Can't get appropriate values from query? - mysql

I want to list companyIds and with the mostly occur commentable type (0,1,2).
This is subquery
select a.companyId, a.commentable, count(1) _count
from article a
group by a.companyId, a.commentable
| companyId | commentable | _count |
|-----------|-------------|--------|
| 1 | 0 | 1 |
| 1 | 1 | 1 |
| 2 | 0 | 7759 |
| 2 | 1 | 7586 |
| 2 | 2 | 7856 |
| 3 | 0 | 7828 |
| 3 | 1 | 7866 |
| 3 | 2 | 7706 |
| 4 | 0 | 7851 |
| 4 | 1 | 7901 |
| 4 | 2 | 7738 |
| 5 | 0 | 7775 |
| 5 | 1 | 7884 |
| 5 | 2 | 7602 |
| 25 | 0 | 7888 |
| 25 | 1 | 7939 |
| 25 | 2 | 7784 |
For example above
Most commentable type occur for companyId=4 is 7901 and commentable type for that is 1. In below query , i see 4-0-7901, but i expected 4-1-7901
SELECT x.companyId, x.commentable, MAX(x._count) _count
FROM
( SELECT a.companyId, a.commentable, COUNT(1) _count
FROM article a
GROUP BY a.companyId, a.commentable
) AS X
GROUP BY x.companyId;
companyId commentable _count
1 0 1
2 0 7856
3 0 7866
4 0 7901
5 0 7884
25 0 7939
Expected result
companyId commentable _count
1 0 1
2 2 7856
3 1 7866
4 1 7901
5 1 7884
25 1 7939
I dont understand 'why is all commentable column is '0' .

You need a big ugly join here. In the query below, you may view the GROUP BY query on the company and comment type the base unit of work. This query appears as itself, aliased as t1. In alias t2, we subquery and aggregate only by commentable, to find the max count for each such comment type. This, we join back to t1 to restrict only the company having the max count.
SELECT
t1.companyId,
t1.commentable,
t1.cnt
FROM
(
SELECT companyId, commentable, COUNT(*) cnt
FROM article
GROUP BY companyId, commentable
) t1
INNER JOIN
(
SELECT companyId, MAX(cnt) max_cnt
FROM
(
SELECT companyId, commentable, COUNT(*) cnt
FROM article
GROUP BY companyId, commentable
) t
GROUP BY companyId
) t2
ON t1.companyId = t2.companyId AND t1.cnt = t2.max_cnt;
By the way, things get somewhat nicer in MySQL 8+, where we can take advantage of analytic functions:
WITH cte AS (
SELECT companyId, commentable, COUNT(*) cnt,
ROW_NUMBER() OVER (PARTITION BY commentable ORDER BY COUNT(*) DESC) rn
FROM article
GROUP BY companyId, commentable
)
SELECT companyId, commentable, cnt
FROM cte
WHERE rn = 1;

You can do this using a having clause:
SELECT a.companyId, a.commentable, COUNT(*) as _count
FROM article a
GROUP BY a.companyId, a.commentable
HAVING COUNT(*) = (SELECT COUNT(*)
FROM article a2
WHERE a2.companyId = a.companyId
GROUP BY a2.commentable
ORDER BY COUNT(*) DESC
LIMIT 1
);
In the event of ties, you will get multiple rows. If you want only one row per company, you can instead use commentable for the comparison in the HAVING:
SELECT a.companyId, a.commentable, COUNT(*) as _count
FROM article a
GROUP BY a.companyId, a.commentable
HAVING a.commentable = (SELECT a2.commentable
FROM article a2
WHERE a2.companyId = a.companyId
GROUP BY a2.commentable
ORDER BY COUNT(*) DESC
LIMIT 1
);
As others have mentioned, your problem is the mis-use of GROUP BY. The unaggregated columns in the SELECT need to match the GROUP BY keys -- and vice versa.

Cause commentable is not one of group by columns. In this case, with ONLY_FULL_GROUP_BY disabled, MySQL is free to choose any one value for this column.
From MySQL doc
If ONLY_FULL_GROUP_BY is disabled, a MySQL extension to the standard SQL use of GROUP BY permits the select list, HAVING condition, or ORDER BY list to refer to nonaggregated columns even if the columns are not functionally dependent on GROUP BY columns. This causes MySQL to accept the preceding query. In this case, the server is free to choose any value from each group, so unless they are the same, the values chosen are nondeterministic, which is probably not what you want.

