I have a table "Report" with relevant columns "Date", "Doctor". Each doctor appears several times throughout the table. The following code is what I have at current:
SET #variable = (SELECT Date FROM Report WHERE Doctor='DocName' ORDER BY Date DESC LIMIT 1)
SELECT DATEDIFF(CURDATE(),#variable) AS DiffDate
This gives me the DATEDIFF for one doctor, without name. Is there any way to loop through the table, find the last row/date for each doctor, then perform a DATEDIFF on each individual doctor outputting a list of doctors with their DATEDIFFs (against current date) next to them?
Thanks in advance!
you can use group by to get only 1 row per doctor and max to select latest date:
select `Doctor`, DATEDIFF(CURDATE(),max(`Date`))
from `Report`
group by `Doctor`
Related
I am using the Graph Reports for the select below. The MySQL database only has the active records in the database, so if no records are in the database from X hours till Y hours that select does not return anything. So in my case, I need that select return Paypal zero values as well even the no activity was in the database. And I do not understand how to use the UNION function or re-create select in order to get the zero values if nothing was recorded in the database in time interval. Could you please help?
select STR_TO_DATE ( DATE_FORMAT(`acctstarttime`,'%y-%m-%d %H'),'%y-%m-%d %H')
as '#date', count(*) as `Active Paid Accounts`
from radacct_history where `paymentmethod` = 'PayPal'
group by DATE_FORMAT(`#date`,'%y-%m-%d %H')
When I run the select the output is:
Current Output
But I need if there are no values between 2016-07-27 07:00:00 and 2016-07-28 11:00:00, then in every hour it should show zero active accounts Like that:
Needed output with no values every hour
I have created such select below , but it not put to every hour the zero value like i need. showing the big gap between the 12 Sep and 13 Sep anyway, but there should be the zero values every hour
(select STR_TO_DATE ( DATE_FORMAT(acctstarttime,'%y-%m-%d %H'),'%y-%m-%d %H')
as '#date', count(paymentmethod) as Active Paid Accounts
from radacct_history where paymentmethod <> 'PayPal'
group by DATE_FORMAT(#date,'%y-%m-%d %H'))
union ALL
(select STR_TO_DATE ( DATE_FORMAT(acctstarttime,'%y-%m-%d %H'),'%y-%m-%d %H')
as '#date', 0 as Active Paid Accounts
from radacct_history where paymentmethod <> 'PayPal'
group by DATE_FORMAT(#date,'%y-%m-%d %H')) ;
I guess, you want to return 0 if there is no matching rows in MySQL. Here is an example:
(SELECT Col1,Col2,Col3 FROM ExampleTable WHERE ID='1234')
UNION (SELECT 'Def Val' AS Col1,'none' AS Col2,'' AS Col3) LIMIT 1;
Updated the post: You are trying to retrieve data that aren't present in the table, I guess in reference to the output provided. So in this case, you have to maintain a date table to show the date that aren't in the table. Please refer to this and it's little bit tricky - SQL query that returns all dates not used in a table
You need an artificial table with all necessary time intervals. E.g. if you need daily data create a table and add all day dates e.g. start from 1970 till 2100.
Then you can use the table and LEFT JOIN your radacct_history. So for each desired interval you will have group item (group by should be based on the intervals table.
I have attendance data for employees stored in the table attendance with the following column names:
emp_id (employee ID)
date
type (leave, absent, etc.)
(there are others but I'm omitting them for the sake of simplicity)
My objective is to retrieve all dates of the given month on which the employee was on leave (type = 'Leave') and the last leave taken in the last month, if any.
It's easy to do it using two queries (I'm using PHP to get process the data), but is there any way this can be done in a single query?
I'm answering my own question so as to close it. As #bpgergo pointed out in the comments, UNION will do the trick here.
SELECT * FROM table_name
WHERE type="Leave" AND
date <= (CURRENT_DATE() - 30)
Select the fields, etc you want then se a combined where clause using mysql's CURRENT_DATE() function. I subtracted 30 for 30 days in a month.
If date is a date column, this will return everyone who left 1 month or longer ago.
Edit:
If you want a specific date, change the 2nd month like this:
date <= (date_number - 30)
I have the following table structure:
ID, User_ID, DateTime
Which stores a user id and datetime of an order purchased. How would I get the average number of orders a day, across every row?
In pseudo code I'm thinking:
Get total number of orders
Get number of days in range (from first row to last row).
Divide 1. by 2. to get average?
So it would return me a value of 50, or 100?
Thanks
Since you know the date range, and you are not guaranteed to have and order on these dates, you can't just subtract the max(date) from min(date), but you know the number of days before you run the query, therefore simply:
select count(*) / <days>
from mytable
where DateTime between <start> and <end>
Where you supply the indicated values because you know them.
select DATEDIFF(NOW(), date_time) as days, AVG(count(*))
from table
group by days
I have not tested the query, its just the idea, I guess it should work.
Assume you have a table with a stock time series on a daily basis.
Now you need to filter one data point per week, because you need weekly data for some analysis. You don't to have weekly averages, since this would leave much of the variation out.
This would be my initial approach, but it's not clear which of the data points falling in a given week is selected.
SELECT date, price from stock_series
GROUP BY WEEK(date)
1 How do I make sure it's always the first data point existing for a given week that gets picked?
EDIT:
2 If the above query stayed the way it is - which data point gets chosen every week? What's the MySQL logic in this case? Or is it just unpredictible?
If you want to have a better control over it, you could try using a subquery :
SELECT date,price
FROM stock_series
WHERE date IN
(
SELECT MIN(inner.date)
FROM stock_series inner
GROUP BY WEEK(inner.date)
) GROUP BY date
I've added GROUP BY date in the main query because you probably have more than one entry per day, otherwise it could be ommited.
EDIT:
or try joining with it:
SELECT date,price
FROM stock_series
JOIN
(
SELECT MIN(date) AS innerdate
FROM stock_series
GROUP BY WEEK(date)
) inner ON date=innerdate;
You can order by date ascending, which should give you just the first result of the WEEK() group.
SELECT date,price from stock_series
GROUP BY WEEK(date)
ORDER BY date
I have a transaction table and I'm looking to generate a dataset to drive a line chart that shows the sum of all the sales happened on each day during a given period. I have never grouped results like that before and am scratching my head.
Let's say the table is called "transactions", the "datetime" field is called timestamp, and the sales amount on each transaction is "txn_amount". I want the result set to include each day: "1/2/10" and the sum of the transaction amounts.
I need to get a book and spend a week learning mysql... Thanks for helping me out.
select sum(txn_amount) ,timestamp from transactions where timestamp in (select distinct timestamp from transactions) group by timestamp
if datatype is datetime,Use this
select sum(amt) ,substring(dt,1,10) from transactions where substring(dt,1,10) in (select distinct substring(dt,1,10) from transactions) group by substring(dt,1,10)