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waleed111's avatar
waleed111
Icon for Helper V rankHelper V
5 years ago
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YEAR function

YEAR function generate wrong values: Year A = Year('All Manufacturings (4)'[Added Time]) it gave me 1905 for all values, what is the problem?  
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  • v-yingjl's avatar
    v-yingjl
    5 years ago

    Hello @waleed111 ,

    Based on the sample data, how did you create this [Time Added] column? Basically the Year() function will return the corresponding year of the column value and works well in my environment if the column is using formula or quote directly.

    y1.pngy2.png

    Maybe you can also try using .year to check:

    Column 3 = 'Calendar'[Date]. [Year]

    In addition, if you use the format() function to get the year value, it will be displayed as the default text type and can only be converted to the sunch number type as integer, decimal number, and so on, the date type is not allowed.

    Like this:

    Column 4 = CONVERT(FORMAT('Calendar'[Date],"YYYY"),DOUBLE)
Column 4 = CONVERT(FORMAT('Calendar'[Date],"YYYY"),INTEGER)

    Best Looks,
    Yingjie Li

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

v-yingjl's avatar
v-yingjl
Icon for Community Support rankCommunity Support
5 years ago

Hello @waleed111 ,

Based on the sample data, how did you create this [Time Added] column? Basically the Year() function will return the corresponding year of the column value and works well in my environment if the column is using formula or quote directly.

y1.pngy2.png

Maybe you can also try using .year to check:

Column 3 = 'Calendar'[Date]. [Year]

In addition, if you use the format() function to get the year value, it will be displayed as the default text type and can only be converted to the sunch number type as integer, decimal number, and so on, the date type is not allowed.

Like this:

Column 4 = CONVERT(FORMAT('Calendar'[Date],"YYYY"),DOUBLE)
Column 4 = CONVERT(FORMAT('Calendar'[Date],"YYYY"),INTEGER)

Best Looks,
Yingjie Li

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