Please let me how to remove double spaces and characters from below string.
String = Test----$$$$19****45@@@@ Nothing
Clean String = Test-$19*45@ Nothing
I have used regex "\s+" but it just removing the double spaces and I have tried other patterns of regex but it is too complex... please help me.
I am using vb.net
What you'll want to do is create a backreference to any character, and then remove the following characters that match that backreference. It's usually possible using the pattern (.)\1+, which should be replaced with just that backreference (once). It depends on the programming language how it's exactly done.
Dim text As String = "Test@@@_&aa&&&"
Dim result As String = New Regex("(.)\1+").Replace(text, "$1")
result will now contain Test@_&a&. Alternatively, you can use a lookaround to not remove that backreference in the first place:
Dim text As String = "Test@@@_&aa&&&"
Dim result As String = New Regex("(?<=(.))\1+").Replace(text, "")
Edit: included examples
. with [^A-Za-z0-9]. This is a character class that matches anything that's not (the ^ negates it) either an uppercase letter, lowercase letter or number. Therefore, it will remove anything else. Your pattern would then look like ([^A-Za-z0-9])\1+.For a faster alternative try:
Dim text As String = "Test@@@_&aa&&&"
Dim sb As New StringBuilder(text.Length)
Dim lastChar As Char
For Each c As Char In text
If c <> lastChar Then
sb.Append(c)
lastChar = c
End If
Next
Console.WriteLine(sb.ToString())
Here is a perl way to substitute all multiple non word chars by only one:
my $String = 'Test----$$$$19****45@@@@ Nothing';
$String =~ s/(\W)\1+/$1/g;
print $String;
output:
Test-$19*45@ Nothing
Here's how it would look in Java...
String raw = "Test----$$$$19****45@@@@ Nothing";
String cleaned = raw.replaceAll("(.)\\1+", "$1");
System.out.println(raw);
System.out.println(cleaned);
prints
Test----$$$$19****45@@@@ Nothing
Test-$19*45@ Nothing
