Finding entry count for each unique id in SQL - sql

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

Related

Link data from sql from 2/3 columns

I don't know if the question title is so clear, but here is my question:
I had table UsersMovements which contains Users along with their movements
UsersMovements:
ID
UserID
MovementID
Comments
Time/Date
I need help looking for a query which would give me if users 1, 2 & 3 had been in a common MovementID, knowing that I don't know what is the MovementID
The real case is that, I want to see if those X users which I would select been in an area (in a limited interval, assuming I had date/Time in the table)
Thank you
if you want to select list of movements which have userid 1,2 and 3 you can use group by with having
select movementid
from usermovements
where userid in(1,2,3)
group by movementid
having count(distinct userid)=3

Need to find duplicates where two columns values added together create a duplicate

Problem: There is a conference. Each class in the conference will run three times (time1, time2, time3). Attendee can only go to one session per class. I'm looking for duplicate class registrations. (User is going to class1: time1 and time2)
I need to write a query to find duplicate registrations with information in 3 different tables. The tables I'm joining are (class, user, registration). I need to find all duplicates that are created by adding the values for two columns. (class.title + user.id#) = duplicate.
I need the query to display only the duplicate rows found plus display additional column information such as id number, title, first, last, status.
For Example: The search would find these results
table
But display only...
results
I'm not sure where to start. The information in class.title + user.id# will vary per row, so I can't search by specific information. (ie: user.id#=45624).
If I understood you right is the DISTINCT-Keyword what you are looking for: http://www.w3schools.com/sql/sql_distinct.asp
An example:
SELECT * FROM (SELECT DISTINCT title, status, first, last FROM your_table)
This will make sure that there will be no duplicate rows returned identified by title, status, first and last.
I may misunderstood your question. In case you are looking for duplicates only, consider the following code. This will return you all duplicate rows identified by title, status, first and last and give you id as an additional information.
SELECT id, title, status, first, last
FROM your_table
GROUP BY title, status, first, last
HAVING COUNT(*) > 1
Which one is the answer you were looking for?
maybe this will work
select * from(
SELECT id, title, status, first, last ,
row_number() over ( partition by id, title order by status) rowid
FROM your_table
) x
where rowid = 2

Find row number in a sort based on row id, then find its neighbours

Say that I have some SELECT statement:
SELECT id, name FROM people
ORDER BY name ASC;
I have a few million rows in the people table and the ORDER BY clause can be much more complex than what I have shown here (possibly operating on a dozen columns).
I retrieve only a small subset of the rows (say rows 1..11) in order to display them in the UI. Now, I would like to solve following problems:
Find the number of a row with a given id.
Display the 5 items before and the 5 items after a row with a given id.
Problem 2 is easy to solve once I have solved problem 1, as I can then use something like this if I know that the item I was looking for has row number 1000 in the sorted result set (this is the Firebird SQL dialect):
SELECT id, name FROM people
ORDER BY name ASC
ROWS 995 TO 1005;
I also know that I can find the rank of a row by counting all of the rows which come before the one I am looking for, but this can lead to very long WHERE clauses with tons of OR and AND in the condition. And I have to do this repeatedly. With my test data, this takes hundreds of milliseconds, even when using properly indexed columns, which is way too slow.
Is there some means of achieving this by using some SQL:2003 features (such as row_number supported in Firebird 3.0)? I am by no way an SQL guru and I need some pointers here. Could I create a cached view where the result would include a rank/dense rank/row index?
Firebird appears to support window functions (called analytic functions in Oracle). So you can do the following:
To find the "row" number of a a row with a given id:
select id, row_number() over (partition by NULL order by name, id)
from t
where id = <id>
This assumes the id's are unique.
To solve the second problem:
select t.*
from (select id, row_number() over (partition by NULL order by name, id) as rownum
from t
) t join
(select id, row_number() over (partition by NULL order by name, id) as rownum
from t
where id = <id>
) tid
on t.rownum between tid.rownum - 5 and tid.rownum + 5
I might suggest something else, though, if you can modify the table structure. Most databases offer the ability to add an auto-increment column when a row is inserted. If your records are never deleted, this can server as your counter, simplifying your queries.

How to produce a distinct count of records that are stored by day by month

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.

SQL Query to find if different values exist for a column

I have a temporary table with three columns
pay_id,
id_client_grp,
id_user
Basically i want to ensure that this table should have all the rows having same client group and same id_user if not i want to know which pay_id is the culprit and throw error to user.
Can somebody help me with a query.
Thanks,
Rishi
When you say 'culprit,' I assume you mean the pay_id(s) that are not like the others, assuming there is a majority.
The problem is all of the pay_id's could potentially become culprits once your SELECT COUNT(DISTINCT id_client_grp, id_user) returns > 1 record, if there is a relatively even distribution. It is difficult to program for this scenario, since you will need to determine what exactly a majority is.
Your best bet will be to return all distinct combinations of those 3 fields, then decide where to go from there based on your business logic.
So could this question be asked like this:
If I wanted to add a unique index on my table across the three columns: client group, id user, pay id, identify those that break the unique condition where we have non unique pay id for a client group and id user??
select a.id_client_grp, a.id_user, a.pay_id , a.count from (
/* this should return 1 row per client group and user, */
/* if the pay id is the same for all */
select id_client_grp, id_user, pay_id, count(1) as count
from table t
group by id_client_grp, id_user ) a
group by a.id_client_grp, a.id_user
/* if we have more than one row per client group and user, then we have a dupe, so report them all */
having count (1) > 1
If you want all the rows to have the same values for some set of columns (your question is not entirely clear to me as t9o what you want to be the same)
Do you know going in WHICH pay_id, id_client_grp all the rows should be? Or do you not care, as long as they are all the same?
If you know the values you are looking for, simply test for rows that are not set to those desired values
Select distinct id_user
From tempTable
Where pay_id <> #PayIdValue
Or id_client_grp <> #ClientGroupIDValue
If you don't care, and just want them all to be the same, and they're not, then you need to specify which of the more than one set of values IS the "culprit" as you said...
If you want some other question answered. please explain more clearly...
Based on yr comment, then, to determine if there is more than one id_client_grp, pay_id
Select Count(Distinct id_client_grp, pay_id)
From tempTable
If this = 1 then every record has the same values for these 2 fields.... Any other value indicates that three is more than one set of distinct values in the table.
SELECT DISTINCT p.pay_id,
t.[count]
FROM rishi_table p
INNER JOIN ( SELECT id_client_grp, id_user, COUNT(*) As 'count'
FROM rishi_table
GROUP BY id_client_grp, id_user
HAVING COUNT(*) > 1 ) t
ON p.id_client_grp = t.id_client_grp AND p.id_user = t.id_user
basically create a set with the dupes, and bounce that against the main table to get your offending list.
SELECT DISTINCT id_client_grp, id_user
should let you do something like
IF ##ROWCOUNT > 1 THEN
...
Or possibly SELECT COUNT(DISTINCT id_client_grp, id_user) ...
but that's more vendor-dependent as to its availability and proper syntax.