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Thiago's avatar
Thiago
Regular Visitor
10 years ago
Solved

Sum hours

Hello everyone!   I need to get the sum of hours in a given field and I know that the total sum that exceeds 24 hours .   I've tried using the code below , but it didn't work: Column = (LEFT([H...
  • AlexChen's avatar
    AlexChen
    10 years ago

    Hi, 

     

    For example, you have a table like below, the “time” column is date type:

     

     

    You can create a measure :

     

    Measure = sumx('table', hour('table'[time])) + TRUNC(sumx('table', MINUTE('table'[time]))/60)  & ":" & mod(sumx('table', MINUTE('table'[time])), 60) & ":00"

     

    The result is what you want:

     

     

     

     

     

Thiago's avatar
Thiago
Regular Visitor
10 years ago

Hi Vvelarde! Thanks for the answer!

 

It didn't work!

I tried to implement, but the result was not expected.

See example below, where the total should be 47 hours and 24 minutes. When I used your code, the result was almost 633 hours!

I believe it has some configuration that I have not learned to use..

 

HE
01:11:00
02:17:00
02:32:00
01:22:00
05:32:00
04:07:00
03:19:00
06:19:00
00:00:00
05:15:00
00:00:00
02:03:00
00:00:00
02:37:00
01:22:00
03:01:00
00:52:00
00:59:00
00:21:00
02:36:00
01:39:00
47:24:00

  • Anonymous's avatar
    Anonymous
    Not applicable
    10 years ago

    Seems quite possible there is a better way to do this, but...  I do think this will "work"

     

    MyMeasure = 24 * sumx(Table1, round(Table1[MyTime],99))

     

    MyTime was a DateTime type for me (not a string), and I'm just using round to force it into a floating point number (where 1.0 is a full day).