Related

First Unique Sql row

I have a MySql table of users order and it has columns such as:
user_id | timestamp | is_order_Parent | Status |
1 | 10-02-2020 | N | C |
2 | 11-02-2010 | Y | D |
3 | 11-02-2020 | N | C |
1 | 12-02-2010 | N | C |
1 | 15-02-2020 | N | C |
2 | 15-02-2010 | N | C |
I want to count number of new custmer per day defined as: a customer who orders non-parent order and his order status is C AND WHEN COUNTING A USER ONCE IN A DAY WE DONT COUNT HIM FOR OTHER DAYS
An ideal resulted table will be:
Timestamp: Day | Distinct values of User ID
10-02-2020 | 1
11-02-2010 | 1
12-02-2010 | 0 <--- already counted user_id = 1 above, so no need to count it here
15-02-2010 | 1
table name is cscart_orders
If you are running MySQL 8.0, you can do this with window functions an aggregation:
select timestamp, sum(timestamp = timestamp0) new_users
from (
select
t.*,
min(case when is_order_parent = 'N' and status = 'C' then timestamp end) over(partition by user_id) timestamp0
from mytable t
) t
group by timestamp
The window min() computes the timestamp when each user became a "new user". Then, the outer query aggregates by date, and counts how many new users were found on that date.
A nice thing about this approach is that it does not require enumerating the dates separately.
You can use two levels of aggregation:
select first_timestamp, count(*)
from (select t.user_id, min(timestamp) as first_timestamp
from t
where is_order_parent = 'N' and status = 'C'
group by t.user_id
) t
group by first_timestamp;

select two tables mysql without join

There are two tables, recharge and purchase.
select * from recharge;
+-----+------+--------+---------------------+
| idx | user | amount | created |
+-----+------+--------+---------------------+
| 1 | 3 | 10 | 2016-01-09 20:16:18 |
| 2 | 3 | 5 | 2016-01-09 20:16:45 |
+-----+------+--------+---------------------+
select * from purchase;
+-----+------+----------+---------------------+
| idx | user | resource | created |
+-----+------+----------+---------------------+
| 1 | 3 | 2 | 2016-01-09 20:55:30 |
| 2 | 3 | 1 | 2016-01-09 20:55:30 |
+-----+------+----------+---------------------+
I want to figure out balance of users which is SUM(amount) - COUNT(purchase.idx). (in this case, 13)
So I had tried
SELECT (SUM(`amount`)-COUNT(purchase.idx)) AS balance
FROM `recharge`, `purchase`
WHERE purchase.user = 3 AND recharge.user = 3
but, it returned error.
If you want an accurate count, then aggregate before doing arithmetic. For your particular case:
select ((select sum(r.amount) from recharge where r.user = 3) -
(select count(*) from purchase p where p.user = 3)
)
To do this for multiple users, move the subqueries to the from clause or use union all and aggregation. The second is safer if a user might only be in one table:
select user, coalesce(sum(suma), 0) - coalesce(sum(countp), 0)
from ((select user, sum(amount) as suma, null as countp
from recharge
group by user
) union all
(select user, null, count(*)
from purchase
group by user
)
) rp
group by user
It is possible to using union like this
SELECT SUM(`amount`-aidx) AS balance
FROM(
SELECT SUM(`amount`) as amount, 0 as aidx
from `recharge` where recharge.user = 3
union
select 0 as amount, COUNT(purchase.idx) as aidx
from `purchase`
WHERE purchase.user = 3 )a

Get second highest values from a table

I have a table like this:
+----+---------+------------+
| id | conn_id | read_date |
+----+---------+------------+
| 1 | 1 | 2010-02-21 |
| 2 | 1 | 2011-02-21 |
| 3 | 2 | 2011-02-21 |
| 4 | 2 | 2013-02-21 |
| 5 | 2 | 2014-02-21 |
+----+---------+------------+
I want the second highest read_date for particular 'conn_id's i.e. I want a group by on conn_id. Please help me figure this out.
Here's a solution for a particular conn_id :
select max (read_date) from my_table
where conn_id=1
and read_date<(
select max (read_date) from my_table
where conn_id=1
)
If you want to get it for all conn_id using group by, do this:
select t.conn_id, (select max(i.read_date) from my_table i
where i.conn_id=t.conn_id and i.read_date<max(t.read_date))
from my_table t group by conn_id;
Following answer should work in MSSQL :
select id,conn_id,read_date from (
select *,ROW_NUMBER() over(Partition by conn_id order by read_date desc) as RN
from my_table
)
where RN =2
There is an intresting article on use of rank functions in MySQL here : ROW_NUMBER() in MySQL
If your table design as ID - date matching (ie a big id always a big date), you can group by id, otherwise do the following:
$sql_max = '(select conn_id, max(read_date) max_date from tab group by 1) as tab_max';
$sql_max2 = "(select tab.conn_id,max(tab.read_date) max_date2 from tab, $sql_max
where tab.conn_id = tab_max.conn_id and tab.read_date < tab_max.max_date
group by 1) as tab_max2";
$sql = "select tab.* from tab, $sql_max2
where tab.conn_id = tab_max2.conn_id and tab.read_date = tab_max2.max_date2";

