Forum Discussion
Trim a string with character then with 5 digits (C#####)
- 3 years ago
Hi on_qq ,
How about this?
Credits to ImkeF.
Here the code in Power Query M that you can paste into the advanced editor (if you do not know, how to exactly do this, please check out this quick walkthrough)
let Source = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("LYwxDoAgEAS/sqG2gQOB8rS2QitCDDG0Fkp8vwTtNrOTiVFIReBpRqhXKXXAokhqEmmIgnXbavfeI+Sz5t/pnzEG8q7gp+AYrSPdsSKNtTT+qdwC1mELLFJ6AQ==", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [Address = _t]), #"ChangedType" = Table.TransformColumnTypes(Source,{{"Address", type text}}), #"CharactersToReplace" = Text.Combine(List.RemoveItems(List.Transform({1..126}, each Character.FromNumber(_)),{"0".."9"})), Convert=Table.AddColumn(ChangedType, "Custom1", each Text.SplitAny( [Address], CharactersToReplace) ), Convert2=Table.AddColumn(Convert, "Custom2", each List.Select( [Custom1], (li) => Text.Length(li) >=5){0} ), #"Added Custom" = Table.AddColumn(Convert2, "Custom", each Text.Range ([Address], Text.PositionOf ([Address], [Custom2]) - 1, 6)) in #"Added Custom"
I suppose the idea is to fetch the postal code that is in the shape of "A12345". Be aware, though, that in case you have a street number with five or more digits, the solution above might not work as expected.Let me know if this helps! ๐
/Tom
https://www.tackytech.blog/
https://www.instagram.com/tackytechtom/
Hi on_qq ,
How about this?
Credits to ImkeF.
Here the code in Power Query M that you can paste into the advanced editor (if you do not know, how to exactly do this, please check out this quick walkthrough)
let
Source = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("LYwxDoAgEAS/sqG2gQOB8rS2QitCDDG0Fkp8vwTtNrOTiVFIReBpRqhXKXXAokhqEmmIgnXbavfeI+Sz5t/pnzEG8q7gp+AYrSPdsSKNtTT+qdwC1mELLFJ6AQ==", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [Address = _t]),
#"ChangedType" = Table.TransformColumnTypes(Source,{{"Address", type text}}),
#"CharactersToReplace" = Text.Combine(List.RemoveItems(List.Transform({1..126}, each Character.FromNumber(_)),{"0".."9"})),
Convert=Table.AddColumn(ChangedType, "Custom1", each Text.SplitAny(
[Address],
CharactersToReplace)
),
Convert2=Table.AddColumn(Convert, "Custom2", each List.Select(
[Custom1],
(li) => Text.Length(li) >=5){0}
),
#"Added Custom" = Table.AddColumn(Convert2, "Custom", each Text.Range ([Address], Text.PositionOf ([Address], [Custom2]) - 1, 6))
in
#"Added Custom"
I suppose the idea is to fetch the postal code that is in the shape of "A12345". Be aware, though, that in case you have a street number with five or more digits, the solution above might not work as expected.
Let me know if this helps! ๐
/Tom
https://www.tackytech.blog/
https://www.instagram.com/tackytechtom/