I've created a query that I'm hoping to use to fill a table with daily budgets at the end of every day. To do this, I just need some simple maths:
monthly budget / number of days left in the month.
Then, at the end of the month, I can SUM all of the rows to calculate an accurate budget.
The query is as follows:
SELECT *,
ROUND(SAFE_DIVIDE(budget, DATETIME_DIFF(CURRENT_DATE(), LAST_DAY(CURRENT_DATE()), DAY)),2) AS daily_budget
FROM `mm-demo-project.marketing_hub.budget_manager`
When executing the query, my results present as negative numbers, which according to the documentation for this function, is likely caused by the result overflowing the result type.
View the results of the query.
I've made a fools guess at rounding the calculation. Needless to say that it did not work at all.
How can I stop my query from returning negative number?
use below
SELECT *,
ROUND(SAFE_DIVIDE(budget, DATETIME_DIFF(LAST_DAY(CURRENT_DATE()), CURRENT_DATE(), DAY)),2) AS daily_budget
FROM `mm-demo-project.marketing_hub.budget_manager`
Related
Im trying to calculate the average time a tool stays on loan. The time a tool stays on loan is the number of days between loan_status_change_date and tool_out_date (table columns). the date type of these 2 columns is ex: 01-SEP-17
whats the best way to approach this?
We can do arithmetic with Oracle dates. It's not clear from the column names which one is the start of the loan and which the end; in the following example I've assumed loan_status_change is when the tool is returned.
select tool
, avg(loan_status_change - tool_out_date) as avg_loan_days
from your_table
group by tool
/
The AVG() function is an aggregate function, so it handles the /ns for us. The substraction is to calculate the length of a particular loan, which is the value you want to average. The result of that substraction already is a number of days, so no further transformation is necessary. If your columns have a time element then the result might not be an integer.
In my table I have 3 colmuns
T1 T2 T3
I want to average the time in all 3.
My query in design view is set as follows:
Score: Avg([swim]+[bike]+[run])
In total, I have it done as an expression.
When I run the query an example result shows as: 8.81406810035843E-02
Any ideas how I can get it to show the average time over all 3 correctly?
I have also changed the format to hh:nn:ss - The results looks better but not correct
I hope you are looking for this
Score: Avg(([swim]+[bike]+[run])/3)
Using the Avg() funciton, you are making average of every time in column. Adding them together, the resulting time is too huge to show for access. When you average something in one row, you have to add it up and then divide by number of each member.
Im trying to make a small report for myself to see how my much time I get inputed in my system every day.
The goal is to have my SQL to sum up the name, Total time worked and Total NG product found for one specific day.
In this order:
1.) Sort out my data for a specific 'date'. I.E 2016-06-03
2.) Present a DISTINCT value for 'operators'
3.) SUM() all time registered at this 'date' and by this 'operator' under 'total_working_time_h'
4.) SUM() all no_of_defects registered at this 'date' and by this 'operator' under 'no_of_defects'
date, operator, total_working_time_h, no_of_defects
Currently I get the data I want by using the Query below. But now I need both the DISTINCT value of the operator and the SUM of the information. Can I use sub-queries for this or should it be done by a loop? Any other hints where I can learn more about how to solve this?
If i run the DISTINCT function I don't get the opportunity to sum my data the way I try.
SELECT date, operator, total_working_time_h, no_of_defects FROM {$table_work_hours} WHERE date = '2016-06-03' "
Without knowing the table structure or contents, the following query is only a good guess. The bits to notice and work with are sum() and GROUP BY. Actually syntax will vary a bit depending on what RDBMS you are using.
SELECT
date
,operator
,SUM(total_working_time_h) AS total_working_time_h
,SUM(no_of_defects) AS no_of_defects
FROM {$table_work_hours}
WHERE date = '2016-06-03'
GROUP BY
date
,operator
(Take out the WHERE clause or replace it with a range of dates to get results per operator per date.)
I'm not sure why you are trying to do DISTINCT. You want to know the data, no of hours, etc for a specific date.
do this....
Select Date, Operator, 'SumWorkHrs'=sum(total_working_time_h),
'SumDefects'=sum(no_ofDefects) from {$table_work_hours}
Where date='2016-06-03'
Try this:
SELECT SUM(total_working_time) as total_working_time,
SUM(no_of_defects) as no_of_defects ,
DISTINCT(operator) AS operator FROM {$table_work_hours} WHERE
date = '2016-06-03'
I have this statement that calculates the difference in time intervals below:
(DateDiff ("s",previous({PROD_TRKG_TRAN.MOD_DATE_TIME}) ,{PROD_TRKG_TRAN.MOD_DATE_TIME}))/60
Now I want to be able to get that average of that datediff(), this is what I believe should work but I'm getting "The remaining text does not appear to be part of the formula". error message:
SELECT ({PROD_TRKG_TRAN.USER_ID}),
({USER_MASTER.USER_NAME}),
Average (diff((DateDiff ("s",previous({PROD_TRKG_TRAN.MOD_DATE_TIME}) ,{PROD_TRKG_TRAN.MOD_DATE_TIME})))/60
You need to group the data so you can get an average for the group. This can be done in your Query but that will remove the details from the results. If you need to see all the detail lines, then leave the datediff formula and return all the results to your report. Then create groups in your report and create a formula that gets the average for the group(s).
I have an SQLite database with the following fields for example:
date (yyyymmdd fomrat)
total (0.00 format)
There is typically 2 months of records in the database. Does anyone know a SQL query to find a weekly average?
I could easily just execute:
SELECT COUNT(1) as total_records, SUM(total) as total FROM stats_adsense
Then just divide total by 7 but unless there is exactly x days that are divisible by 7 in the db I don't think it will be very accurate, especially if there is less than 7 days of records.
To get a daily summary it's obviously just total / total_records.
Can anyone help me out with this?
You could try something like this:
SELECT strftime('%W', thedate) theweek, avg(total) theaverage
FROM table GROUP BY strftime('%W', thedate)
I'm not sure how the syntax would work in SQLite, but one way would be to parse out the date parts of each [date] field, and then specifying which WEEK and DAY boundaries in your WHERE clause and then GROUP by the week. This will give you a true average regardless of whether there are rows or not.
Something like this (using T-SQL):
SELECT DATEPART(w, theDate), Avg(theAmount) as Average
FROM Table
GROUP BY DATEPART(w, theDate)
This will return a row for every week. You could filter it in your WHERE clause to restrict it to a given date range.
Hope this helps.
Your weekly average is
daily * 7
Obviously this doesn't take in to account specific weeks, but you can get that by narrowing the result set in a date range.
You'll have to omit those records in the addition which don't belong to a full week. So, prior to summing up, you'll have to find the min and max of the dates, manipulate them such that they form "whole" weeks, and then run your original query with a WHERE that limits the date values according to the new range. Maybe you can even put all this into one query. I'll leave that up to you. ;-)
Those values which are "truncated" are not used then, obviously. If there's not enough values for a week at all, there's no result at all. But there's no solution to that, apparently.