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aggregate query without group by


You can use a correlated subquery to compute the aggregated values:

SELECT JNumber,
       (SELECT SUM(Quantity * Weight) /
SUM(Quantitiy)
        FROM SPJ
        JOIN Part USING (PNumber)
        WHERE JNumber = Project.JNumber
       ) AS AvgParts
FROM Project

This is typically just as fast as a join with GROUP BY, but slightly more complex.


Categories : SQL

Related to : aggregate query without group by
invalid in select list because it is not contained in either an aggregate function or the GROUP BY clause
You need to remove qualitystd.qualitystd from the Select statement, you can only use those fields in select statement that are used in group by clause, so if you need to counts of different qualitystd.qualitystd_id you can use only qualitystd.qualitystd_id and count(qualitystd.qualitystd_id) as you are grouping by the data on qualitystd.qualitystd_id. So the sql statement will be something like t

Categories : Sql Server
MySQL aggregate function with GROUP BY to avoid NULL's overwriting desired field value
Just aggregate your results using MAX: SELECT lesson_number, MAX(CASE WHEN dayofweek = 0 THEN lesson_date END) AS 'MONDAY', MAX(CASE WHEN dayofweek = 1 THEN lesson_date END) AS 'TUESDAY', MAX(CASE WHEN dayofweek = 2 THEN lesson_date END) AS 'WEDNESDAY', MAX(CASE WHEN dayofweek = 3 THEN lesson_date END) AS 'THURSDAY', MAX(CASE WHEN dayofweek = 4 THEN lesson_date END) AS 'FR

Categories : Mysql
Column is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause
Well, you have columns in the select list that are not in the group by. This is generally not allowed. Try either: select SalesOrderID, salesorderdetailid, SUM(OrderQty * UnitPrice) as price from sales.SalesOrderDetail group by SalesOrderID, salesorderdetailid or select SalesOrderID, SUM(OrderQty * UnitPrice) as price from sales.SalesOrderDetail group by SalesOrderID or select salesorderd

Categories : SQL
Column 'city.POPULATION' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause
You should include c.population in the group by, since you use that column outside the count aggregate function: COUNT( NB_births_M+ NB_births_F)/C.population Use this (added D.NB_deaths_F too since it is necessary too): select D.ID_city , D.ID_year , D.NB_deaths , D.NB_births_M , D.NB_births_F , D.NB_population_AS , D.NB_population_ACL , COUNT( NB_births_M+

Categories : SQL
group MongoDB collection by month and aggregate revenue for churn/revenue chart
If I understand correctly you want the revenue by customer and by month. The $group operator in an aggregate query is exactly what you're looking for here. db.orders.aggregate([ { $group : { _id: { customer_id: "$customer_id", month: {$month: "$order_date"} }, revenue: {$sum: "$net_revenue"} } }], ... ); W

Categories : Mongodb
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