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Frequent Visitor

Average Time between Repair Work orders

Hello All,

I have a table where cloumn1 = Porduct Name , Cloumn 2 = Serial Numbers of products , Cloumn 3 = Repair Work Order Numbers , Cloumn 4 = Work order Created date , Column 5 = Repair Work Order Closed Date . Now i have to find a time difference between last closed repair Work order date w.r.t previous repair closed date . Please suggest

1 ACCEPTED SOLUTION
Community Support

Hi @SAU1111 ,

If the Repair Work Order Numbers is random like below:

You can create a column to rank the Closed Date for each Product Name and Serial Numbers of products, like this:

``````Rank =
RANKX (
FILTER (
'Table',
'Table'[Porduct Name] = EARLIER ( 'Table'[Porduct Name] )
&& 'Table'[Serial Numbers of products]
= EARLIER ( 'Table'[Serial Numbers of products] )
),
'Table'[Repair Work Order Closed Date],
,
ASC,
DENSE
)
``````

Then create a measure.

``````Average Diff =
VAR _T =
'Table',
"Diff",
DATEDIFF (
MAXX (
FILTER (
'Table',
'Table'[Rank]
= EARLIER ( 'Table'[Rank] ) - 1
&& 'Table'[Porduct Name] = EARLIER ( 'Table'[Porduct Name] )
&& 'Table'[Serial Numbers of products]
= EARLIER ( 'Table'[Serial Numbers of products] )
),
'Table'[Repair Work Order Closed Date]
),
'Table'[Repair Work Order Closed Date],
DAY
)
)
RETURN
AVERAGEX ( _T, [Diff] )
``````

Get the result for each Product Name and Serial Numbers of products.

I attach my sample below for reference.

Best Regards,
Community Support Team _ kalyj

If this post helps, then please considerAccept it as the solution to help the other members find it more quickly.

3 REPLIES 3
Frequent Visitor

Sorry I forgot to mention that for 1 Serial numbers there can be N numbers of Work orders and there closed On date . Now problem is that more than 1 serial numbers has same repair work order closed on date . SO WOrk order numbers also be random cannot be arranged as on same day two Work order of a sequence like 1 and 2 can be for different serial numbers . Now First how to arrange by Serial Numbers all the WO and its closed on Date so that i can arrange the dates in ascending order and calculate a difference .

Community Support

Hi @SAU1111 ,

If the Repair Work Order Numbers is random like below:

You can create a column to rank the Closed Date for each Product Name and Serial Numbers of products, like this:

``````Rank =
RANKX (
FILTER (
'Table',
'Table'[Porduct Name] = EARLIER ( 'Table'[Porduct Name] )
&& 'Table'[Serial Numbers of products]
= EARLIER ( 'Table'[Serial Numbers of products] )
),
'Table'[Repair Work Order Closed Date],
,
ASC,
DENSE
)
``````

Then create a measure.

``````Average Diff =
VAR _T =
'Table',
"Diff",
DATEDIFF (
MAXX (
FILTER (
'Table',
'Table'[Rank]
= EARLIER ( 'Table'[Rank] ) - 1
&& 'Table'[Porduct Name] = EARLIER ( 'Table'[Porduct Name] )
&& 'Table'[Serial Numbers of products]
= EARLIER ( 'Table'[Serial Numbers of products] )
),
'Table'[Repair Work Order Closed Date]
),
'Table'[Repair Work Order Closed Date],
DAY
)
)
RETURN
AVERAGEX ( _T, [Diff] )
``````

Get the result for each Product Name and Serial Numbers of products.

I attach my sample below for reference.

Best Regards,
Community Support Team _ kalyj

If this post helps, then please considerAccept it as the solution to help the other members find it more quickly.

Community Support

Hi @SAU1111 ,

According to your descrption, I create a sample.

Here's my solution, create a measure.

``````Average Diff =
VAR _T =
'Table',
"Diff",
DATEDIFF (
MAXX (
FILTER (
'Table',
'Table'[Repair Work Order Numbers]
= EARLIER ( 'Table'[Repair Work Order Numbers] ) - 1
&& 'Table'[Porduct Name] = EARLIER ( 'Table'[Porduct Name] )
&& 'Table'[Serial Numbers of products]
= EARLIER ( 'Table'[Serial Numbers of products] )
),
'Table'[Repair Work Order Closed Date]
),
'Table'[Repair Work Order Closed Date],
DAY
)
)
RETURN
AVERAGEX ( _T, [Diff] )
``````

Get the correct result.

I attach my sample below for reference.

Best Regards,
Community Support Team _ kalyj

If this post helps, then please considerAccept it as the solution to help the other members find it more quickly.

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