I have a table like this:
MyTable:
id: pk
numero: varchar
data_modif: timestamp
...
I have multiple records with same value in numero and I need to return each distinct numero record with oldest data_modif. How can I do this?
This sounds like aggregation:
select numero, min(data_modif)
from mytable
group by numero;
If you want the entire row, then window functions are one method:
select t.*
from (select t.*,
row_number() over (partition by numero order by data_modif asc) as seqnum
from mytable t
) t
where seqnum = 1;
EDIT:
In an old version of MySQL, you woudl use:
select t.*
from t
where t.data_modif = (select min(t2.data_modif)
from t t2
where t2.numero = t.numero
);
Related
So i have a table named "log" with the following columns,
id, endpoint ,response ,group
SAMPLE DATA.
1. endpoint1 ,{"last_name":"data here"} ,1234
2. endpoint2 ,{"first_name":"data here"} ,1234
3. endpoint3 ,{"dob":"12-21-2301"} ,1234
what I want to achieve is to write a query that can generate a record grouped by the "
group" column and the final output should be something like this.
{"last_name","data here","first_name":"data here","dob":"12-21-2301"}
for each record with each key been a column.
Thank you
WITH RECURSIVE
cte1 AS ( SELECT response,
`group`,
ROW_NUMBER() OVER (PARTITION BY `group`) rn
FROM log ),
cte2 AS ( SELECT response,
`group`,
rn
FROM cte1
WHERE rn = 1
UNION ALL
SELECT JSON_MERGE_PRESERVE(cte1.response, cte2.response),
cte1.`group`,
cte1.rn
FROM cte2
JOIN cte1 USING (`group`)
WHERE cte2.rn + 1 = cte1.rn )
SELECT DISTINCT
FIRST_VALUE(response) OVER (PARTITION BY `group` ORDER BY rn DESC) responses,
`group`
FROM cte2;
https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=913b1923d7d5dbc7e42baeefb6e6ec86
I need to get my data set as this table
I am trying to get eligible set like this, need to group_concat pinged set also
x.id IN (SELECT MAX(x.id) FROM x WHERE ping rider id IS NULL GROUP BY orderId)
You can assign a group based on the cumulative number of non-null values in eligible_riders. Then aggregate and take the last value:
select og.*
from (select order_id, grp, max(eligible_riders) as eligible_riders,
group_concat(rider_id) as riders,
row_number() over (partition by order_id order by min(id) desc) as seqnum
from (select t.*,
sum(eligible_riders <> '') over (partition by order_id order by id) as grp
from t
) t
group by order_id, grp
) og
where seqnum = 1;
Hmmm . . . You could also do this with a correlated subquery, which might look a bit simpler:
select order_id, max(eligible_riders) as eligible_riders,
group_concat(rider_id) as riders
from t
where t.id >= (select max(t2.id)
from t t2
where t2.order_id = t.order_id and
t2.eligible_riders <> ''
)
group by order_id;
For performance, you want an index on (order_id, eligible_riders).
I have a table with following structure
Date
train
time1
train1
time2
train2
time3
train1
time4
train2
I want to create a new table and keeping only the latest record of each distinct train
Date
train
time3
train1
time4
train2
How should I achieve so?
One method is for selecting the most recent rows is:
select t.*
from releng_retry_test_phases t
where t.date = (select max(t2.date) from releng_retry_test_phases t2 where t2.train = t.train);
If you actually want to modify the table and delete the older rows;
delete t
from releng_retry_test_phases t join
(select t2.train, max(date) as max_date
from releng_retry_test_phases t2
group by t2.train
) t2
using (train)
where t.date < t2.max_date;
You can use ROW_NUMBER() to identify the rows you want:
select date, train
from (
select *,
row_number() over(partition by train order by date desc) as rn
) x
where rn = 1
WITH temp As(
SELECT *, Row_Number() over (PARTITION BY train ORDER BY date DESC ) as
rowNumber FROM table
)
SELECT date, train FROM temp WHERE rowNumber = 1
You can use row_number() method.
This is my query... SELECT * FROM comments WHERE content_id in (525, 537) LIMIT 60 This is the SS of result:
here content_id = 537 is selected 5 times.
(comment_id is UNIQUE key )..
My question is: How to limit selected rows by 2, where values of content_id is same...
Maximum two duplicate records for each content_id... like in this picture:
If you are running MySQL 8.0, you can do this with row_number():
select comment_id, content_id
from (
select t.*, row_number() over(partition by content_id order by comment_id) rn
from mytable t
) t
where rn <= 2
In earlier versions, one solution is a correlated subquery:
select t.*
from mytable t
where (
select count(*)
from mytable t1
where t1.content_id = t.content_id and t1.comment_id < t.comment_id
) < 2
I need to have the last price for each product for each client . I am not really good with SQL and I don't understand how I can do it.
Data :
What I want :
It is possible to have this data with a SQL request ?
Use window function ROW_NUMBER(), if available in your RDBMS:
SELECT product, price, date, client
FROM (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY product, client ORDER BY date DESC) rn
FROM mytable t
) x
WHERE rn = 1
In MySQL < 8.0:
SELECT product, price, date, client
FROM mytable t
WHERE NOT EXISTS (
SELECT 1
FROM mytable t1
WHERE t1.client = t.client AND t1.product = t.product AND t1.date > t.date
)
One option could be a correlated subquery
SELECT product, price, date, client
FROM tablename a where date =
(select max(date) from tablename b where a.product=b.product)