Can I turn a dictionary into two separate columns for a report in AL?

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Posted on by jdrco

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Say I have a dictionary: { 'a':0, 'b':2, 'c':3 }. Can this be turned into two columns where one column are the keys, and the values for the other?
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YUN ZHU 95,597 Super User 2025 Season 2 on at

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Hi, you can try to use Foreach method.

Using this method you can get the key and value separately. Then insert them into the columns you want to use.

For example:

pageextension 50100 MyExtension extends "Customer List" { trigger OnOpenPage() begin Test(); end; local procedure Test() var Dict: Dictionary of [Text, Integer]; KeyText: Text; IntValue: Integer; begin Dict.Add('a', 0); Dict.Add('b', 2); Dict.Add('c', 3); foreach KeyText in Dict.Keys do begin Message(Format(KeyText)); Message(Format(Dict.Get(KeyText))); end; end; }

Hope this helps.

Thanks.

ZHU

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jdrco 481 on at

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HI yzhums , I am now in the process of trying to insert the keys and values into two columns. Not sure how to go about that.

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YUN ZHU 95,597 Super User 2025 Season 2 on at

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Hi, 'm not sure what your specific needs are, but here's a simple example that hopefully will give you some hints.

page 50100 TestPageForDictionary { Caption = 'Test Page For Dictionary'; PageType = List; UsageCategory = Lists; ApplicationArea = All; Editable = false; SourceTable = TestTableForDictionary; layout { area(Content) { repeater(Group) { field(DictionaryKey; Rec.DictionaryKey) { ApplicationArea = All; } field(DictionaryValue; Rec.DictionaryValue) { ApplicationArea = All; } } } } trigger OnOpenPage() begin InsertColumns(); end; local procedure InsertColumns() var Dict: Dictionary of [Text, Integer]; KeyText: Text; IntValue: Integer; begin Dict.Add('a', 0); Dict.Add('b', 2); Dict.Add('c', 3); foreach KeyText in Dict.Keys do begin Rec.Init(); Rec.DictionaryKey := KeyText; Rec.DictionaryValue := Format(Dict.Get(KeyText)); Rec.Insert(); end; end; } table 50100 TestTableForDictionary { DataClassification = CustomerContent; fields { field(1; DictionaryKey; Text[50]) { DataClassification = CustomerContent; } field(2; DictionaryValue; Text[50]) { DataClassification = CustomerContent; } } keys { key(Key1; DictionaryKey) { Clustered = true; } } }

Thanks.

ZHU

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