Mysql - Select at least one or select none

I have a table as so...
----------------------------------------
| id | name | group | number |
----------------------------------------
| 1 | joey | 1 | 2 |
| 2 | keidy | 1 | 3 |
| 3 | james | 2 | 2 |
| 4 | steven | 2 | 5 |
| 5 | jason | 3 | 2 |
| 6 | shane | 3 | 3 |
----------------------------------------
I'm running a select like so:
SELECT * FROM table WHERE number IN (2,3);
The problem im trying to solve is that I want to only grab get results from groups that have 1 or more rows of each number. For instance the above query is returning id's 1-2-3-5-6, when I'd like the results to exclude id 3 since the group of '2' can only return 1 result for the number of '2' and not for BOTH 2 and 3, since there's no row with the number 3 for the group 2 i'd like it to not even select id 3 at all.
Any help would be great.
Try it this way
SELECT *
FROM table1 t
WHERE number IN(2, 3)
AND EXISTS
(
SELECT *
FROM table1
WHERE number IN(2, 3)
AND `group` = t.`group`
GROUP BY `group`
HAVING MAX(number = 2) > 0
AND MAX(number = 3) > 0
)
or
SELECT *
FROM table1 t JOIN
(
SELECT `group`
FROM table1
WHERE number IN(2, 3)
GROUP BY `group`
HAVING MAX(number = 2) > 0
AND MAX(number = 3) > 0
) q
ON t.`group` = q.`group`;
or
SELECT *
FROM table1
WHERE `group` IN
(
SELECT `group`
FROM table1
WHERE number IN(2, 3)
GROUP BY `group`
HAVING MAX(number = 2) > 0
AND MAX(number = 3) > 0
);
Sample output (for both queries):
| ID | NAME | GROUP | NUMBER |
|----|-------|-------|--------|
| 1 | joey | 1 | 2 |
| 2 | keidy | 1 | 3 |
| 5 | jason | 3 | 2 |
| 6 | shane | 3 | 3 |
Here is SQLFiddle demo
On this, you can approach from a fun way with multiple joins for what you WANT qualified, OR, apply a prequery to get all qualified groups as others have suggested, but readability is a bit off for me..
Anyhow, here's an approach going through the table once, but with joins
select DISTINCT
T.id,
T.Name,
T.Group,
T.Number
from
YourTable T
Join YourTable T2
on T.Group = T2.Group AND T2.Group = 2
Join YourTable T3
on T.Group = T3.Group AND T3.Group = 3
where
T.Number IN ( 2, 3 )
So on the first record, it is pointing to by it's own group to the T2 group AND the T2 group is specifically a 2... Then again, but testing the group for the T3 instance and T3's group is a 3.
If it cant complete the join to either of the T2 or T3 instances, the record is done for consideration, and since indexes work great for joins like this, make sure you have one index for your NUMBER criteria, and another index on the (GROUP, NUMBER) for those comparisons and the next query sample...
If doing by more than this simple 2, but larger group, prequery qualified groups, then join to that
select
YT2.*
from
( select YT1.group
from YourTable YT1
where YT1.Number in (2, 3)
group by YT1.group
having count( DISTINCT YT1.group ) = 2 ) PreQualified
JOIN YourTable YT2
on PreQualified.group = YT2.group
AND YT2.Number in (2,3)
Maybe this,if I understand you
SELECT id FROM table WHERE `group` IN
(SELECT `group` FROM table WHERE number IN (2,3)
GROUP BY `group`
HAVING COUNT(DISTINCT number)=2)
SQL Fiddle
This will return all ids where BOTH numbers exist in a group.Remove DISTINCT if you want ids for groups where just one numbers is in.

Sort data before using GROUP BY?

I have read that grouping happens before ordering, is there any way that I can order first before grouping without having to wrap my whole query around another query just to do this?
Let's say I have this data:
id | user_id | date_recorded
1 | 1 | 2011-11-07
2 | 1 | 2011-11-05
3 | 1 | 2011-11-06
4 | 2 | 2011-11-03
5 | 2 | 2011-11-06
Normally, I'd have to do this query in order to get what I want:
SELECT
*
FROM (
SELECT * FROM table ORDER BY date_recorded DESC
) t1
GROUP BY t1.user_id
But I'm wondering if there's a better solution.
Your question is somewhat unclear but I have a suspicion what you really want is not any GROUP aggregates at all, but rather ordering by date first, then user ID:
SELECT
id,
user_id,
date_recorded
FROM tbl
ORDER BY date_recorded DESC, user_id ASC
Here would be the result. Note reordering by date_recorded from your original example
id | user_id | date_recorded
1 | 1 | 2011-11-07
3 | 1 | 2011-11-06
2 | 1 | 2011-11-05
5 | 2 | 2011-11-06
4 | 2 | 2011-11-03
Update
To retrieve the full latest record per user_id, a JOIN is needed. The subquery (mx) locates the latest date_recorded per user_id, and that result is joined to the full table to retrieve the remaining columns.
SELECT
mx.user_id,
mx.maxdate,
t.id
FROM (
SELECT
user_id,
MAX(date_recorded) AS maxdate
FROM tbl
GROUP BY user_id
) mx JOIN tbl t ON mx.user_id = t.user_id AND mx.date_recorded = t.date_recorded
Iam just using the technique
"Using order clause before group by inserting it in group_concat clause"
SELECT SUBSTRING_INDEX(group_concat(cast(id as char)
ORDER BY date_recorded desc),',',1),
user_id,
SUBSTRING_INDEX(group_concat(cast(`date_recorded` as char)
ORDER BY `date_recorded` desc),',',1)
FROM data
GROUP BY user_id