i want to select something from table while one condition is true,
SELECT * FROM (SELECT * FROM`table1` `t1` ORDER BY t1.date) `t2` WHILE t2.id!=5
when while condition comes to false it stop selecting next rows.
Please help me, I have already search a lot and many similars in stackoverflow but I can't get it.
please don't tell me about where , i want solution in sql not in php or anything other
OK the real problem is here
SELECT *,(SELECT SUM(t2.amount) FROM (select * from transaction as t1 order by t1.date) `t2`) as total_per_transition FROM transaction
here i want to calculate total balance on each transaction
First find the first date where the condition fails, so where id=5:
SELECT date
FROM table1
WHERE id = 5
ORDER BY date
LIMIT 1
Then make the above a derived table (we call it lim) and join it to the original table to get all rows with previous dates: t.date < lim.date
SELECT t.*
FROM table1 AS t
JOIN
( SELECT date
FROM table1
WHERE id = 5
ORDER BY date
LIMIT 1
) AS lim
ON t.date < COALESCE(lim.date, '9999-12-31') ;
The COALESCE() is for the case when there are no rows at all with id=5 - and in that case we want all rows from the table.
Related
This question is part of a bigger mySQL query I have. So I have a table of 'playerIds', 'dates', 'scores' and 'problems'. It's Table T0 in the image attached. I am running a SQL query on it to get the most-recent row for all players where the 'date' is <= (2020-08-14 - 7days). Not all players will have a row with a date that satisfies that condition, so naturally those playerId rows will not appear in the resulting table (Table T1 in the pic).
Now what I want to do is to include those missing rows with 0 values for 'score' and 'problems' in the resulting table (See Table T2 in the pic). I am totally at a loss as to how to go about it since I am very new to SQL queries.
Here's the part of the SQL query which is producing Table T1 from T0, but I want to modify it such that it produces Table T2 from T0:
select *
from (
select *, row_number() over (partition by playerId order by date desc) as ranking
from player
where date<=date_add(date('2020-08-14'),interval -7 day)
) t
where t.ranking = 1
One option uses a subquery to list all the players, and then brings your current resultset with a left join:
select p.playerId, t.date, coalesce(t.score, 0) score, coalesce(t.problem, 0) problem
from (select distinct playerId from player) p
left join (
select p.*, row_number() over (partition by playerId order by date desc) as rn
from player p
where date <= '2020-08-14' - interval 7 day
) t on t.playerId = p.playerId and t.rn = 1
If you have a referential table for all players, you can just replace the select distinct subquery with that table.
I have mysql table t1 like this :
What i want to do is do calculations between all rows and save the value in new coloumn called diff
TICKETID| DATENEW | DIFF
16743 12:36:46 0
16744 12:51:25 15. minute
16745 12:57:25 6.5 minute
..........
.......
etc
i know there are similar questions ,but ive tried all of the solutions
posted here with no success,so how to solve this query ???
To get the time difference in minutes between the current and previous row, you can use timestampdiff on datenow and the previous time, which you can get via subquery:
select ticketid, datenew,
timestampdiff(minute,datenew,(select datenew from mytable t2
where t2.ticketid < t1.ticketid order by t2.ticketid desc limit 1)) as diff
from mytable t1
Update
Here's another way using a variable to store the previous datenew value that might be faster:
select ticketid, datenew, timestampdiff(minute,datenew,prevdatenew)
from (
select ticketid, datenew, #prevDateNew as prevdatenew,
#prevDateNew := datenew
from mytable order by ticketid
) t1
select
t1.*
,coalesce(timestampdiff(MINUTE,t2.dt,t1.dt),0) as tdiff
from t t1 left join t t2
on t1.id = t2.id+1
order by t1.id
As you are only looking for a difference between the current row and the next, you can join on the next row and calculate the time difference in minutes.
Note: This assumes there are no missing id's in the table. You might have to change the join condition if there were missing id's.
