"\:(.*)$"
Hi all i am using above expression to remove all the string before : (colon), but it is giving me all the string before this. how can i do this. Thanks a lot.
My string is:
This is text: Hi here we go
I am getting: This is text
I want : Hi here we go
Updated code
Sub Main()
Dim input As String = "This is text with : far too much "
Dim pattern As String = "\:(.*)$"
Dim replacement As String = " "
Dim rgx As New Regex(pattern)
Dim result As String = rgx.Replace(input, replacement)
Console.WriteLine("Original String: {0}", input)
' MsgBox("Original String: {0}")
Console.WriteLine("Replacement String: {0}", result)
MsgBox("Original String: {0}")
End Sub
Try this pattern. This will help you to match string after colon
/?:(.)/
or
/: (.+)/
It should be:
Dim pattern As String = "(.*)\:"
' in vb if above one doesn't work, then try this one
' Dim pattern As String = "^(.*)\:"
' also i don't think we need to use any brackets here as well.
This regex means, anything before the colon(:), Where you were using anything after the colon(:) in your example.
If you are not dead set on RegEx then you can also use
Dim result As String
result = Strings.Split(Input, ":", 2)(1)
This splits the input into an array with two elements. First element is the text before the first ":", the second element is the text after.
Related
This is the text:
"{\"step_naAme\": \"\", \"nonce_code\": \"HGtDXNmPhb\", \"user_id\": 1545816852, \"is_stateless\": false}"
I just want to extract HGtDXNmPhb and 1545816852 without the rest. I tried:
dim text as string = regex.match(text=fromstring,"""{\""step_name\"":\""\"", \""nonce_code\"": \""(.*?)\"", \""user_id\"": (.*?),\""is_stateless\"": false}""").groups(1).value
the response was empty every time. please help, thanks.
Here's one way, if you know for fact that the values will always be in the same place in the array (make sure you escape the original string properly first of course):
Dim strString As String = """{\""step_naAme\"": \""\"", \""nonce_code\"": \""HGtDXNmPhb\"", \""user_id\"": 1545816852, \""is_stateless\"": false}"""
Dim strSearch() As String = Regex.Split(strString, "\W+")
Debug.Writeline(strSearch(3) & " " & strSearch(5))
Result:
HGtDXNmPhb 1545816852
I have this text:
2|#Favo|Name||26.0000|50.10000|_GRE|||||City|Road||||
I want to capture anything between those special chars: ||
For example, I want to capture "Name" only or I want to capture "City"
I've spent many hours and all I came up with is this regex:
([^|].*[$|])\w+
Here are the required values:
How can I capture one of them?
Thank you.
You may split the string with | removing empty entries and also all those that are blank or consisting only of digits:
Dim strng As String = "2|#Favo|Name||26.0000|50.10000|_GRE|||||City|Road||||"
Dim reslt As List(Of String) = strng.Split(New String() {"|"}, StringSplitOptions.RemoveEmptyEntries).Where(
Function(m) m.All(AddressOf Char.IsDigit) = False And String.Equals(m.Trim(), String.Empty) = False).ToList()
Console.Write(String.Join(", ", reslt))
Output:
My application stores the path of the userdata.dll in a String.
I need to convert this string: C:\Applications\User\userdata.dll
into this: C:\\Applications\\User\\userdata.dll
All \ will need to be duplicated, independent on how many \ the path have.
Something like:
Dim defaultPath As String = "C:\Applications\User\userdata.dll"
' Regex
Dim r As Regex = New Regex( ... )
' This is the replacement string
Dim Replacement As String = " ... $1 ... "
' Replace the matched text in the InputText using the replacement pattern
Dim modifiedPath As String = r.Replace(defaultPath,Replacement)
Any help on this? I am trying to follow this question:
How to replace some part of this string with vb.net?
But cant find out how to make this Regex...
