I understand why the first query needs a GROUP BY, as it doesn't know which date to apply the sum to, but I don't understand why this is the case with the second query. The value that ultimately is the max amount is already contained in the table - it is not calculated like SUM is. thank you
-- First Query
select
sum(OrderSales),OrderDates
From Orders
-- Second Query
select
max(FilmOscarWins),FilmName
From tblFilm
It is not the SUM and MAX that require the GROUP BY, it is the unaggregated column.
If you just write this, you will get a single row, for the maximum value of the FilmOscarWins column across the whole table:
select
max(FilmOscarWins)
From
tblFilm
If the most Oscars any film won was 12, that one row will say 12. But there could be multiple films, all of which won 12 Oscars, so if we ask for the FilmName alongside that 12, there is no single answer.
By adding the Group By, we fundamentally change the query: instead of returning one number for the whole table, it will return one row for each group - which in this case, means one row for each film.
If you do want to get a list of all those films which had the maximum 12 Oscars, you have to do something more complicated, such as using a sub-query to first find that single number (12) and then find all the rows matching it:
select
FilmOscarWins,
FilmName
From
tblFilm
Where FilmOscarWins = (
select
max(FilmOscarWins)
From
tblFilm
)
If you want the film with the most Oscar wins, then use select top:
select top (1) f.*
From tblFilm f
order by FilmOscarWins desc;
In an aggregation query, the select columns need to be consistent with the group by columns -- the unaggregated columns in the select must match the group by.
Related
Q1: After using the Group By function, why does it only output one row of each group at most? Does this mean that having is supposed to filter the group rather than filter the records in each group?
Q2: I want to find the records in each group whose ages are greater than the average age of that group. I tried the following, but it returns nothing. How should I fix this?
SELECT *, avg(age) FROM Mytable Group By country Having age > avg(age)
Thanks!!!!
You can calculate the average age for each country in a subquery and join that to your table for filtering:
SELECT mt.*, MtAvg.AvgAge
FROM Mytable mt
inner join
(
select mtavgs.country
, avg(mtavgs.age) as AvgAge
from Mytable mtavgs
group by mtavgs.country
) MTAvg
on mtavg.country=mt.country
and mt.Age > mtavg.AvgAge
GROUP BY returns always 1 row per unique combination of values in the GROUP BY columns listed (provided that they are not removed by a HAVING clause). The subquery in our example (alias: MTAvg) will calculate a single row per country. We will use its results for filtering the main table rows by applying the condition in the INNER JOIN clause; we will also report that average by including the calculated average age.
GROUP BY is a keyword that is called an aggregate function. Check this out here for further reading SQL Group By tutorial
What it does is it lumps all the results together into one row. In your example it would lump all the results with the same country together.
Not quite sure what exactly your query needs to be to solve your exact problem. I would however look into what are called window functions in SQL. I believe what you first need to do is write a window function to find the average age in each group. Then you can write a query to return the results you need
Depending on your dbms type and version, you may be able to use a "window function" that will calculate the average per country and with this approach it makes the calculation available on every row. Once that data is present as a "derived table" you can simply use a where clause to filter for the ages that are greater then the calculated average per country.
SELECT mt.*
FROM (
SELECT *
, avg(age) OVER(PARTITION BY country) AS AvgAge
FROM Mytable
) mt
WHERE mt.Age > mt.AvgAge
On a report builder 3.0, i retreived some items and counted them using a Count aggregate. Now i want to order them from highest to lowest. How do i use the ORDER BY function on the aggregated column? The picture below show the a column that i want to ORDER BY it, it is ticked.
Pic
The code is vers simple as shown bellow:
SELECT DISTINCT act_id,NameOfAct,
FROM Acts
Your picture indicates you also want a Total row at the bottom:
SELECT
COALESCE(NameOfAct,'Total') NameOfAct,
COUNT(DISTINCT act_id) c
FROM Acts
GROUP BY ROLLUP(NameOfAct)
ORDER BY
CASE WHEN NameOfAct is null THEN 1 ELSE 0 END,
c DESC;
Result of example data:
NameOfAct count
-------------- -------
Act_B 3
Act_A 2
Act_Z 1
Total 6
Try it with example rows at: http://sqlfiddle.com/#!18/dbd6c/2
I looked at the Pic. So you might have duplicate acts with the same name. And you want to know the number of acts that have the same unique name.
You might want to group the results by name:
GROUP BY NameOfAct
And include the act names and their counts in the query results:
SELECT NameOfAct, COUNT(*) AS ActCount
(Since the act_id column is not included in the groups, you need to omit it in the SELECT. The DISTINCT is also not necessary anymore, since all groups are unique already.)
Finally, you can sort the data (probably descending to get the acts with the largest count on top):
ORDER BY ActCount DESC
Your complete query would become something like this:
SELECT NameOfAct, COUNT(*) AS ActCount
FROM Acts
GROUP BY NameOfAct
ORDER BY ActCount DESC
Edit:
By the way, you use field "act_id" in your SELECT clause. That's somewhat confusing. If you want to know counts, you want to look at either the complete table data or group the table data into smaller groups (with the GROUP BY clause). Then you can use aggregate functions to get more information about those groups (or the whole table), like counts, average values, minima, maxima...
Single record information, like an act's ID in your case, is typically not important if you want to use statistic/aggregate methods on grouped data. Suppose your query returns an act name which is used 10 times. Then you have 10 records in your table, each with a unique act_id, but with the same name.
If you need just one act_id that represents each group / act name (and assuming act_id is an autonumbering field), you might include the latest / largest act_id value in the query using the MAX aggregate function:
SELECT NameOfAct, COUNT(*) AS ActCount, MAX(act_id) AS LatestActId
(The rest of the query remains the same.)