SQL Fiddle: http://www.sqlfiddle.com/#!9/4dcae/15
I have a PHP program w/MySQL database which contains many records. Two columns of particular relevance are incidentnumber and date. These both move forward only. However, sometimes a user enters data which is out of sequence; eg:
Incident Date
1 Jan 1 2000
2 Jan 1 2010
3 Jan 1 2002
It appears that incident 2 was entered with the wrong date, it should be Jan 1 2001.
Is there any way to query for records where the date is out of sequence? Or do I have to iterate through all records tracking last date to find the error?
ADDED NOTE: The incidents are not sequential (they might go 1,3,6,123, etc). Nor are the dates sequential. And these are columns in the same table.
This command selects any records for which there exists in the same table a record with a lower Incident number but a higher Date.
SELECT * FROM TableName T1 WHERE EXISTS
(SELECT * FROM TableName T2
WHERE T2.Incident < T1.Incident AND T2.Date > T1.Date)
This slightly more complex command will find only records for which are out of order in "both directions", meaning they have an later dated record earlier in the file and an earlier dated record later in the file. This avoids the situation in which making a mistake in a very early record in the file will make all the subsequent records appear out of order. However, it will not catch a problem in the two records with the lowest or highest incident numbers.
SELECT * FROM TableName T1 WHERE EXISTS
(SELECT * FROM TableName T2
WHERE T2.Incident < T1.Incident AND T2.Date > T1.Date)
AND EXISTS
(SELECT * FROM TableName T2
WHERE T2.Incident > T1.Incident AND T2.Date < T1.Date)
Finally, as ruakh points out in the comments, the above query gives you ALL the out-of-order records. Although that is, technically, what you wanted it makes it difficult to find the "point of breakage" in the chain of dates. The following query will give you only the records where the chain gets messed up, does not require IncidentID to increase monotonically, and allows deletions of incidents.
SELECT * FROM TableName T1 WHERE
Date < (SELECT Date FROM TableName T2 WHERE T2.IncidentID =
(SELECT MAX(IncidentID) FROM TableName T3 WHERE T3.IncidentID < T1.IncidentID))
OR Date > (SELECT Date FROM TableName T2 WHERE T2.IncidentID =
(SELECT MAX(IncidentID) FROM TableName T3 WHERE T3.IncidentID > T1.IncidentID))
(Not tested, since I don't have a copy of MySQL handy).
select * from yourtable t1
inner join yourtable t2
on t1.incident=t2.incident-1
and t1.date>t2.date
This selects all of the ids where the date is greater than the next records date. That should tell you which ones are out of order.
SELECT Incident FROM table a
WHERE a.Date > (SELECT b.Date FROM table b WHERE b.Incident = (a.Incident + 1))
In case that the IncidentID column is always in a regular incremental sequence:
SELECT c.IncidentID AS cincID, p.IncidentID AS pincID,
c.Date AS cDate, p.Date AS pDate,
DATEDIFF(c.Date, p.Date)
FROM Incident c, Incident p
WHERE c.IncidentID = (p.IncidentID + 1)
AND datediff(c.Date, p.Date) < 1
I have a mySQl db (name "stocks") with 50 tables, each tables with
id, symbol, date, time, open, high, low, close, volume as columns (9 columns).
I would like to know what is the last record for each table, ordered for date then time.
Should I have to ORDER BY all data for each table or there is a better way to just know last record?
I am asking help for a query that just return only last record for each table in db.
Thanks
PS For last record I mean most recent as Date then Time
There are two options how to do that:
-- I would use this only if you need more than one records
SELECT * FROM table ORDER BY date DESC LIMIT 1;
-- Way to go:
SELECT * FROM table WHERE date = (SELECT MAX(date) FROM table) LIMIT 1;
Don't forget to add index on date. If it's possible you add lot's of records at the same time you will have to add:
ORDER BY id DESC -- In case that date is highest for records for last records
ORDER BY time DESC -- Every other case
To the end of query
I am going to make the assumption that the record with the largest ID is the "last" (assuming strictly increasing sequential IDs that are unique within a table). If you have a better definition of "last" that could make a difference.
