Couldn't really explain my problem with words, but with an example I can show it clearly:
I have a table like this:
id num val
0 3 10
1 5 12
2 7 12
3 11 15
And I want to go through all the rows, and calculate the increase of the "num", and multiply that difference with the "val" value. And when I calculated all of these, I want to add these results together.
This is the mathematical equation, that I want to run on the table:
Result = (3-0)*10 + (5-3)*12 + (7-5)*12 + (11-7)*15
138 = Result
Thank you.
You can do with mysql variables, but you will still get one record for each entry.
select
#lastTotal := #lastTotal + ( (yt.num - #lastNum) * yt.val ) thisLineTotal,
#lastNum := yt.num as saveForNextRow,
yt.id
from
yourTable yt,
( select #lastTotal := 0,
#lastNum := 0 ) sqlvars
order by
id
This SHOULD give you what you want to confirm the calculations to each record basis.
Now, to get the one record and one column result, you can wrap it such as
select
pq.thisLineTotal
from
(above entire query ) as pq
order by
pq.id DESC
limit 1
Assuming the IDs are consecutive as your sample data suggests, just join the table to itself:
select sum((t1.num-ifnull(t2.num,0))*t1.val) YourValue
from YourTable t1
left join YourTable t2
on t2.id = t1.id - 1;
http://www.sqlfiddle.com/#!2/40b9f/12
This will give you the total. Make sure to order in the order you wish - I have ordered by id
SET #runtot:=0;
SET #prevval:=0;
select max(rt) as total FROM (
SELECT
q.val,
q.num,
(#runtot := #runtot + (q.num- #prevval) * q.val) AS rt,
(#prevval := q.num) AS pv
FROM thetable q
ORDER by ID) tot
If you want to see the details of the calculation, leave out the outer select as so:
SET #runtot:=0;
SET #prevval:=0;
SELECT
q.val,
q.num,
(#runtot := #runtot + (q.num- #prevval) * q.val) AS rt,
(#prevval := q.num) AS pv
FROM thetable q
ORDER by ID
If it is possible to have negative numbers for your column values, using max(rt) won't work for the total. You should then use:
SET #runtot:=0;
SET #prevval:=0;
select #runtot as total FROM (
SELECT
q.val,
q.num,
(#runtot := #runtot + (q.num- #prevval) * q.val) AS rt,
(#prevval := q.num) AS pv
FROM thetable q
ORDER by ID) tot LIMIT 1
Related
I have a table like this:
01-Jul-17 100
02-Jul-17 100
03-Jul-17 300
04-Jul-17 300
05-Jul-17 500
06-Jul-17 500
07-Jul-17 300
08-Jul-17 400
09-Jul-17 100
10-Jul-17 100
What I want to output is (in this order) by eliminating the continuous duplicates but not all duplicates:
100
300
500
300
400
100
I cannot select Distinct, as it will eliminate the second instances of 300, 100. Is there a way to achieve this result in MySQL?
Thanks!
You want to get the previous value. If the dates really have no gaps or duplicates, just do:
select t.*
from t left join
t tprev
on t.col1 = date_add(tprev.col1, interval 1 day)
where tprev.col2 is null or tprev.col2 <> t.col2;
EDIT:
If the dates don't meet these conditions, then you can use variables:
select t.*
from (select t.*,
(#rn := if(#v = col2, #rn + 1,
if(#v := col2, 1, 1)
)
) as rn
from t cross join
(select #v := 0, #rn := 0) params
order by t.col1
) t
where rn = 1;
Note that MySQL does not guarantee the order of evaluation of expressions in the SELECT. So variables should not be assigned in one expression and then used in another -- they should be assigned in a single expression.
One way to handle this problem is by using session variables to track the changes of the values as ordered by your date column. In the query below, we keep track of the value, ordered by date, and assign a row number to each group of identical value. Then, only the first value in each group is retained. Note that this approach is robust to any number of duplicates. It is also robust with respect to there being gaps in your dates, so long as each record can be ordered by date.
