I have this Table
I want to run subqueries first then add them together grouped by date
Expected Result should be like this:
I am running this query
(
SELECT DATE_FORMAT(dd1.modified_datetime,'%Y-%m-%d') as date, (v1+v2) as value FROM
(SELECT modified_datetime, Sum(data->"$.amount") as v1
FROM transactions
GROUP BY modified_datetime) as dd1 ,
(SELECT modified_datetime, MAX(data->"$.amount") as v2
FROM transactions
GROUP BY modified_datetime) as dd2
GROUP BY dd1.modified_datetime, value
)
and getting this result:
Use JOIN between subqueries and every next one:
(SELECT modified_datetime, Sum(data->"$.amount") as v1
FROM transactions
GROUP BY modified_datetime) as dd1 JOIN
(SELECT modified_datetime, MAX(data->"$.amount") as v2
FROM transactions
GROUP BY modified_datetime) as dd2 ON dd1.modified_datetime=dd2.modified_datetime
If I followed you correctly, you can use union all and aggregation:
select date_format(dt, '%Y-%m-%d') dt_day, sum(amount) value
from (
select modified_datetime dt, data ->> '$.amount' amount from transactions
union all
select created_datetime, data ->> '$.amount' from transactions
) t
group by dt_day
order by dt_day
Related
Is it possible to order when the data comes from many select and union it together? Such as
In this statement, the vouchers data is not showing in the same sequence as I saved on the database, I also tried it with "ORDER BY v_payments.payment_id ASC" but won't be worked
( SELECT order_id as id, order_date as date, ... , time FROM orders WHERE client_code = '$searchId' AND order_status = 1 AND order_date BETWEEN '$start_date' AND '$end_date' ORDER BY time)
UNION
( SELECT vouchers.voucher_id as id, vouchers.payment_date as date, v_payments.account_name as name, ac_balance as oldBalance, v_payments.debit as debitAmount, v_payments.description as descriptions,
vouchers.v_no as v_no, vouchers.v_type as v_type, v_payments.credit as creditAmount, time, zero as tax, zero as freightAmount FROM vouchers INNER JOIN v_payments
ON vouchers.voucher_id = v_payments.voucher_id WHERE v_payments.client_code = '$searchId' AND voucher_status = 1 AND vouchers.payment_date BETWEEN '$start_date' AND '$end_date' ORDER BY v_payments.payment_id ASC , time )
UNION
( SELECT return_id as id, return_date as date, ... , time FROM w_return WHERE client_code = '$searchId' AND w_return_status = 1 AND return_date BETWEEN '$start_date' AND '$end_date' ORDER BY time)
Wrap the sub-select queries in the union within a SELECT
SELECT id, name
FROM
(
SELECT id, name FROM fruits
UNION
SELECT id, name FROM vegetables
)
foods
ORDER BY name
If you want the order to only apply to one of the sub-selects, use parentheses as you are doing.
Note that depending on your DB, the syntax may differ here. And if that's the case, you may get better help by specifying what DB server (MySQL, SQL Server, etc.) you are using and any error messages that result.
You need to put the ORDER BY at the end of the statement i.e. you are ordering the final resultset after union-ing the 3 intermediate resultsets
To use an ORDER BY or LIMIT clause to sort or limit the entire UNION result, parenthesize the individual SELECT statements and place the ORDER BY or LIMIT after the last one. See link below:
ORDER BY and LIMIT in Unions
(SELECT a FROM t1 WHERE a=10 AND B=1)
UNION
(SELECT a FROM t2 WHERE a=11 AND B=2)
ORDER BY a LIMIT 10;
i have a mysql table with colums: id(primary), name(varchar), TIME(timestamp)
ID , NAME , TIME
i want to get just first and last log for each day
example if i have data like this
1,name,2018-20-21 12:35:00
2,name,2018-20-21 13:38:00
3,name,2018-20-21 14:25:00
4,name,2018-20-21 15:39:00
5,name,2018-20-21 21:48:00
6,name,2018-20-22 13:25:00
7,name,2018-20-22 14:39:00
8,name,2018-20-22 19:48:00
i want to get in just this
1,name,2018-20-21 12:35:00
5,name,2018-20-21 21:48:00
6,name,2018-20-22 13:25:00
8,name,2018-20-22 19:48:00
Try this:
SELECT name, MAX(time), MIN(time) FROM Table GROUP BY DATE(time);
You could use the union for the min and the max time group by date
and join this with your table
select * from my_table
inner join (
select * from (
select min(time) my_time
from my_table
group by date(time)
union
select max(time)
from my_table
group by date(time)
) t on t.my_time = my_table.time
order by my_table.time
Hope this helps.
