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DateDiff and Hours

I have a datediff thats not working as I expected, probably just something I've done wrong so sharing here for input

Let's say the difference between the two times is 4 hrs 55 minutes, instead of returning 4 hours (which i assumed the rounddown would do, my formula is returning 5 hours. Who here knows what I am doing wrong?

RoundDown(DateDiff(ThisItem.CLOCKIN,theclock,Hours),0)

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Resident Rockstar

Hey @TheRobRush, I think that using DateDiff with Hours will return something a bit different as it seems to round the hours to the nearest integer value all on its own and a bit differently than any of the Round functions would. I think it may have to do with the portion of the day that is used or something like that but it doesn't seem to depend solely on the difference in hours.

For instance, in some testing using the Now() function, I found a difference of 4.7 hours would return 4 hours but 4.8 would return 5 but later in the day, 4.7 hours returned 5. So instead you might try this:

```RoundDown(
DateDiff(
ThisItem.CLOCKIN,
theclock,
Minutes
) / 60,
0
)```

Basically, this converts the difference in minutes into a decimal (non-integer) version of the hours and then rounds that down to the lowest hour. Feel free to let me know if this works out or not.

5 REPLIES 5
Resident Rockstar

Hey @TheRobRush, I think that using DateDiff with Hours will return something a bit different as it seems to round the hours to the nearest integer value all on its own and a bit differently than any of the Round functions would. I think it may have to do with the portion of the day that is used or something like that but it doesn't seem to depend solely on the difference in hours.

For instance, in some testing using the Now() function, I found a difference of 4.7 hours would return 4 hours but 4.8 would return 5 but later in the day, 4.7 hours returned 5. So instead you might try this:

```RoundDown(
DateDiff(
ThisItem.CLOCKIN,
theclock,
Minutes
) / 60,
0
)```

Basically, this converts the difference in minutes into a decimal (non-integer) version of the hours and then rounds that down to the lowest hour. Feel free to let me know if this works out or not.

Community Support

Hi @TheRobRush ,

The reason why you get 4 hours instead of 4 hrs 55 minutes is that DateDiff will only return time of one kind of unit.

For example:

DateDiff(....,Hours) will only return how many hours.

DateDiff(....,Minutes) will only return how many minutes.

DateDiff(....,Seconds) will only return how many seconds.

If you want to get result that is combination of hour and minute, you need to transfer.

Try this formula:

`Time(DateDiff(ThisItem.CLOCKIN,theclock,Hours),     DateDiff(ThisItem.CLOCKIN,theclock,Minutes)-DateDiff(ThisItem.CLOCKIN,theclock,Hours)*60,     0)`

Best regards,

Community Support Team _ Phoebe Liu

Community Support Team _ Phoebe Liu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
@wyotim Will try this tomorrow thank you
_____________________________________________________________________________________
Like my answer? - Hit that Thumbs Up. Resolved the Issue? - Hit Accept as Solution.
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Super User
@v-yutliu-msft
Think you misread my question. Not trying to return 4 hrs 55 minutes, trying to round down properly.
_____________________________________________________________________________________
Like my answer? - Hit that Thumbs Up. Resolved the Issue? - Hit Accept as Solution.
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Community Support

Hi @TheRobRush ,

Sorry that I did not fully understand your issue.

The reason why you did not get the right answer is that DateDiff function will return approximate number.

Since 4h55min is closer to 5 than 4, so the result will return 5.

I agree with the formula that @wyotim posted.

The most accurate hour result is ：
DateDiff(...,....,Minutes)/60

So, I suggest you try this:

RoundDown(DateDiff(...,....,Minutes)/60,0)

Best regards,

Community Support Team _ Phoebe Liu

Community Support Team _ Phoebe Liu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

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