To get one "last" record, you could do:
Select * from table_1 where id = (select max(id) from table_1);
To get the results of all 50 tables into a single result set, you could do:
Select * from table_1 where id = (select max(id) from table_1)
union
Select * from table_2 where id = (select max(id) from table_2)
union
Select * from table_3 where id = (select max(id) from table_3)
union...
A MySQL-specific solution could be
Select * from table_1 order by id desc limit 1
union
Select * from table_2 order by id desc limit 1
union
Select * from table_3 order by id desc limit 1
union...
Based on your edit (where you actually define what you mean by "last"):
Select * from table_1 order by date desc, time desc, id desc limit 1
union
Select * from table_2 order by date desc, time desc, id desc limit 1
union
Select * from table_3 order by date desc, time desc, id desc limit 1
union...
Here is one way to do it without sorting the table:
select * from tab1
where time = (select max(time)
from tab1
where date = (select max(date) from tab1))
and date = (select max(date) from tab1)
It should be very fast, like, O(c), provided that both columns are indexed, otherwise the time will simply be O(n)
I've searched around SO and can't seem to find a question with an answer that works fine for me. I have a table with almost 2 million rows in, and each row has a MySQL Date formatted field.
I'd like to work out (in seconds) how often a row was inserted, so work out the average difference between the dates of all the rows with a SQL query.
Any ideas?
-- EDIT --
Here's what my table looks like
id, name, date (datetime), age, gender
If you want to know how often (on average) a row was inserted, I don't think you need to calculate all the differences. You only need to sum up the differences between adjacent rows (adjacent based on the timestamp) and divide the result by the number of the summands.
The formula
((T1-T0) + (T2-T1) + … + (TN-TN-1)) / N
can obviously be simplified to merely
(TN-T0) / N
So, the query would be something like this:
SELECT TIMESTAMPDIFF(SECOND, MIN(date), MAX(date)) / (COUNT(*) - 1)
FROM atable
Make sure the number of rows is more than 1, or you'll get the Division By Zero error. Still, if you like, you can prevent the error with a simple trick:
SELECT
IFNULL(TIMESTAMPDIFF(SECOND, MIN(date), MAX(date)) / NULLIF(COUNT(*) - 1, 0), 0)
FROM atable
Now you can safely run the query against a table with a single row.
Give this a shot:
select AVG(theDelay) from (
select TIMESTAMPDIFF(SECOND,a.date, b.date) as theDelay
from myTable a
join myTable b on b.date = (select MIN(x.date)
from myTable x
where x.date > a.date)
) p
The inner query joins each row with the next row (by date) and returns the number of seconds between them. That query is then encapsulated and is queried for the average number of seconds.
EDIT: If your ID column is auto-incrementing and they are in date order, you can speed it up a bit by joining to the next ID row rather than the MIN next date.
select AVG(theDelay) from (
select TIMESTAMPDIFF(SECOND,a.date, b.date) as theDelay
from myTable a
join myTable b on b.date = (select MIN(x.id)
from myTable x
where x.id > a.id)
) p
EDIT2: As brilliantly commented by Mikael Eriksson, you may be able to just do:
select (TIMESTAMPDIFF(SECOND,(MAX(date),MIN(date)) / COUNT(*)) from myTable
There's a lot you can do with this to eliminate off-peak hours or big spans without a new record, using the join syntax in my first example.
Try this:
select avg(diff) as AverageSecondsBetweenDates
from (
select TIMESTAMPDIFF(SECOND, t1.MyDate, min(t2.MyDate)) as diff
from MyTable t1
inner join MyTable t2 on t2.MyDate > t1.MyDate
group by t1.MyDate
) a