I have the following query:
SELECT a.*
FROM `user_subscription_history` a
LEFT JOIN `transaction` b ON b.tran_date=a.date_time
b.tran_date and a.date_time are both timestamps. I'd like the join to return the closest matching. Currently only returns if matching.
How do i solve this?
You can use ROW_NUMBER() to identify the closest timestamp of another table:
select *
from (
select a.*,
row_number() over(
order by abs(timestampdiff(microsecond, b.tran_date, a.date_time))
) as rn
from user_subscription_history a
cross join `transaction` b
) x
where rn = 1
Related
I am trying to join two tables and get the count and grouped by specific field. However, it outputs same count values even if the other table consist only two rows. How should I fix this?
Here's my code:
SELECT tbl1.preferredDay, COUNT(tbl1.preferredDay) as count_1, COUNT(tbl2.preferredDay) as count_2
FROM tblschedule as tbl1
LEFT JOIN tblappointments as tbl2 ON (tbl1.preferredDay = tbl2.preferredDay)
WHERE tbl1.preferredDay = tbl2.preferredDay
GROUP BY preferredDay;
Here is the output but it should be [15, 0][3, 3]
Your query is based on left join it will return the same count().
This is a working query for Mysql 8:
with tbl1 as (
SELECT preferredDay, count(1) as count_1
FROM tblschedule
GROUP BY preferredDay
),
tbl2 as (
SELECT preferredDay, count(1) as count_2
FROM tblappointments
GROUP BY preferredDay
)
select t1.preferredDay, t1.count_1, t2.count_2
from tbl1 t1
inner join tbl2 t2 on t1.preferredDay = t2.preferredDay
There are two WITHs to get separately the count and then an INNER JOIN to join those results
For Mysql 5.7 and lower :
select t1.preferredDay, t1.count_1, t2.count_2
from (
SELECT preferredDay, count(1) as count_1
FROM tblschedule
GROUP BY preferredDay
) as t1
inner join (
SELECT preferredDay, count(1) as count_2
FROM tblappointments
GROUP BY preferredDay
) as t2 on t1.preferredDay = t2.preferredDay
I have 3 tables:
table user
table activity a
table activity b
I want to retrieve all the data in the user table and all the last data in the table of activities a and b, like this :
how do i do sql writing?
In MySQL, I would recommend using row_number() and left join:
select u.*, a.*, b.*
from user u left join
(select a.*,
row_number() over (partition by id_user order by date_a desc) as seqnum
from a
) a
on a.id_user = u.id_user and a.seqnum = 1 left join
(select b.*,
row_number() over (partition by id_user order by date_b desc) as seqnum
from b
) b
on b.id_user = u.id_user and b.seqnum = 1;
I want to query above picture.
Left picture is original data, right picture is query data.
select distinct ID, Nickname, Revision
from test_table
This query do not show above picture.
How to avoid duplicate data?
If SQL Server, using window function ROW_NUMBER in subquery:
select t.id, t.nickname, t.revision
from (
select t.*, row_number() over (
partition by t.id order by t.revision desc
) rn
from your_table t
) t
where rn = 1;
Or using TOP with ties with ROW_NUMBER:
select top 1 with ties *
from your_table
order by row_number() over (
partition by id order by revision desc
)
If MySQL:
select t.*
from your_table t
inner join (
select id, MAX(revision) revision
from your_table
group by id
) t1 on t.id = t1.id
and t.revision = t1.revision;
Another trick using TOP 1 with TIES
SELECT Top 1 with ties *
FROM your_table t
Order by row_number() over (partition BY t.id order by t.revision DESC)
select distinct ID, Nickname, MAX(Revision)
from test_table
group by ID
I am new to SQL and I want to fetch 5th highest value without using LIMIT. Here is the code I am trying but it is not working properly. It is showing the 5th lowest value instead of 5th highest value.
SELECT a . * FROM user AS a
WHERE 5 =
(SELECT count( DISTINCT b.id ) FROM user AS b WHERE b.id >= a.id ORDER BY a.id DESC)
Can anyone help me with this?
You could also do:
SET #nth := 5;
SELECT
a.*
FROM jos_modules AS a
WHERE #nth = (
SELECT
COUNT(b.id)
FROM user AS b
WHERE
a.id >= b.id
);
Try this
SELECT a . *
FROM user AS a
WHERE 5 = (
SELECT count( DISTINCT b.id )
FROM user AS b
WHERE a.id >= b.id ORDER BY a.id )
select * from (
select a.* , row_number() over (order by id asc) as RANK
from a ) where RANK=5 ;
If you are using a Teradata DB , you can use qualify statement :
select * from a
qualify row_number () over(order by id asc)=5;
I have a table with lots of fields in mysql
I need a query to return (in the same raw!) the top last 3 dates (dates can have large gaps between them)
ie:
2012/01/20
2012/01/18
2012/01/12
2012/01/10
2012/01/04
etc...
Any help will be appreciated
I must get them in the same row!
This is the query I am trying to use with no success:
SELECT a.id, a.thedate, b.id AS id1, b.thedate AS thedate1, c.id AS id2, c.thedate as thedate2
FROM mytable AS a INNER JOIN mytable AS b ON a.id = b.id INNER JOIN mytable AS c ON b.id=c.id
WHERE c.thedate = SELECT MAX(thedate)
EDIT :
SELECT group_concat(date) FROM (SELECT date FROM my_table ORDER BY date DESC LIMIT 3) AS temp
Corrected-
SELECT group_concat(date) FROM ( select date from table_name order by date desc limit 3) as a
SELECT GROUP_CONCAT(a.date )
FROM (
SELECT date
FROM my_table
ORDER BY date DESC
LIMIT 3
) AS a