How can i count unique text in rows which occur more than 1 time

[HarryHKE]

Hello,

I have a question and i seems to figure this one out, in my database i have two rows. 1 is a Stamordernumber row and the other a ordernumber row when there is a project which has no additional cost the stamorder row and the ordernumber row have the same number. However when i make a addional order which is link to the Stamorder number the stamorder number will stay the same and the ordernumber will change. See below axample.

Stamorder number   -   Ordernumber

SO32100060                  SO32200060   (This is a order with no additional cost on is, so this is the main and only order)

SO32100105                  SO32200105   (This is a order with additional cost on is, so this is the main order)

SO32100105                  SO32200106   (and this is the additonal order with the addition cost on it)

Now i would lik to have a colum count in front of the stamorder number which counts the amount of rows which has the same stamorder number, see below example.

Count    -   Stamorder number   -   Ordernumber

1                SO32100060                  SO32200060   (This is a order with no additional cost on is, so this is the main and only order)

2                SO32100105                  SO32200105   (This is a order with additional cost on is, so this is the main order)

2                SO32100105                  SO32200106   (and this is the additonal order with the addition cost on it)

I tried everthing with CALCULATE and DISTINCTCOUNT but i just don't seem to get a correct count, could anyone help me please.

The table with the stamordernumber in it is a text table.

See also below a sample of the databse, 

Thank You,

Harry

[HarryHKE]

Tank you !!

[v-stephen-msft · Accepted solution]

Hi @HarryHKE ,

From what I am understanding, you want to count group by Stamorder number.

Please try this calculated column. 

Count = CALCULATE(COUNT('Table'[Stamorder number]),ALLEXCEPT('Table','Table'[Stamorder number]))

Best Regards,

Stephen Tao

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