I have a table with several "ticket" records in it. Each ticket is stored by day (i.e. 2011-07-30 00:00:00.000) I would like to count the unique records in each month by year I have used the following sql statement
SELECT DISTINCT
YEAR(TICKETDATE) as TICKETYEAR,
MONTH(TICKETDATE) AS TICKETMONTH,
COUNT(DISTINCT TICKETID) AS DAILYTICKETCOUNT
FROM
NAT_JOBLINE
GROUP BY
YEAR(TICKETDATE),
MONTH(TICKETDATE)
ORDER BY
YEAR(TICKETDATE),
MONTH(TICKETDATE)
This does produce a count but it is wrong as it picks up the unique tickets for every day. I just want a unique count by month.
Try combining Year and Month into one field, and grouping on that new field.
You may have to cast them to varchar to ensure that they don't simply get added together. Or.. you could multiple through the year...
SELECT
(YEAR(TICKETDATE) * 100) + MONTH(TICKETDATE),
count(*) AS DAILYTICKETCOUNT
FROM NAT_JOBLINE GROUP BY
(YEAR(TICKETDATE) * 100) + MONTH(TICKETDATE)
Presuming that TICKETID is not a primary or unique key, but does appear multiple times in table NAT_JOBLINE, that query should work. If it is unique (does not occur in more than 1 row per value), you will need to select on a different column, one that uniquely identifies the "entity" that you want to count, if not each occurance/instance/reference of that entity.
(As ever, it is hard to tell without working with the actual data.)
I think you need to remove the first distinct. You already have the group by. If I was the first Distict I would be confused as to what I was supposed to do.
SELECT
YEAR(TICKETDATE) as TICKETYEAR,
MONTH(TICKETDATE) AS TICKETMONTH,
COUNT(DISTINCT TICKETID) AS DAILYTICKETCOUNT
FROM NAT_JOBLINE
GROUP BY YEAR(TICKETDATE), MONTH(TICKETDATE)
ORDER BY YEAR(TICKETDATE), MONTH(TICKETDATE)
From what I understand from your comments to Phillip Kelley's solution:
SELECT TICKETDATE, COUNT(*) AS DAILYTICKETCOUNT
FROM NAT_JOBLINE
GROUP BY TICKETDATE
should do the trick, but I suggest you update your question.
I am running the following queries against a database:
CREATE TEMPORARY TABLE med_error_third_party_tmp
SELECT `med_error_category`.description AS category, `med_error_third_party_category`.error_count AS error_count
FROM
`med_error_category` INNER JOIN `med_error_third_party_category` ON med_error_category.`id` = `med_error_third_party_category`.`category`
WHERE
year = 2003
GROUP BY `med_error_category`.id;
The only problem is that when I create the temporary table and do a select * on it then it returns multiple rows, but the query above only returns one row. It seems to always return a single row unless I specify a GROUP BY, but then it returns a percentage of 1.0 like it should with a GROUP BY.
SELECT category,
error_count/SUM(error_count) AS percentage
FROM med_error_third_party_tmp;
Here are the server specs:
Server version: 5.0.77
Protocol version: 10
Server: Localhost via UNIX socket
Does anybody see a problem with this that is causing the problem?
Standard SQL requires you to specify a GROUP BY clause if any column is not wrapped in an aggregate function (IE: MIN, MAX, COUNT, SUM, AVG, etc), but MySQL supports "hidden columns in the GROUP BY" -- which is why:
SELECT category,
error_count/SUM(error_count) AS percentage
FROM med_error_third_party_tmp;
...runs without error. The problem with the functionality is that because there's no GROUP BY, the SUM is the SUM of the error_count column for the entire table. But the other column values are completely arbitrary - they can't be relied upon.
This:
SELECT category,
error_count/(SELECT SUM(error_count)
FROM med_error_third_party_tmp) AS percentage
FROM med_error_third_party_tmp;
...will give you a percentage on a per row basis -- category values will be duplicated because there's no grouping.
This:
SELECT category,
SUM(error_count)/x.total AS percentage
FROM med_error_third_party_tmp
JOIN (SELECT SUM(error_count) AS total
FROM med_error_third_party_tmp) x
GROUP BY category
...will gives you a percentage per category of the sum of the categories error_count values vs the sum of the error_count values for the entire table.
another way to do it - without the temp table as seperate item...
select category, error_count/sum(error_count) "Percentage"
from (SELECT mec.description category
, metpc.error_count
FROM med_error_category mec
, med_error_third_party_category metpc
WHERE mec.id = metpc.category
AND year = 2003
GROUP BY mec.id
);
i think you will notice that the percentage is unchanging over the categories. This is probably not what you want - you probably want to group the errors by category as well.
tell some big, diff between order by and group by,
like sort columns data=>order by
group it by similar data used for aggregation , order by could be used inside the grouped items ,
please Tell 5 diff
The order by clause is used to order your data set. For example,
select *
from customers
order by customer_id asc
will give you a list of customers in order of customer id from lowest to highest.
The group by clause is used to aggregate your data. For example,
select customer_id, sum(sale_price), max(sale_price)
from customers
group by customer_id
order by customer_id asc
will give you each customer along with their total sales and maximum sale, again ordered by customer id.
In other words, grouping allows you to combine multiple rows from the database into a single output row, based on some criteria, and select functions of those fields not involved in the grouping (minimum, maximum, total, average and so on).
group by groups data by one or more columns, and order by orders the data by one or more columns? i don't really get the question?
using group by is similar to select distinct in the aspect that only unique values for the given values will be returned. furthermore you can use aggregate functions to calculate e.g. the sum for each group.
what do you want to hear? tell me five differences between apples and oranges?