I am trying to create a query that will return results for the number of distinct users who have accessed something by date. Right now I have a query that will display 2 columns, the first being date and the second being user name. It will list all the distinct users who accessed the application on a certain date but they will each have their own distinct row. Here is the query that does that:
SELECT DISTINCT logdate, User AS ReportUser
FROM table
WHERE appname='abcd1234' AND logdate >=DATE-30
I have tried putting COUNT() around User but it says selected non-aggregate values must be part of the associated group.
Any idea how I can get this query to just show like row for the past 30 days and the count of distinct users?
This will be the right approach for that.
SELECT logdate, Count(User) AS ReportUser
FROM table
WHERE appname='abcd1234' AND logdate >=DATE-30
GROUP BY 1
Never use DINTINCT in Teradata. It always slows down your query performance. Use GROUP BYinstead.
CORRECTION In Teradata 13 the optimizer is able to determines which version is more efficient, based on statistics. This can be found in the Release Summary for Teradata 13 under "Group By and DISTINCT Performance Equivalence". - http://www.info.teradata.com/edownload.cfm?itemid=083440012 (PDF)
use "GROUP BY" after WHERE clause
SELECT logdate, COUNT (User) AS ReportUser
FROM table
WHERE appname='abcd1234' AND logdate >=DATE-30
GROUP BY logdate
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
I have a requirement to create a report that counts a total from 2 date fields into one. A simplified example of the table I'm querying is:
ID, FirstName, LastName, InitialApplicationDate, UpdatedApplicationDate
I need to query the two date fields in a way that creates similar output to the following:
Date | TotalApplications
I would need the date output to include both InitialApplicationDate and
UpdatedApplicationDate fields and the TotalApplications output to be a count of the total for both types of date fields. Originally I thought maybe a Union would work however that returns 2 separate records for each date. Any ideas how I might accomplish this?
The simplest way, I think, is to unpivot using apply and then aggregate:
select v.thedate, count(*)
from t cross apply
(values (InitialApplicationDate), (UpdatedApplicationDate)) v(thedate)
group by v.thedate;
You might want to add where thedate is not null if either column could be NULL.
Note that the above will count the same application twice, once for each date. That appears to be your intention.
I am trying to select all records in a time-variant Account table for each account with a change in an associated value (e.g. the maturity date). A change in the value will result in the most recent record for an account being end-dated and a new record (containing a new effective date of the following day) being created. The most recent records for accounts in this table have an end-date of 12/31/9000.
For instance, in the below illustration, account 44444444 would not be included in my query result set since it hasn't had a change in the value (and thus also has no additional records aside from the original); however, the other accounts have multiple changes in values (and multiple records), so I would want to see those returned.
I've tried using the row_num function, as well as a reflexive join, but for some reason I'm not getting the expected results. What are some ways to obtain the results I need?
Note: The primary key for this table includes the acct_id and eff_dt. Also, I'm using PostgreSQL in a Greenplum environment.
Here are two types of queries I tried to use but which produced problematic results:
Query 1
Query 2
If you want only the accounts, use aggregation:
select acct_id
from t
group by acct_id
having min(value) <> max(value);
Based on your description, you could also use count(*) >.
If you want the original records, you can use window functions:
select t.*
from (select t.*, count(*) over (partition by acct_id) as cnt
from t
) t
where cnt > 1;
I have a table with some search results. The search results maybe repeated because each result may be found using a different metric. I want to then query this table select only the distinct results using the ID column. So to summarize I have a table with an ID column but the IDs may be repeated and I want to select only one of each ID with MS Access SQL, how should I go about doing this?
Ok I have some more info after trying a couple of the suggestions. The Mins, and Maxes won't work because the column they are operating on cannot be shown. I get an error like You tried to execute a query that does not include the specified expression... I now have all my data sorted, here is what it looks like
ID|Description|searchScore
97 test 1
97 test .95
120 ball .94
97 test .8
120 ball .7
so the problem is that since the rows were put into the table using different search criteria I have duplicated rows with different scores. What I want to do is select only one of each ID sorted by the searchScore descending. Any ideas?
SELECT DISTINCT ID
FROM Search_Table;
Based on the last update to your question, the following query seems appropriate.
SELECT ID, [Description], Max(searchScore)
FROM Search_Table
GROUP BY ID, [Description];
However that's nearly the same as Gordon's suggestion from yesterday, so I'm unsure whether this is what you want.
Here is a way where you can get one of the search criteria:
select id, min(search_criteria)
from t
group by id
This will always return the first one alphabetically. You can also easily get the last one using max().
You could also use:
select id, first(search_criteria)
from t
group by id
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.