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Power Platform Community / Forums / Power Apps / Build string using a l...
Power Apps
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Build string using a loop and pass the string as a parameter to the Table() function

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I want to build a string in the format required by the Table function.

 

 

This string will build uniquely for the current user facing record.

 

Note: Below code is rough draft. I need to accomplish something along those lines using Powerapps.

 

//Code

Set(BuildString, "");

Set(CurrentIndex, 0);

Set(TestString, "a,b,c,d");

Set(OccurancesOfDelimiter, ....); //Also need a way to retrieve occurances of a character in a TestString

Set(Counter, 0); DoUntil(OccurancesOfDelimiters, Set(Counter, Counter + 1); Set(CurrentValue, Mid(TestString, CurrentIndex, Find(",", TestString) - 1)); Set(BuildTable, BuildTable&"{Text:" &Char(34)&CurrentValue&Char(34)&","&"Val:"&Counter&"}"&", ") Set(CurrentIndex, Find(",", TestString, CurentIndex + 1)) ); Set(BuiltTable, Table(BuildTable))

There are two things I need from Powerapps to accomplish this.

 

1.) Find occurance of a character in a string. e.x For String x = "a,b,c,d"; If i did an operation to find how many occurances of commas there exists on X, then the return value should be 3.

 

2.) A way to loop X amount of times

 

3.) To be able to build a string and it pass as a parameter to Table() function.

 

Appreciate the help. 

 

- Syed

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I have the same question (0)
  • Verified answer
    v-xida-msft Profile Picture
    on at

    Hi @Anonymous,

     

    Could you please share a bit more about your scenario?

     

    For your first question, do you want to retrieve the occurances of commas existed on a string (e.g "a,b,c,d")?

     

    If you want to retrieve the occurances of commas existed on a string (e.g "a,b,c,d"), I think the Split function could achieve your needs. I have made a test on my side, please take a try with the following workaround:1.JPG

     

    In order to retrieve the number of commas existed within a string, set the Text property of the Label control to following formula:

    CountRows(Split(TextInput1.Text,","))-1

    On your side, you should type the following formula:

    CountRows(Split(TestString,","))-1

    Note: The TextString represents the String X that you mentioned.

     

    For your second question, do you want to loop the character separated via a comma within the TestString (X)? 

     

    If you want to loop the character separated via a comma within the TestString (X), I think the ForAll function (the only way to achieve a loop within PowerApps currently) could achieve your needs.

     

    More details about ForAll function in PowerApps, please check the following article:

    ForAll function

     

    For your third question, do you want to build a string (e.g. {Text: "a", Val:1} ) using a loop and psss it as a parameter to Table() function?

     

    In addition, If you want to achieve your needs using Table function, I afraid that there is no way to achieve your needs within PowerApps currently. Because the Set(Counter, Counter + 1) formula could not invoked within the ForAll function.

     

    As an alternative solution, you could consider take a try to use a Collection to achieve your needs instead of Table function. I have made a test on my side, please take a try with the following workaround:2.JPG

     

    Set the OnSelect property of the "Build String" button to following formula:

    Clear(TestString);
    ForAll(
    Split(TextInput1.Text,","),Collect(TestString,{Text:Result,Val:CountRows(TestString)+1})
    )

    On your side, you should type the following formula:

    Clear(TestString);
    ForAll(
    Split(X,","),Collect(TestString,{Text:Result,Val:CountRows(TestString)+1})
    )

    Note: The X represents the variable X that you mentioned.

     

     

    Best regards,

    Kris

     

     

  • Community Power Platform Member Profile Picture
    on at

    Thanks Kris!

     

    I did not know the Split function parsed a string into a table of strings, Result. This works perfectly with the ForAll function. I tried your solution on my end and it works. I marked your post as the solution.

     

    - Syed

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