid, u.username,
(SELECT substring_index(substring_index(group_concat(taskid order by taskid desc
), ',', 50), ',', -1)
from task t
WHERE t.userid = u.userid AND t.tasktype = 'task1'
) + 0 as taskid50
FROM user u
WHERE u.usertype = 'utype'
) ut join
task t
on ut.userid = t.userid and
ut.taskid50 >= t.taskid and t.tasktype = 'task1'
group by ut.userid, ut.username;
try this
SELECT u.userid, u.username, AVG(t.score ) AS avgscore
FROM user u
INNER JOIN task t
ON t.userid = u.userid
WHERE u.usertype = 'utype' AND t.tasktype = 'task1'
GROUP BY u.userid
ORDER BY t.taskid DESC LIMIT 50;

Sub Query counting character strings in MySQL

LEFT JOIN
(
SELECT user_id, review, COUNT(user_id) totalCount
FROM reviews
GROUP BY user_id
) b ON b.user_id= b.user_id
I am trying to fit WHERE LENGTH(review) > 100 in this somewhere but every I put it, it gives me problems.
The sub-query above counts all total reviews by user_id. I simply want to add one more qualification. Only count reviews greater than 100 length.
On a side note, I've seen the function CHAR_LENGTH -- not sure if that i what I need either.
EDIT:
Here is complete query working perfectly as expected for my needs:
static public $top_users = "
SELECT u.username, u.score,
(COALESCE(a.totalCount, 0) * 4) +
(COALESCE(b.totalCount, 0) * 5) +
(COALESCE(c.totalCount, 0) * 1) +
(COALESCE(d.totalCount, 0) * 2) +
(COALESCE(u.friend_points, 0)) AS totalScore
FROM users u
LEFT JOIN
(
SELECT user_id, COUNT(user_id) totalCount
FROM items
GROUP BY user_id
) a ON a.user_id= u.user_id
LEFT JOIN
(
SELECT user_id, COUNT(user_id) totalCount
FROM reviews
GROUP BY user_id
) b ON b.user_id= u.user_id
LEFT JOIN
(
SELECT user_id, COUNT(user_id) totalCount
FROM ratings
GROUP BY user_id
) c ON c.user_id = u.user_id
LEFT JOIN
(
SELECT user_id, COUNT(user_id) totalCount
FROM comments
GROUP BY user_id
) d ON d.user_id = u.user_id
ORDER BY totalScore DESC LIMIT 25;";
LENGTH() returns the length of the string measured in bytes. You probably want CHAR_LENGTH() as it will give you the actual characters.
SELECT user_id, review, COUNT(user_id) totalCount
FROM reviews
WHERE CHAR_LENGTH(review) > 100
GROUP BY user_id, review
You're also not using GROUP BY correctly.
See the documentation
The query that you want is:
LEFT JOIN
(
SELECT user_id, COUNT(user_id) totalCount,
sum(case when length(review) > 100 then 1 else 0 end
) as NumLongReviews
FROM reviews
GROUP BY user_id
) b ON b.user_id= b.user_id
This counts both the reviews and the "long" reviews. That count is done using a case statement nested in a sum() function.