SELECT id, tmp.name, tmp.time FROM
(SELECT id, name, min(time) as time FROM table1 GROUP BY DATE(time)
UNION ALL
(SELECT id, name, max(time) as time FROM table1 GROUP BY DATE(time)) tmp
ORDER BY tmp.time
You can try selecting the min and max for each day, since you want the entire line, a join is needed
and to filter out the actual min and max day, a aub query is needed
SELECT id, name, time
FROM
(
SELECT t2.*, MIN(DATE(t.time)) As min0 MAX(DATE(t.time)) As max0
FROM
table t
INNER JOIN table t2 ON t.id = t2.id
GROUP BY
DATE (t.time),
min0,
max0
) a
SELECT
l.id,l.name,l.time
FROM
log l
LEFT JOIN
(SELECT
max(time) as maxTime
FROM
log
GROUP BY date(time)) l1 ON l.time = l1.maxTime
LEFT JOIN
(SELECT
min(time) as minTime
FROM
log
GROUP BY date(time)) l2 ON l.time = l2.minTime
WHERE
(maxTime IS NOT NULL
OR minTime IS NOT NUll);
SELECT * from stack.log;
I have a simple group by query:
SELECT timestamp, COUNT(users)
FROM my_table
GROUP BY users
How do I add a sum_each_day column that will sum the users count of each row and will aggregate it forward to the next row and so on
The output should be like this:
timestamp | users | sum_each_day
2015-11-27 1 1
2015-11-28 5 6
2015-11-29 3 9
2015-11-30 7 16
Thanks in advance
You could use a sub-query, like this:
SELECT timestamp,
num_users,
(SELECT COUNT(users)
FROM my_table
WHERE timestamp <= main.timestamp) sum_users
FROM (
SELECT timestamp,
COUNT(users) num_users
FROM my_table
GROUP BY timestamp
) main
If you really need this in mysql it'll cost some performance but i believe a sub query with a count will solve it:
SELECT t1.timestamp, count (), select count () from my_table t2 where t2.timestamp <= t1.timestamp From my_table t1 Group by users
If you display this data through a scripting language like PHP it would be easier to keep a counter and display the aggregate per row.
I would do this using variable:
SET #total := 0;
SELECT timestamp, DayCount, (#total := #total + DayCount) AS Total
FROM
(SELECT timestamp, COUNT(users) AS DayCount
FROM my_table
GROUP BY timestamp) AS t1
Fiddler: I am not using your table structure here, but you can get idea
If I understand correclty, this will work:
set #c=0;
SELECT `timestamp`,sum(`users`),(select #c:=#c+sum(`users`))
FROM `my_table`
group by `timestamp`;
I have two SELECT statements that give the values "TotalSales" and "VendorPay_Com". I want to be able to subtract VendorPay_Com from TotalSales within the one MySQL statement to get the value "Outstanding_Funds" but I haven't found a reliable way to do so.
These are my two statements:
Query 1:
SELECT SUM(Price) AS TotalSales
FROM PROPERTY
WHERE Status = 'Sold';
Query 2:
SELECT SUM(AC.Amount) AS VendorPay_Comm
FROM (
SELECT Amount FROM lawyer_pays_vendor
UNION ALL
SELECT CommissionEarned AS `Amount` FROM COMMISSION AS C WHERE C.`status` = 'Paid') AS AC
Any help on this matter would be greatly appreciated :)
You can do it as follows :
select (select ...) - (select ...)
