Sort + Distinct + Search

(0) Share

Report

Posted on by Community Power Pla... Microsoft Employee

I try to retrieve a list of customer from a gallery with a search in which I would like to use a distinct.

I did not found an answer about it. And , I don't know if it's possible to do it like that

Sort(

Distinct(

Search(

Table1;

REC_BASE_CLIENT_SAISIE.Text; "CLIENT_NOM"

);

Table1

);

CLIENT_NOM

)

Thank you for your help

Categories:

Building Power Apps

I have the same question (0)

All responses (9)

Answers (1)

Sort by

KrishnaV 5,023 on at

Like (0)

Report
Copy link

Link copied!
Hi @Anonymous ,

Try this:
Sort(Distinct(Search(Table1,REC_BASE_CLIENT_SAISIE.Text, "CLIENT_NOM"),CLIENT_NOM),CLIENT_NOM,Ascending)

I hope this resolved your issue if you see any challenge/need further help please let me know I am always happy to do it for my community.

Regards,
Krishna
If this post helps you give a 👍 and if it solved your issue consider Accept it as the solution to help the other members find it more.

Was this reply helpful? Yes No
Community Power Pla... Microsoft Employee on at

Like (0)

Report
Copy link

Link copied!
Thank you for your email

I had to replace , by ; but it did not work

Sort( Distinct( Search( Table1,REC_BASE_CLIENT_SAISIE.Text, "CLIENT_NOM" ),CLIENT_NOM ),CLIENT_NOM,Ascending )

Was this reply helpful? Yes No
v-xida-msft Microsoft Employee on at

Like (0)

Report
Copy link

Link copied!
Hi @Anonymous ,

Based on the formula that you mentioned, I think there is something wrong with it.

Please consider modify your formula as below:

Sort( Distinct( Search( Table1; REC_BASE_CLIENT_SAISIE.Text; "CLIENT_NOM" ); CLIENT_NOM ); Result; // Type Result formula here Ascending )

Please consider take a try with above solution, then check if the issue is solved.

Best regards,

Was this reply helpful? Yes No
KrishnaV 5,023 on at

Like (0)

Report
Copy link

Link copied!
Hi @Anonymous ,

I am able see the result without any issues for you I have I have a created working model with the below formula:
Sort( Distinct( Search( SampleList, TextInput21.Text, "Title" ), Title ), Result, Ascending )
Working model:

I hope this resolved your issue if you see any challenge/need further help please let me know I am always happy to do it for my community.

Regards,
Krishna
If this post helps you give a 👍 and if it solved your issue consider Accept it as the solution to help the other members find it more.

Was this reply helpful? Yes No
Community Power Pla... Microsoft Employee on at

Like (0)

Report
Copy link

Link copied!

Hi, Sorry for the delay. Thank you for your email.
The request works but I can't retrieve the value in my form

I suppose I need to use .Result but it seems I forget something.

Was this reply helpful? Yes No
Verified answer

v-xida-msft Microsoft Employee on at

Like (0)

Report
Copy link

Link copied!
Hi @Anonymous ,

Have you try the following formula within your Gallery Items property?

Sort( Distinct( Search( Table1; REC_BASE_CLIENT_SAISIE.Text; "CLIENT_NOM" ); CLIENT_NOM ); Result; // Type Result formula here Ascending )

According to the issue that you mentioned, I think the above formula could achieve your needs. Please set the Items property of the Gallery to above formula.

Within your Gallery, please set the Text property of the Label to following:

ThisItem.Result

Note: Because, you have applied the Distinct function to your Search result, you could only reference the Result column value inside your Gallery.

If you also want to display related "TELEPHONE" column value in your Gallery, please consider try the following workaround:

Add another Label inside your Gallery, set the Text property to following:

Concat( Filter(Table1; CLIENT_NOM = ThisItem.Result); TELEPHONE & "; " )

Please try above solution, check if the issue is solved.

Best regards,

Was this reply helpful? Yes No
Community Power Pla... Microsoft Employee on at

Like (0)

Report
Copy link

Link copied!
Hi
Thank you for your answer. It works perferctly and I understand better the process.
Regarding a dropdown list, I need to show the record which is true or false

Concat( Filter(Table1; CLIENT_NOM = Text(Gallery2_1.Selected.Result));LOYALTY_NUMBER )

What do I need to modify for a boolean value to show ?

Was this reply helpful? Yes No
Community Power Pla... Microsoft Employee on at

Like (0)

Report
Copy link

Link copied!

Hi,

I would like to know how I can get only one result in the datacardvalue from the concat formula.

Thank you

Was this reply helpful? Yes No
KrishnaV 5,023 on at

Like (0)

Report
Copy link

Link copied!
Hi @Anonymous ,

try this:
Contact(First(Filter(Tabel1,Client_Nom=Text(Gallery2.Selected.result)),Telephone)

I hope this resolved your issue if you see any challenge/need further help please let me know I am always happy to do it for my community.

Regards,
Krishna
If this post helps you give a 👍 and if it solved your issue consider Accept it as the solution to help the other members find it more.

Was this reply helpful? Yes No