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.
Related
I have a sales data table with cust_ids and their transaction dates.
I want to create a table that stores, for every customer, their cust_id, their last purchased date (on the basis of transaction dates) and the count of times they have purchased.
I wrote this code:
SELECT
cust_xref_id, txn_ts,
DENSE_RANK() OVER (PARTITION BY cust_xref_id ORDER BY CAST(txn_ts as timestamp) DESC) AS rank,
COUNT(txn_ts)
FROM
sales_data_table
But I understand that the above code would give an output like this (attached example picture)
How do I modify the code to get an output like :
I am a beginner in SQL queries and would really appreciate any help! :)
This would be an aggregation query which changes the table key from (customer_id, date) to (customer_id)
SELECT
cust_xref_id,
MAX(txn_ts) as last_purchase_date,
COUNT(txn_ts) as count_purchase_dates
FROM
sales_data_table
GROUP BY
cust_xref_id
You are looking for last purchase date and count of distinct transaction dates ( like if a person buys twice, it should be considered as one single time).
Although you mentioned you want count of dates but sample data shows you want count of distinct dates - customer 284214 transacted 9 times but distinct will give you 7.
So, here is the SQL you can use to get your result.
SELECT
cust_xref_id,
MAX(txn_ts) as last_purchase_date,
COUNT(distinct txn_ts) as count_purchase_dates -- Pls note distinct will count distinct dates
FROM sales_data_table
GROUP BY 1
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
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.
The following table has Key and StartTime and EndTime. I want to my query to return, my key, the number of records with this Key and the total minutes of all records with the same Key. I cant run with out StartTime and EndTime in my group by unfortunately this groups each row with a different start or stop time.
SELECT sn.Key,
COUNT(*) as SessonNoteCount,
sum( dbo.fnCalcTime( sn.StartTime, sn.EndTime)) as min
FROM SessionNote sn
group by sn.Key, sn.StartTime, sn.EndTime
order by sn.Key
You should group by all non-aggregated selected columns, which in your case is just sn.Key:
group by sn.Key
As an aside, IMHO the group by clause should be optional because it is entirely determined by the column selected columns and could easily be generated by the query parser, but that boat has sailed...
let's say my table schema is like bellow (it's only a simplified example):
MyTable (table name)
ID - int (unique, auto increment)
Message - string
Timestamp - Datetime
I want to select the number of ID, group them by message and order them by timestamp, so I'll do something like this:
SELECT count (ID), Message FROM MyTable
GROUP BY (Message)
ORDER BY Timestamp desc
However, SQL Server management studio throws me this error:
Column 'Timestamp ' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
The problem is that if I put Timestamp in the Group By statement with Message, it messes up my grouping. The other suggestion to put Timestamp in an aggregate function doesn't make sense (ordering by say, count(Timestamp) doesn't mean anything...)
Any idea on how to do this?
Thanks a lot!
Looking for something like this?
SELECT Message, count (ID), max(Timestamp) as maxDate FROM MyTable
GROUP BY (Message)
ORDER BY maxDate desc
When you do aggregation, you are GROUPing rows together based on certain criteria. This mean that each row of your result set actually represents multiple rows in the raw data.
When you want to ORDER BY Timestamp, there will be MULTIPLE timestamp values for each row in the result set, since each one of those rows represents several rows of data.
So, you need to decide which timestamp you want for each set. The MAX? The MIN? You will need to aggregate that field as well to get accurate or meaningful results.
Let's say the same message is in your table multiple times:
1|the mackerel likes frying|1/1/1917
2|at night all cats are grey|12/15/1956
3|the mackerel likes frying|2/2/1918
And you want to group by the message-string, counting the number of times the message appears in the table:
the mackerel likes frying|2
at night all cats are grey|1
The timestamp column is NOT part of the aggregation aka the grouping, but is part of the detail row. It CANNOT appear in the grouping, because timestamp is not "it" (singular) but timestamps, they, plural. There are two different timestamps in the example above for the mackerel message. Which one would you choose? How would the query know which one it was? All you have at your disposal are the aggregate functions:
min(timestamp)
max(timestamp)
count(timestamp)
and if it were other than a datetime, you'd also have AVG(timestamp).
If you want to order the messages based on the max timestamp within the Group then try:
SELECT count (ID), Message
FROM MyTable
GROUP BY (Message)
ORDER BY MAX(Timestamp) DESC
The problem here is that you probably have multiple messages that are the same, but with different timestamps, because you're grouping by message. If you have two messages 'hello' with different timestamps, which should it use for the order by?
This is one way. You could also do a trick with cross apply or row_number.
SELECT count(ID), Message FROM MyTable
GROUP BY (Message)
ORDER BY Max(Timestamp) desc