Using the below example, I want to get a count of records grouped by Person and Product, for each day.
The example on the left is the data in the SQL Server 2008 database. The data on the right is my desired query result.
I could just substring off the time portion of the date values, then just do a Group By... but ideally the Max(date) would contain the full timestamp...
Is this what you want?
select t.person, t.product, count(*) as cnt, max(t.date) as date
from table t
group by t.person, t.product, cast(t.date as date);
Related
I have made an table which has 3 columns
Table
In this above table there are 3 columns i.e Trandsaction, EmpID, Date having some data. I need to form a table using some queries so that in the result table I will get how many transactions done in each month by a particular EmpID.
So the result table must be like:
So how to get month wise number of transactions per EmpID?
Get all employees (I assume you have an employee table).
Get all months (you could get them from the transaction table).
Cross join the two in order to get all combinations, because you want to show these in your result.
Outer join the transaction counts per month and employee (an aggregation subquery). SQL Server's date to string conversion is a bit awkward compared to other DBMS. You need to convert to a predefined format and use a substring of that.
Use COALESCE to turn null (for no count for the employee and month) to zero.
The query:
select m.month, e.empid, coalesce(t.cnt, 0) as transaction_count
from employees e
cross join (select distinct convert(varchar(7), t.date, 126) as month from transactions) m
left join
(
select convert(varchar(7), t.date, 126) as month, empid, count(*) as total
from transactions
group by convert(varchar(7), t.date, 126), empid
) t on t.empid = e.empid and t.month = m.month
order by m.month, e.empid;
If you don't want all employees, but only those that have at least one transaction in some month, then replace from employees e with from (select distinct empid from transactions) e.
SELECT Date, EmpID, COUNT(transaction)
FROM your_table
GROUP BY Date, EmpID
To get the date like "YYYY-MM" use
for MySQL (replace the String with the date column):
DATE_FORMAT("2019-11-25 10:49:30.000", "%Y - %m")
for oracle:
to_char(date, 'YYYY-MM')
I have an employee table with the hire_date column.
I am stuck with one query related to the date function, where I have used data type 'DATE' to insert date of hiring and using DATE_FORMAT fun. to retrieve no. of employees hired in every month, but in SQL-server it is not supporting the date_format function.
I'm using SQL -server
Query: - list of the no.of employee hired every month in ascending order.
select date_format(hire_date,'%b') month, count(*)
from employee
group by DATE_FORMAT(hire_date,'%b')
order by month
date_format(hire_date,'%b') in MySQL will return abbreviated monthname. However, you can still have this functionality by combining MONTHNAME with LEFT in SQL Server.
select LEFT(DATENAME(MONTH,hire_date),3) month, count(*)
from employee
group by LEFT(DATENAME(MONTH,hire_date),3)
order by month
select Month(Hire_Date),count('x') 'count' from employee
group by Month(Hire_Date)
order by Month(Hire_Date) asc
Instead, you can directly use:
select MONTH(hire_date), count(*)
from employee
group by MONTH(hire_date)
order by MONTH(hire_date)
or
select hire_date.MONTH, count(*)
from employee
group by hire_date.MONTH
order by hire_date.MONTH
I have a table with 3 columns: id, date and amount, but I would like to get accumulated SUM for each date (Last column).
Do you have an easy solution how to add this column?
I am trying with this:
SELECT date, sum(amount) as accumulated
FROM table group by date
WHERE max(date);
Should I user OVER() for this?
Use a window function to the total for each day:
SELECT date,
amount,
sum(amount) over (partition by date) as accumulated
FROM the_table;
However this will only work, if your dates all have the same time part (in Oracle a DATE column also contains a time). To make sure you ignore the time part, use trunc() to make sure all time parts are normalized to 00:00:00
SELECT date,
amount,
sum(amount) over (partition by trunc(date)) as accumulated
FROM the_table;
Use This:
SELECT T.ID, T.DATE, T.AMOUNT, (SELECT SUM(S.AMOUNT) FROM TABLE S WHERE S.DATE=T.DATE) ACCUMULATED
from
table T
This will give you the records from the table with a sum for all records for the date.
I need to put together the results of these two queries into a single return with the following structure:
"date", avg(selic."Taxa"), avg(titulos."puVenda")
Partial structure of tables:
selic
"dtFechamento" date,
"pTaxa" real
titulos
"dtTitulo" date,
"puVenda" real,
"nomeTitulo" character(30)
Query table selic:
select to_char("dtFechamento", 'YYYY-MM') as data, avg("pTaxa")
from "selic"
group by data
order by data
Query table titulos:
select to_char("dtTitulo", 'YYYY-MM') as data, avg("puVenda")
from "titulos"
where "nomeTitulo" = 'LFT010321'
group by data
order by data
I tried a subquery, but it returned the fields next to each other and can not muster.
select *
from (select to_char("dtFechamento", 'YYYY-MM') as data, avg("pTaxa")
from "selic"
group by data
order by data) as selic,
(select to_char("dtTitulo", 'YYYY-MM') as data, avg("puVenda")
from "titulos"
where "nomeTitulo" = 'LFT010321'
group by data
order by data) as LFT010321;
Assuming you want to return one row per month where either of your two queries returns a row. And pad missing values from the other query with NULL.
Use a FULL [OUTER] JOIN:
SELECT to_char(mon, 'YYYY-MM') AS data, s.avg_taxa, t.avg_venda
FROM (
SELECT date_trunc('month', "dtFechamento") AS mon, avg("pTaxa") AS avg_taxa
FROM selic
GROUP BY 1
) s
FULL JOIN (
SELECT date_trunc('month', "dtTitulo") AS mon, avg("puVenda") AS avg_venda
FROM titulos
WHERE "nomeTitulo" = 'LFT010321'
GROUP BY 1
) t USING (mon)
ORDER BY mon;
It is substantially faster to join after aggregating than before (fewer join operations).
It is also faster to GROUP BY, JOIN and ORDER on timestamp values than on a text rendition. Typically also cleaner and less error prone (although text is unambiguous in this particular case). That's why I use date_trunc() instead of to_char() on lower levels.
If the format for the month is not important, you can just return the timestamp value. Else you can format any way you like after you are done processing.
Similar case with more explanation:
PostgreSQL merge two queries with COUNT and GROUP BY in each
This should get what you need. The inner "PQ" (PreQuery) does a union all between each possible date, but also adds a flag column to identify which average it was associated with. Each part is grouped by date. So now, the outer query will AT MOST have 2 records for a given date... one for tax, the other be Venda. So now you dont need any full outer join, nor need to build some dynamic calendar data basis to get the details for all possible dates.
So, it is possible for only a Tax average OR a Venda average OR BOTH.
SELECT
PQ.Data,
SUM( CASE when PQ.SumType = 'T' then PQ.TypeAvg else 0 end ) as AvgTax,
SUM( CASE when PQ.SumType = 'V' then PQ.TypeAvg else 0 end ) as AvgVenda
from
( select
to_char( dtFechamento, 'YYYY-MM') as data,
'T' as sumtype,
avg( pTaxa ) as TypeAvg
from
selic
group by
to_char( dtFechamento, 'YYYY-MM') as data
UNION ALL
select
to_char( dtTitulo, 'YYYY-MM') as data,
'V' as sumType,
avg( puVenda ) as TypeAvg
from
titulos
where
nomeTitulo = 'LFT010321'
group by
to_char( dtTitulo, 'YYYY-MM') ) PQ
group by
PQ.Data
order by
PQ.Data
I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)