The two columns in the image are part of a table and I'm trying to get the AVG Sal depending on specific date. For example, the Avg sal for 1981. Because of the date, I can't get it. How do I do this? Thanks!
2 Columns
Try this option:
SELECT AVG(SAL) AS AVG_SAL
FROM yourTable
WHERE EXTRACT(YEAR FROM HIREDATE) = 1981;
More generally, to find averages for all years:
SELECT
EXTRACT(YEAR FROM HIREDATE) AS HIRE_YEAR,
AVG(SAL) AS AVG_SAL
FROM yourTable
GROUP BY
EXTRACT(YEAR FROM HIREDATE);
How about
SELECT AVG(SAL)
FROM MyTable
WHERE HIREDATE BETWEEN '1/1/1981' AND '12/31/1981'
Here we're using the BETWEEN operator which is pretty handy in SQL and is not limited to dates. You might also look through the Oracle documentation for some handy date and time functions.
Related
I have a table called Table1. I am trying to get the weekly average, but I only have daily data. My table contains the following attributes: caseID, date, status and some other (irrelevant) attributes. With the following query, I made the following table which comes close to what I want:
However, I would like to add a average per week of the number of cases. I have look everywhere, but I am not sure how to include that. Has anybody any clues for how to add that.
Thanks.
To expand on #luuk's answer...
SELECT
date,
COUNT(id) as countcase,
EXTRACT(WEEK FROM date) AS weeknbr,
AVG(COUNT(id)) OVER (PARTITION BY EXTRACT(WEEK FROM date)) as weeklyavg
FROM table1
GROUP BY date, weeknbr
ORDER BY date, weeknbr
This is possible as the Aggregation / GROUP BY is applied before the window/analytic function.
select
date,
countcase,
extract(week from date) as weeknbr,
avg(countcase) over (partition by extract(week from date)) as weeklyavg
from table1;
I am trying to figure out the aggregate functions in SQL SSRS to give me to sum of total sales for the given information by YEAR. I need to combine the year, the months within that year and provide the total sum of sales for that year. For example: for 2018 I need to combine month's 2-12 and provide the total sum, for 2019 combine 1-12 and provide total sum and so on.
enter image description here
I'm not sure where to begin on this one as I am new to SQL SSRS. Any help would be appreciated!
UPDATE:
Ideally I want this to be the end result:
id Year Price
102140 2019 ($XXXXX.XX)
102140 2018 ($XXXXX.XX)
102140 2017 ($XXXXX.XX)
And so on.
your query:
Select customer_id
, year_ordered
--, month_ordered
--, extended_price
--, SUM(extended_price) OVER (PARTITION BY year_ordered) AS year_total
, SUM(extended_price) AS year_total
From customer_order_history
Where customer_id = '101646'
Group By
customer_id
, year_ordered
, extended_price
--, month_ordered
Provides this:
enter image description here
multiple "years_ordered" because it is still using each month and that months SUM of price.
There are two approaches.
Do this in your dataset query:
SELECT Customer_id, year_ordered, SUM(extended_price) AS Price
FROM myTable
GROUP BY Customer_id, year_ordered
This option is best when you will never need the month values themselves in the report (i.e. you don't intend to have a drill down to the month data)
Do this in SSRS
By default you will get a RowGroup called "Details" (look under the main design area and you will row groups and column groups).
You can right-click this and add grouping for both customer_id and year_ordered. You can then change the extended_price textbox's value property to =SUM(Fields!extended_price.Value)
You could use a window function in your SQL:
select [year], [month], [price], SUM(PRICE) OVER (PARTITION BY year) as yearTotal
from myTable
can someone help me to write a query?
I have for example columns:
Date
product_key
category_code
In one day I expect to have same category_code for one product, but I want to check this with SQL.
Thank you.
If you want to find the day, the product_key and the category_code that doubles, You should use query like this:
SELECT
date,
product_key,
category_code,
count(1)
FROM your_table
GROUP BY date, product_key, category_code
HAVING count(1) > 1;
You can group your results by date and product, and use count and distinct to find if there is more than one category code for a product. You can then filter rows having more than 1 distinct category in the group.
SELECT
Date, product_key, count(distinct category_code) AS categories
FROM
my_table
GROUP BY
Date, product_key
HAVING
count(distinct category_code) > 1
I have a requirement to display spend estimation for last 30 days. SpendEstimation is calculated multiple times a day. This can be achieved using simple SQL query:
SELECT DISTINCT ON (date) date(time) AS date, resource_id , time
FROM spend_estimation
WHERE
resource_id = '<id>'
and time > now() - interval '30 days'
ORDER BY date DESC, time DESC;
Unfortunately I can't seem to be able to do the same using SQLAlchemy. It always creates select distinct on all columns. Generated query does not contain distinct on.
query = session.query(
func.date(SpendEstimation.time).label('date'),
SpendEstimation.resource_id,
SpendEstimation.time
).distinct(
'date'
).order_by(
'date',
SpendEstimation.time
)
SELECT DISTINCT
date(time) AS date,
resource_id,
time
FROM spend
ORDER BY date, time
It is missing ON (date) bit. If I user query.group_by - then SQLAlchemy adds distinct on. Though I can't think of solution for given problem using group by.
Tried using function in distinct part and order by part as well.
query = session.query(
func.date(SpendEstimation.time).label('date'),
SpendEstimation.resource_id,
SpendEstimation.time
).distinct(
func.date(SpendEstimation.time).label('date')
).order_by(
func.date(SpendEstimation.time).label('date'),
SpendEstimation.time
)
Which resulted in this SQL:
SELECT DISTINCT
date(time) AS date,
resource_id,
time,
date(time) AS date # only difference
FROM spend
ORDER BY date, time
Which is still missing DISTINCT ON.
Your SqlAlchemy version might be the culprit.
Sqlalchemy with postgres. Try to get 'DISTINCT ON' instead of 'DISTINCT'
Links to this bug report:
https://bitbucket.org/zzzeek/sqlalchemy/issues/2142
A fix wasn't backported to 0.6, looks like it was fixed in 0.7.
Stupid question: have you tried distinct on SpendEstimation.date instead of 'date'?
EDIT: It just struck me that you're trying to use the named column from the SELECT. SQLAlchemy is not that smart. Try passing in the func expression into the distinct() call.
How to get the total count for particular date by selecting different dates.
For Example:
Record contains from '2014-04-01' to till date. Each date contains multiple records with different IST time.
How to get the total count from each date?
depending on your table structure and result you want, Query should look somewhat like this
SELECT DATE(date_column), COUNT(*)
FROM tablename
WHERE date_column IN (your_date_list)
GROUP BY date(date_column);
Have a look at the following sql (I have not tested)
SELECT date_column, COUNT(*)
FROM tablename
WHERE date_column BETWEEN date_column AND current_date
GROUP BY to_date(date_column::text,'YYYY-MM-DD')