I have 2 columns i need to get the different period with days and hours
Example
2019-10-22 13:22:59 min getdate()
I need result to be
3 days 7 hours
Related
I'm struggling with this.
I have a column in Snowflake called DURATION, it is VARCHAR type.
The values include basically number in days, hours, minutes, seconds. The value could include either just the number with one unit of time (day or hour or minute or second) such as 3 hours or 14 minutes or 3 seconds or it could include the combination of either all units of time or a few such as 1 day 3 hours 35 minutes or 1 hour 9 minutes or 45 minutes 1 second.
The value could also be blank or invalid such as text or it could be indicating day, hour or minute but without a number (see the last 3 rows in the table below).
I would greatly appreciate it if you guys could help me with the following:
in SNOWFLAKE, convert all valid values to number type and normalize them to minutes (e.g. the resulted value for 7 Hours and 13 Minutes would be 433).
Thanks a lot, guys!
DURATION
1 Second
10 Seconds
1 Minute
3 Minutes
20 Minutes
1 Hour
2 Hours
7 Hours 13 Minutes
1 Hour 1 Minute
1 Day
1 Day 1 Hour
1 Day 1 Hour 1 Minute
1 Day 10 Hours
2 Days 1 Hour
3 Days 9 Hours
1 Day 3 Hours 45 Minutes
Duration (invalid)
Days
Day Minute
Minutes
I tried many things using regex_substr, try_to_number, coalesce functions in CASE statements but I'm getting either 0s or NULL for all values. Very frustrating
I think you would want to use STRTOK_TO_ARRAY in a CTE subquery or put into a temp table. Then you could use ARRAY_POSITION to find the labels and the index one less than the label should be the value. Those values could be put into separate columns with a case for each label pulling the found values. The case statements could be computed columns if you insert the results of the first query into a table. From there you can concatenate colons and cast to a time type and use datediff, or do the arithmetic to calculate the minutes.
I'm currently stucked on an issue. I have daily data and I want to SUM all the data for the next 30 days over a year.
Date
Views
28-01-2021
1
29-01-2021
5
30-01-2021
1
31-01-2021
5
And I want to have the number of views starting the 28th for the next 30 days, then the next 30 days etc... over a year (or twelve times)
So basically what I want to see is something like this, Series being the series of 30 days (first 30 days, second 30 days etc...)
Series
Views
1 (from 28-01 to 28-02)
250
2 (from 01-03 to 30-03)
200
3 (from 31-04 to 29-04)
300
4 (from 30-04 to 29-05)
550
Thank you if anyone can help.
Regards
Assuming you have one row for each day, you can use a window function:
select t.*,
sum(views) over (order by date
rows between current row and 29 following
) as views_30days
from t;
Note: This interprets "next 30 days" as really being "today plus the next 29 days". If you don't want the current day, then the window frame would be:
rows between 1 following and 29 following
how to calculate the 2 days or 3 days total amounts in MySQL.
a sum of 3 days amounts and sum of 2 days amounts
how to write a query in MySql
I don't know how your relation looks like but if I get it right you want to SUM one column over the past 2 or 3 days which will be something like:
SELECT FORMAT(SUM(yourColumnName)
FROM tableName
WHERE entry.date >= CURDATE() - INTERVAL 2 days
For 3 days you have to set INTERVAL 3 days
I have a query to retrieve a set of non null records from a column x consisting of DATE format.
If count(x) = 35 then i need to display the value as 1 Month & 5 days
If 369 days then 1 year & 4 days or If 400 days then 1 year 1 month 5 days respectively
Query: In the above instance,unfortunately i am neglecting 0.25 days but How to tweak my actual requirement in such a way that i don't end up neglecting days and handle leap year logic too
How to solve this issue?
it is not clear if you need of a time's generic computing in years, months and days, based on averages of number of days of a month, etc.. or if you want an exact compute of the number of the months that pass starting froma adate.. for example if your total sum of 400 days start from 15 march, then the month is counted by 1 after, let's say, 15 days, and so the remaining days are 20.. I don't know if I explained..
In the first hypotesis, you may use the following pseudo-coded solution, that is a very approximative method..
however, if you know a start date, it is possible to compute exactly how many "bissextile" days are comprehended between your interval of days starting from your start date
let's say to have output variables Years, Months, Days .. and an input totalDaysdays assigned with your data retrieved from the db, then:
(pseudocode)
Years = trunc((totalDays / 365)
bissextileDays = trunc((totalDays / 365) / 4)
numDaysOffset = (totalDays Mod 365) + bissextileDays
Months = trunc(numDaysOffset / 30)
Days = numDaysOffset Mod 30
Actually i found something that will suit my requirement.
https://community.oracle.com/thread/2587161?start=0&tstart=0
select days,
floor(days / 365.25) years,
floor(mod(days,365.25) / (365.25 / 12)) months,
round(mod(days,365.25 / 12)) days
from periods
So this can produce expected output when number is given. This produces output as years,months and remaining days
Have table: item (int) and timestamp (datetime).
Need to know if there are any records with timestamp after last 6 AM.
Example:
At 5 AM it should check if there are any records from after 6 AM
yesterday. AT 7 AM it should check if there are any records from
after 6 AM today
This could be done of course by making a variable with datepart as:
if time now is < 6 AM datepart should be yesterday if time now
is >= 6 AM datepart should be today
but there must be a simpler way ?
This expression should always return the most recent 06:00 in the past:
select DATEADD(hour,
(DATEDIFF(hour,'2014-01-01T06:00:00',CURRENT_TIMESTAMP)/24)*24,
'2014-01-01T06:00:00')
It works by asking how many hours have happened since some arbitrary, known, 6AM. We then round this number down to the closest multiple of 24 (by dividing and multiplying with integer maths), and add this number back onto the same, arbitrary, 6AM