How can I declare a field as Identity field?

[YonghunLee]

How can I declare a field as Identity field as picture below? 

[HalloWorld]

- You can go to the visualisation pane, right click on the field and choose not to aggregate. Then it will be spread at least across all distinct values of that column.

- If you create an identity column already in TSQL or if you make it in the Power Query Editor (by the way, you can make the new column an index column in Power Query), you can choose "no aggregation" for that column in the visualisation pane as well, and by that, you will spread the table across all rows.

- If you have a column that already is an identity (unique in each value of the column), then you can just as well choose in the visualisation pane "not to aggregate" that, without the need to make a new identity column.

Here is how you set a column to "do not aggregate" in German:

[serhii_stefaniv]

Hey there. I have the same issue. Have you found the solution?

[polymathy2017]

Ugh! I"m going to have to find another community. Can't seem to find actual answers in here very often. Frustrating.

[v-juanli-msft · Accepted solution]

Hi @YonghunLee 

If you means to create unique id in power bi, it is possible as long as for the same name(different people), there are different values in other columns.

You could create index column with DAX as below:

https://community.powerbi.com/t5/Desktop/Create-unique-ID/td-p/677570

For example,create a column below:

Final Rank =
RANKX (
    Table,
    RANKX ( Table, [Primary column],, ASC )
        + DIVIDE (
            RANKX ( Table, [Secondary column],, ASC ),
            ( COUNTROWS (Table ) + 1 )
        )
)

rank as unique id = 
RANKX (
    'Table (2)',
    RANKX ( 'Table (2)', [name],, ASC )
        + DIVIDE (
            RANKX ( 'Table (2)', [column1],, ASC ),
            ( COUNTROWS ( 'Table (2)' ) + 1 )
        ),,ASC,Dense
)

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