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using a year drop down to filter a date

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Posted on by Markswan20

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Is it possible to have a drop down box with the year 2021, 2022, 2023 etc where when the user selects the relevant year and the gallery will filter between a set date i'e 01/01/2023 - 31-12-2023 automatically?

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Markswan20 121 on at

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Re: using a year drop down to filter a date

However i've just stumbled on another problem, when using this code i can't navigate to the selected records as it's not recognising the gallery, i use the delegable version and it's fine, any ideas
Markswan20 121 on at

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Re: using a year drop down to filter a date

I've sorted it using the following for other users

With( { tt: Sort(Filter( AddColumns( LogDataALL, "yrs", Year('Log Date') ), yrs = Year(Dropdown4.Selected.Value),NewCreatedBy=varUN ),'Log Date',SortOrder.Descending) }, tt )
Markswan20 121 on at

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Re: using a year drop down to filter a date

Sorry for the delay, the code none delegable works perfectly thank you, i need to SORT the data with the newest 1st "fieldname" ('Log Date') where do i need to put this within the code to achieve this?
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v-qiaqi@microsoft.com on at

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Re: using a year drop down to filter a date

Hi @Markswan20,

Have you solved your problem?

I just add something to @Rajkumar_404

Remember to set your year dropdown with following formula:

["2021", "2022", "2023" ]

Then set your Gallery items with @Rajkumar_404 's formula.

Check it and see if it works.
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Rajkumar_404 6 on at

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Re: using a year drop down to filter a date

Hi @Markswan20

Try this formula, this is delegation free:

With(
{
tt: Filter(
AddColumns(
YourDataSource,,
"yrs",
Year(YourDateField)
),
yrs = Year(Dropdown1.Selected.Value)
)
},
tt
)

Thanks!

If my response has been helpful in resolving your issue, I kindly request that you consider clicking "Accept as solution" and "giving it a thumbs up" as a token of appreciation.
Rajkumar_404 6 on at

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Re: using a year drop down to filter a date

Hi @Markswan20

Try this,

Filter( YourDataSource, Year(YourDateField) = Year(DropdownYear.Selected.Value) )

Thanks!

If my response has been helpful in resolving your issue, I kindly request that you consider clicking "Accept as solution" and "giving it a thumbs up" as a token of appreciation.