SET #rn = 1;
SET #val = NULL;
SELECT t.val
FROM
(
SELECT
#rn:=CASE WHEN #val = val THEN #rn+1 ELSE 1 END rn,
#val:=val AS val,
dt
FROM yourTable
ORDER BY dt
) t
WHERE t.rn = 1
ORDER BY t.dt;
Output:
Demo here:
Rextester
You can make use of lag and lead functions.
select y from (select y , lag(y,1,0) over (order by x) as prev_y from t1) where y <> prev_y;
Let's say there are millions of records in my_table.
Here is my query to extract rows with a specific name from list:
SELECT * FROM my_table WHERE Name IN ('name1','name2','name3','name4')
How do I limit the returned result per name1, name2, etc?
The following query would limit the whole result (to 100).
SELECT * FROM my_table WHERE Name IN ('name1','name2','name3','name4') LIMIT 100
I need to limit to 100 for each name.
This is a bit of a pain in MySQL, but the best method is probably variables:
select t.*
from (select t.*,
(#rn := if(#n = name, #rn + 1,
if(#n := name, 1, 1)
)
) as rn
from my_table t cross join
(select #n := '', #rn := 0) params
order by name
) t
where rn <= 100;
If you want to limit this to a subset of the names, then add the where clause to the subquery.
Note: If you want to pick certain rows -- such as the oldest or newest or biggest or tallest -- just add a second key to the order by in the subquery.
Try
SELECT * FROM my_table WHERE Name IN ('name1','name2','name3','name4') FETCH FIRST 100 ROWS ONLY
Is it possible to get specific row in query using like SUM?
Example:
id tickets
1 10 1-10 10=10
2 35 11-45 10+35=45
3 45 46-90 10+35+45=90
4 110 91-200 10+35+45+110=200
Total: 200 tickets(In SUM), I need to get row ID who have ticket with number like 23(Output would be ID: 2, because ID: 2 contains 11-45tickets in SUM)
You can do it by defining a local variable into your select query (in form clause), e.g.:
select id, #total := #total + tickets as seats
from test, (select #total := 0) t
Here is the SQL Fiddle.
You seem to want the row where "23" fits in. I think this does the trick:
select t.*
from (select t.*, (#total := #total + tickets) as running_total
from t cross join
(select #total := 0) params
order by id
) t
where 23 > running_total - tickets and 23 <= running_total;
SELECT
d.id
,d.tickets
,CONCAT(
TRIM(CAST(d.RunningTotal - d.tickets + 1 AS CHAR(10)))
,'-'
,TRIM(CAST(d.RunningTotal AS CHAR(10)))
) as TicketRange
,d.RunningTotal
FROM
(
SELECT
id
,tickets
,#total := #total + tickets as RunningTotal
FROM
test
CROSS JOIN (select #total := 0) var
ORDER BY
id
) d
This is similar to Darshan's answer but there are a few key differences:
You shouldn't use implicit join syntax, explicit join has more functionality in the long run and has been a standard for more than 20 years
ORDER BY will make a huge difference on your running total when calculated with a variable! if you change the order it will calculate differently so you need to consider how you want to do the running total, by date? by id? by??? and make sure you put it in the query.
finally I actually calculated the range as well.
And here is how you can do it without using variables:
SELECT
d.id
,d.tickets
,CONCAT(
TRIM(d.LowRange)
,'-'
,TRIM(
CAST(RunningTotal AS CHAR(10))
)
) as TicketRange
,d.RunningTotal
FROM
(
SELECT
t.id
,t.tickets
,CAST(COALESCE(SUM(t2.tickets),0) + 1 AS CHAR(10)) as LowRange
,t.tickets + COALESCE(SUM(t2.tickets),0) as RunningTotal
FROM
test t
LEFT JOIN test t2
ON t.id > t2. id
GROUP BY
t.id
,t.tickets
) d
I am using a modified version of a query similiar to another question here:Convert SQL Server query to MySQL
Select *
from
(
SELECT tbl.*, #counter := #counter +1 counter
FROM (select #counter:=0) initvar, tbl
Where client_id = 55
ORDER BY ordcolumn
) X
where counter >= (80/100 * #counter);
ORDER BY ordcolumn
tbl.* contains the field 'client_id' and I am attempting to get the top 20% of the records for each client_id in a single statement. Right now if I feed it a single client_id in the where statement it gives me the correct results, however if I feed it multiple client_id's it simply takes the top 20% of the combined recordset instead of doing each client_id individually.
