Get position/index of item in a collection

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Posted on by AntDim

Hi Community,

I simplify my scenario a little bit, but imagine I have a collection:

ColNames = [Bob, Jack, John, James]

and a label that displays one of these 4 different names.

What formula can I use to get the position of the name displayed in the collection?

For example, I want:

when the label reads "Jack", position_index = 2

when the label reads "John", position_index = 3

when the label reads "James", position_index = 4

etc. (my collection is much larger in reality and everything is dynamic so hard-coding is not an option)

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Sundeep_Malik 6,484 on at

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Re: Get position/index of item in a collection

Hey @AntDim

Found a previous post which might help:

https://powerusers.microsoft.com/t5/Building-Power-Apps/Find-Element-in-Array-and-get-Position-ID/td-p/1491919

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AntDim 68 on at

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Re: Get position/index of item in a collection

I tried this formula but I don't think it's working 😞

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AntDim 68 on at

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Re: Get position/index of item in a collection

I've got a Gallery where the title field is text from a collection text column. I want to set this up so that when I click on this title field, I get the position of this text in the collection

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plinkplink 2 on at

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Re: Get position/index of item in a collection

ClearCollect(ColNames, {Value: "Bob"}, {Value: "Jack"}, {Value: "John"}, {Value: "James"}); Clear(ColNamesIndexed); ForAll( ColNames, Collect(ColNamesIndexed, {Value: Value, Index: CountRows(ColNamesIndexed) + 1}) );

Output: ColNamesIndexed
Index Value
1 Bob
2 Jack
3 John
4 James

Here's how it works:

ClearCollect(ColNames, {Value: "Bob"}, {Value: "Jack"}, {Value: "John"}, {Value: "James"});
This just recreates the collection from your example.

Clear(ColNamesIndexed);
It's a good idea to clear the collection at the start so you don't keep appending to it if this formula gets called again.

ForAll( // Loop ColNames, // The original collection Collect( ColNamesIndexed, // The new collection with indexes { Value: Value, // This is the data from the original collection Index: CountRows(ColNamesIndexed) + 1 // Current number of rows in the collection + 1 } ) );

The part that does the trick is
CountRows(ColNamesIndexed)
This is counting itself on each iteration and returning the current number of rows at each item.

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siwfy Microsoft Employee on at

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Re: Get position/index of item in a collection

This is quite an innovative way of solving the problem. It may cause some performance issue at some point if the dataset is too large, but I think this solution is the best and simplest so far.

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AntDim 68 on at

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Re: Get position/index of item in a collection

thanks @plinkplink for your response. Where and how do you initialise the ColNamesIndexed collection? Using this formula in the Gallery Items property gave me an error in that the ColNamesIndexed collection is not recognized.

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siwfy Microsoft Employee on at

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Re: Get position/index of item in a collection

@AntDim The ColNamesIndexed collection is initialized with Clear(colNamesIndexed). And the ForAll() interates through your orginial collection, and then use Collect() to add an extra collumn of the index number and store all the records in this new collection colNamesIndexed row by row. I am guessing you are using the functions directly in the "Items" property of your gallery, which is why you are getting the error. If that's the case, you need to put the function in a behavior control (like a button), click on the button to generate this "colNamesIndexed" collection first, and then you can use it in your gallery.

By the way, I am a little curious about why you need to add this index to the collection. If you are extracting items from specific positions in a collection, you can use Index() function.

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BenDonahue 932 on at

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Get position/index of item in a collection

@AntDim, did you see @plinkplink's response? It is the best/fastest indexing code I have seen so far; if I had asked this question, I would have awarded them the Solution based on the effectiveness of the code they provided.

For my part, TY plinkplink!!! 'the horns'

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