I have a table users that lists all users who have entered orders in to the system.
Each order has a order_number and user amongst a ton of other columns.
Each week I am looking to get a list of the total orders that each user has entered in to the system, I guess this will need a subquery. I have looked at grouping and subquery but am really lost.
The idea is to SELECT a count of orders entered that week, entry_date between sysdate and sysdate-5 etc. which I don't have a problem doing, but I don't understand how to then count it per user.
So for e.g. Jane entered 150 orders, Steve entered 450 orders etc.
Can someone point me in the right direction here please?
I don't really think you need a sub-query. Won't a GROUP BY solve your problem?
SELECT USER, COUNT(*)
FROM ORDERS
WHERE ENTRY_DATE BETWEEN SYSDATE - 5 AND SYSDATE
GROUP BY USER
Use GROUP BY. I made up some column names.
SELECT u.user, COUNT(*) FROM orders o, users u
WHERE o.user = u.user AND o.entry_date BETWEEN o.sysdate - 5 AND o.sysdate
GROUP BY u.user
Related
I have an SQL database that shows the amount of times a person submits an entry. I want to count how many time each person who owns a unique id makes a claim. Each unique i.d can make mulpile entries into the table and I want to find out how many everyone has made.
I also want to filter the people based on the amount of entries they have made. For example 10.
select id, entry, COUNT(ID) from Table where COUNT(entry) <=10 GROUP BY ID
This is my thinking so far but I havent had much success. If anyone could help I would greatly appreciate it.
This is all you need
select id, COUNT(*) cnt
from Table
GROUP BY ID
having COUNT(*) <= 10
order by 2 desc -- optional descending count
Optionally, add Order by 2. desc or asc
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
I want to show how many orders have been shipped 10 days before the required date in Access SQL, but I can't seem to get the syntax right.
SELECT COUNT(*)
FROM Orders
WHERE DATEDIFF(day,RequiredDate,ShippedDate)=10;
SELECT Count(*)
FROM Orders
WHERE (((DateDiff("d",[Orders].[RequiredDate],[Orders].[ShippedDate]))=10));
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 have two tables, Users and DoctorVisit
User
- UserID
- Name
DoctorsVisit
- UserID
- Weight
- Date
The doctorVisit table contains all the visits a particular user did to the doctor.
The user's weight is recorded per visit.
Query: Sum up all the Users weight, using the last doctor's visit's numbers. (then divide by number of users to get the average weight)
Note: some users may have not visited the doctor at all, while others may have visited many times.
I need the average weight of all users, but using the latest weight.
Update
I want the average weight across all users.
If I understand your question correctly, you should be able to get the average weight of all users based on their last visit from the following SQL statement. We use a subquery to get the last visit as a filter.
SELECT avg(uv.weight) FROM (SELECT weight FROM uservisit uv INNER JOIN
(SELECT userid, MAX(dateVisited) DateVisited FROM uservisit GROUP BY userid) us
ON us.UserID = uv.UserId and us.DateVisited = uv.DateVisited
I should point out that this does assume that there is a unique UserID that can be used to determine uniqueness. Also, if the DateVisited doesn't include a time but just a date, one patient who visits twice on the same day could skew the data.
This should get you the average weight per user if they have visited:
select user.name, temp.AvgWeight
from user left outer join (select userid, avg(weight)
from doctorsvisit
group by userid) temp
on user.userid = temp.userid
Write a query to select the most recent weight for each user (QueryA), and use that query as an inner select of a query to select the average (QueryB), e.g.,
SELECT AVG(weight) FROM (QueryA)
I think there's a mistake in your specs.
If you divide by all the users, your average will be too low. Each user that has no doctor visits will tend to drag the average towards zero. I don't believe that's what you want.
I'm too lazy to come up with an actual query, but it's going to be one of these things where you use a self join between the base table and a query with a group by that pulls out all the relevant Id, Visit Date pairs from the base table. The only thing you need the User table for is the Name.
We had a sample of the same problem in here a couple of weeks ago, I think. By the "same problem", I mean the problem where we want an attribute of the representative of a group, but where the attribute we want isn't included in the group by clause.
I think this will work, though I could be wrong:
Use an inner select to make sure you have the most recent visit, then use AVG. Your User table in this example is superfluous: since you have no weight data there and you don't care about user names, it doesn't do you any good to examine it.
SELECT AVG(dv.Weight)
FROM DoctorsVisit dv
WHERE dv.Date = (
SELECT MAX(Date)
FROM DoctorsVisit innerdv
WHERE innerdv.UserID = dv.UserID
)
If you're using SQL Server 2005 you don't need the sub query on the GROUP BY.
You can use the new ROW_NUMBER and PARTION BY functionality.
SELECT AVG(a.weight) FROM
(select
ROW_NUMBER() OVER(PARTITION BY dv.UserId ORDER BY Date desc) as ID,
dv.weight
from
DoctorsVisit dv) a
WHERE a.Id = 1
As someone else has mentioned though, this is the average weight across all the users who have VISITED the doctor. If you want the average weight across ALL of the users then anyone not visiting the doctor will give a misleading average.
Here's my stab at the solution:
select
avg(a.Weight) as AverageWeight
from
DoctorsVisit as a
innner join
(select
UserID,
max (Date) as LatestDate
from
DoctorsVisit
group by
UserID) as b
on a.UserID = b.UserID and a.Date = b.LatestDate;
Note that the User table isn't used at all.
This average omits entirely users who have no doctors visits at all, or whose weight is recorded as NULL in their latest doctors visit. This average is skewed if any users have more than one visit on the same date, and if the latest date is one of those date where the user got wighed more than once.