What would be the best paterrn for regex to leave only digits, letters, spaces and new lines?
I’m guessing it should be:
System.Text.RegularExpressions.Regex.Replace(myStr, “pattern”,“”) but I don’t know the right “pattern”.
This myStr come from OCR (scan) and there could be a lot of characters like . , & ^ % and many others.
Hi @Yameso …You can try the below pattern.
Regex.Replace(Input,“[\~#%&*{}/:<>?|"-]”,“”)
Please add all the character with in square brackets…
Use the below Regex to only retain Digits, Letters, Spaces and NewLines:
[^A-Za-z0-9\s\r?\n|\r]
This identifies any other character than ones mentioned above. You can replace it using null string.
Check as below
I add the : , remaining it will identify
Use Regex.Replace(Your String, @“[^0-9a-zA-Z:,]+”, “”)
Hope this may help you
Thanks
There’s a double back slash before s. It should be two \s before the letter s. It is being automatically adjusted to one while typing the expression here.
I think that would work [^A-Za-z0-9\s]+
But I require to leave also letters with accents like “Ę” “ę” and other polish characters. is there other way to leave them too except this:
[^A-Za-z0-9\sĘꥹŻżŹźŁłŃńÓóŚśĆć]+
Check this:
Replace:
System.Text.RegularExpressions.Regex.Replace(your_String,"[^[\p{L}|\p{N}|\s]","")
Here you can use ASCII encoding instead of Regex.
This will simply replace any Es with normal E, and same goes for other characters too.
Use in a Assign:
System.Text.Encoding.UTF8.GetString(System.Text.Encoding.GetEncoding(“ISO-8859-8”).GetBytes(string.ToString))
I used:
System.Text.Encoding.UTF8.GetString(System.Text.Encoding.GetEncoding(“ISO-8859-8”).GetBytes(string.ToString))
and this one leaves marks such as " ? * , . ( ) and few others, and change letters from “Ę” to “E”.
I don’t want to replace them for characters without accent. I’d like to ignore them and clean others.
Pattern [~#%&*{}/:<>?|"-] will not get sth like this:
With this one [[1] those dots are also considered as thing to replace and letters with accents stays :
Thanks for help. @Adrian_Star nailed it 
\p{L}|\p{N}|\s ↩︎
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