MySQL INNER JOIN select only one row from second table

I have a users table and a payments table, for each user, those of which have payments, may have multiple associated payments in the payments table. I would like to select all users who have payments, but only select their latest payment. I'm trying this SQL but i've never tried nested SQL statements before so I want to know what i'm doing wrong. Appreciate the help
SELECT u.*
FROM users AS u
INNER JOIN (
SELECT p.*
FROM payments AS p
ORDER BY date DESC
LIMIT 1
)
ON p.user_id = u.id
WHERE u.package = 1
You need to have a subquery to get their latest date per user ID.
SELECT u.*, p.*
FROM users u
INNER JOIN payments p
ON u.id = p.user_ID
INNER JOIN
(
SELECT user_ID, MAX(date) maxDate
FROM payments
GROUP BY user_ID
) b ON p.user_ID = b.user_ID AND
p.date = b.maxDate
WHERE u.package = 1
SELECT u.*, p.*
FROM users AS u
INNER JOIN payments AS p ON p.id = (
SELECT id
FROM payments AS p2
WHERE p2.user_id = u.id
ORDER BY date DESC
LIMIT 1
)
Or
SELECT u.*, p.*
FROM users AS u
INNER JOIN payments AS p ON p.user_id = u.id
WHERE NOT EXISTS (
SELECT 1
FROM payments AS p2
WHERE
p2.user_id = p.user_id AND
(p2.date > p.date OR (p2.date = p.date AND p2.id > p.id))
)
These solutions are better than the accepted answer because they work correctly when there are multiple payments with same user and date. You can try on SQL Fiddle.
SELECT u.*, p.*, max(p.date)
FROM payments p
JOIN users u ON u.id=p.user_id AND u.package = 1
GROUP BY u.id
ORDER BY p.date DESC
Check out this sqlfiddle
SELECT u.*
FROM users AS u
INNER JOIN (
SELECT p.*,
#num := if(#id = user_id, #num + 1, 1) as row_number,
#id := user_id as tmp
FROM payments AS p,
(SELECT #num := 0) x,
(SELECT #id := 0) y
ORDER BY p.user_id ASC, date DESC)
ON (p.user_id = u.id) and (p.row_number=1)
WHERE u.package = 1
You can try this:
SELECT u.*, p.*
FROM users AS u LEFT JOIN (
SELECT *, ROW_NUMBER() OVER(PARTITION BY userid ORDER BY [Date] DESC) AS RowNo
FROM payments
) AS p ON u.userid = p.userid AND p.RowNo=1
There are two problems with your query:
Every table and subquery needs a name, so you have to name the subquery INNER JOIN (SELECT ...) AS p ON ....
The subquery as you have it only returns one row period, but you actually want one row for each user. For that you need one query to get the max date and then self-join back to get the whole row.
Assuming there are no ties for payments.date, try:
SELECT u.*, p.*
FROM (
SELECT MAX(p.date) AS date, p.user_id
FROM payments AS p
GROUP BY p.user_id
) AS latestP
INNER JOIN users AS u ON latestP.user_id = u.id
INNER JOIN payments AS p ON p.user_id = u.id AND p.date = latestP.date
WHERE u.package = 1
#John Woo's answer helped me solve a similar problem. I've improved upon his answer by setting the correct ordering as well. This has worked for me:
SELECT a.*, c.*
FROM users a
INNER JOIN payments c
ON a.id = c.user_ID
INNER JOIN (
SELECT user_ID, MAX(date) as maxDate FROM
(
SELECT user_ID, date
FROM payments
ORDER BY date DESC
) d
GROUP BY user_ID
) b ON c.user_ID = b.user_ID AND
c.date = b.maxDate
WHERE a.package = 1
I'm not sure how efficient this is, though.
SELECT U.*, V.* FROM users AS U
INNER JOIN (SELECT *
FROM payments
WHERE id IN (
SELECT MAX(id)
FROM payments
GROUP BY user_id
)) AS V ON U.id = V.user_id
This will get it working
Matei Mihai given a simple and efficient solution but it will not work until put a MAX(date) in SELECT part so this query will become:
SELECT u.*, p.*, max(date)
FROM payments p
JOIN users u ON u.id=p.user_id AND u.package = 1
GROUP BY u.id
And order by will not make any difference in grouping but it can order the final result provided by group by. I tried it and it worked for me.
My answer directly inspired from #valex very usefull, if you need several cols in the ORDER BY clause.
SELECT u.*
FROM users AS u
INNER JOIN (
SELECT p.*,
#num := if(#id = user_id, #num + 1, 1) as row_number,
#id := user_id as tmp
FROM (SELECT * FROM payments ORDER BY p.user_id ASC, date DESC) AS p,
(SELECT #num := 0) x,
(SELECT #id := 0) y
)
ON (p.user_id = u.id) and (p.row_number=1)
WHERE u.package = 1
This is quite simple do The inner join and then group by user_id and use max aggregate function in payment_id assuming your table being user and payment query can be
SELECT user.id, max(payment.id)
FROM user INNER JOIN payment ON (user.id = payment.user_id)
GROUP BY user.id
If you do not have to return the payment from the query you can do this with distinct, like:
SELECT DISTINCT u.*
FROM users AS u
INNER JOIN payments AS p ON p.user_id = u.id
This will return only users which have at least one record associated in payment table (because of inner join), and if user have multiple payments, will be returned only once (because of distinct), but the payment itself won't be returned, if you need the payment to be returned from the query, you can use for example subquery as other proposed.