In your example, simply :
select
(
SELECT SUM(Price) AS TotalSales
FROM PROPERTY
WHERE Status = 'Sold'
)
-
(
SELECT SUM(AC.Amount) AS VendorPay_Comm
FROM (
SELECT Amount FROM lawyer_pays_vendor
UNION ALL
SELECT CommissionEarned AS `Amount` FROM COMMISSION AS C WHERE C.`status` = 'Paid') AS AC
) AS Outstanding_Funds
Try
SELECT TotalSales-VendorPay_Comm AS Outstanding_Funds
FROM
(SELECT SUM(Price) AS TotalSales
FROM PROPERTY
WHERE Status = 'Sold') t1,
(SELECT SUM(Amount) AS VendorPay_Comm
FROM (SELECT Amount FROM lawyer_pays_vendor
UNION ALL
SELECT CommissionEarned AS Amount
FROM COMMISSION
WHERE Status = 'Paid') t0) t2
Here is sqlfiddle
I have a table called receiving with 4 columns:
id, date, volume, volume_units
The volume units are always stored as a value of either "Lbs" or "Gals".
I am trying to write an SQL query to get the sum of the volumes in Lbs and Gals for a specific date range. Something along the lines of: (which doesn't work)
SELECT sum(p1.volume) as lbs,
p1.volume_units,
sum(p2.volume) as gals,
p2.volume_units
FROM receiving as p1, receiving as p2
where p1.volume_units = 'Lbs'
and p2.volume_units = 'Gals'
and p1.date between "2012-01-01" and "2012-03-07"
and p2.date between "2012-01-01" and "2012-03-07"
When I run these queries separately the results are way off. I know the join is wrong here, but I don't know what I am doing wrong to fix it.
SELECT SUM(volume) AS total_sum,
volume_units
FROM receiving
WHERE `date` BETWEEN '2012-01-01'
AND '2012-03-07'
GROUP BY volume_units
You can achieve this in one query by using IF(condition,then,else) within the SUM:
SELECT SUM(IF(volume_units="Lbs",volume,0)) as lbs,
SUM(IF(volume_units="Gals",volume,0)) as gals,
FROM receiving
WHERE `date` between "2012-01-01" and "2012-03-07"
This only adds volume if it is of the right unit.
This query will display the totals for each ID.
SELECT s.`id`,
CONCAT(s.TotalLbsVolume, ' ', 'lbs') as TotalLBS,
CONCAT(s.TotalGalVolume, ' ', 'gals') as TotalGAL
FROM
(
SELECT `id`, SUM(`volume`) as TotalLbsVolume
FROM Receiving a INNER JOIN
(
SELECT `id`, SUM(`volume`) as TotalGalVolume
FROM Receiving
WHERE (volume_units = 'Gals') AND
(`date` between '2012-01-01' and '2012-03-07')
GROUP BY `id`
) b ON a.`id` = b.`id`
WHERE (volume_units = 'Lbs') AND
(`date` between '2012-01-01' and '2012-03-07')
GROUP BY `id`
) s
this is a cross join with no visible condition on the join, i don't think you meant that
if you want to sum quantities you don't need to join at all, just group as zerkms did
You can simply group by date and volume_units without self-join.
SELECT date, volume_units, sum(volume) sum_vol
FROM receving
WHERE date between "2012-01-01" and "2012-03-07"
GROUP BY date, volume_units
Sample test:
select d, vol_units, sum(vol) sum_vol
from
(
select 1 id, '2012-03-07' d, 1 vol, 'lbs' vol_units
union
select 2 id, '2012-03-07' d, 2 vol, 'Gals' vol_units
union
select 3 id, '2012-03-08' d, 1 vol, 'lbs' vol_units
union
select 4 id, '2012-03-08' d, 2 vol, 'Gals' vol_units
union
select 5 id, '2012-03-07' d, 10 vol, 'lbs' vol_units
) t
group by d, vol_units