I'm aware of how to do this in most databases, but the logic in MySQL is eluding me. I get the feeling it involves some ranking and partitioning.
Sample data is pretty straight forward.
Client_id rate
1 1
1 2
1 3
(etc to rate = 100)
2 1
2 2
2 3
(etc to rate = 100)
Actual values aren't that clean, but it works.
As an added bonus...there is also a date field associated to these records and 1 to 100 exists for this client for multiple dates. I need to grab the top 20% of records for each client_id, year(date),month(date)
You need to do the enumeration for each client:
SELECT *
FROM (SELECT tbl.*, #counter := #counter +1 counter
(#rn := if(#c = client_id, #rn + 1,
if(#c := client_id, 1, 1)
)
)
FROM (select #c := -1, #rn := 0) initvar CROSS JOIN tbl
ORDER BY client_id, ordcolumn
) t cross join
(SELECT client_id, COUNT(*) as cnt
FROM tbl
GROUP BY client_id
) tt
where rn >= (80/100 * tt.cnt);
ORDER BY ordcolumn;
Using Gordon's answer as a starting point, I think this might be closer to what you need.
SELECT t.*
, (#counter := #counter+1) AS overallRow
, (#clientRow := if(#prevClient = t.client_id, #clientRow + 1,
if(#prevClient := t.client_id, 1, 1) -- This just updates #prevClient without creating an extra field, though it makes it a little harder to read
)
) AS clientRow
-- Alteratively (for everything done in clientRow)
, #clientRow := if(#prevClient = t.client_id, #clientRow + 1, 1) AS clientRow
, #prevClient := t.client_id AS extraField
-- This may be more reliable as well; I not sure if the order
-- of evaluation of IF(,,) is reliable enough to guarantee
-- no side effects in the non-"alternatively" clientRow calculation.
FROM tbl AS t
INNER JOIN (
SELECT client_id, COUNT(*) AS c
FROM tbl
GROUP BY client_id
) AS cc ON tbl.client_id = cc.client_id
INNER JOIN (select #prevClient := -1, #clientRow := 0) AS initvar ON 1 = 1
WHERE t.client_id = 55
HAVING clientRow * 5 < cc.c -- You can use a HAVING without a GROUP BY in MySQL
-- (note that clientRow is derived, so you cannot use it in the `WHERE`)
ORDER BY t.client_id, t.ordcolumn
;
This type of question is asked every now and then. The queries provided works, but it affects performance.
I have tried the JOIN method:
SELECT *
FROM nbk_tabl
INNER JOIN (
SELECT ITEM_NO, MAX(REF_DATE) as LDATE
FROM nbk_tabl
GROUP BY ITEM_NO) nbk2
ON nbk_tabl.REF_DATE = nbk2.LDATE
AND nbk_tabl.ITEM_NO = nbk2.ITEM_NO
And the tuple one (way slower):
SELECT *
FROM nbk_tabl
WHERE REF_DATE IN (
SELECT MAX(REF_DATE)
FROM nbk_tabl
GROUP BY ITEM_NO
)
Is there any other performance friendly way of doing this?
EDIT: To be clear, I'm applying this to a table with thousands of rows.
Yes, there is a faster way.
select *
from nbk_table
order by ref_date desc
limit <n>
Where is the number of rows that you want to return.
Hold on. I see you are trying to do this for a particular item. You might try this:
select *
from nbk_table n
where ref_date = (select max(ref_date) from nbk_table n2 where n.item_no = n2.item_no)
It might optimize better than the "in" version.
Also in MySQL you can use user variables (Suppose nbk_tabl.Item_no<>0):
select *
from (
select nbk_tabl.*,
#i := if(#ITEM_NO = ITEM_NO, #i + 1, 1) as row_num,
#ITEM_NO := ITEM_NO as t_itemNo
from nbk_tabl,(select #i := 0, #ITEM_NO := 0) t
order by Item_no, REF_DATE DESC
) as x where x.row_num = 1;