You can use
Dim pattern As String = "\\"
Dim rgx As New Regex(pattern)
Dim input As String = "C:\Applications\User\userdata.dll"
Dim result As String = rgx.Replace(input, "\\")
Console.WriteLine(result)
Ideone Demo
If you mean to say that replace any number of \ to \\, then you can use
Dim pattern As String = "\\+"
Dim rgx As New Regex(pattern)
Dim input As String = "C:\\\\Applications\User\userdata.dll"
Dim result As String = rgx.Replace(input, "\\")
Ideone Demo
Just have a list of words, such as:
gram (g)
kilogram (kg)
pound (lb)
just wondering how I would get the words within the brackets for example get the "g" in "gram (g)" and dim it as a new string.
Possibly using regex?
Thanks.
Use split function ..
strArr = str.Split("(") ' splitting 'gram (g)' returns an array ["gram " , "g)"] index 0 and 1
strArr2 = strArr[1].Split(")") ' splitting 'g)' returns an array ["g " ..]
the string is in
strArr2[0]
Edit
you want getAbbrev and getAbbrev2 to be arrays
try
Dim getAbbrev As String() = Str.Split("(")
Dim getAbbrev2 as String() = getAbbrev[1].Split(")")
To do it without declaring arrays you can do
"gram (g)".Split("(")[1].Split(")")[0]
but that's unreadable
Edit
You have some very trivial errors. I would suggest you strengthen your understanding on objects and declarations first. Then you can look into invoking methods. I rather have you understand it than give it to you. Re-read the book you have or look for a basic tutorial.
Dim unit As String = 'make sure this is the actual string you are getting, not sure where you are supposed to get the string value from => ie grams (g)
Dim getAbbrev As String() = unit.Split("(") 'use unit not Str - Str does not exist
Dim getAbbrev2 As String() = getAbbrev[1].Split(")") 'As no as - case sensitive
for the last line reference getAbbrev2 instead of the unknown abbrev2
Fun with Regular Expressions (I'm really not an expert here, but tested and works)
Imports System.Text.RegularExpressions
.....
Dim charsToTrim() As Char = { "("c, ")"c }
Dim test as String = "gram (g)" + Environment.NewLine +
"kilogram (kg)" + Environment.NewLine +
"pound (lb)"
Dim pattern as String = "\([a-zA-Z0-9]*\)"
Dim r As Regex = new Regex(pattern, RegexOptions.IgnoreCase)
Dim m As Match = r.Match(test)
While(m.Success)
System.Diagnostics.Debug.WriteLine("Match" + "=" + m.Value.ToString())
Dim tempText as String = m.Value.ToString().Trim(charsToTrim)
System.Diagnostics.Debug.WriteLine("String Trimmed" + "=" + tempText)
m = m.NextMatch()
End While
You can split at the space and remove the parens from the second token (by replacing them with an empty string).
A regex is also an option, and is very simple, its pattern is
\w+\s+\((\w+)\)
Which means, a word, then at least one space, then opening parens, then in real regex parens you search for a word, and, eventually a closing paren. The inner parentheses are capturing parentheses, which make it possible to refer to the unit g, kg, lb.
I am trying to separate numbers from a string which includes %,/,etc for eg (%2459348?:, or :2434545/%). How can I separate it, in VB.net
you want only the numbers right?
then you could do it like this
Dim theString As String = "/79465*44498%464"
Dim ret = Regex.Replace(theString, "[^0-9]", String.Empty)
hth
edit:
or do you want to split by all non number chars?
then it would go like this
Dim ret = Regex.Split(theString, "[^0-9]")
You could loop through each character of the string and check the .IsNumber() on it.
This should do:
Dim test As String = "%2459348?:"
Dim match As Match = Regex.Match(test, "\d+")
If match.Success Then
Dim result As String = match.Value
' Do something with result
End If
Result = 2459348
Here's a function which will extract all of the numbers out of a string.
Public Function GetNumbers(ByVal str as String) As String
Dim builder As New StringBuilder()
For Each c in str
If Char.IsNumber(c) Then
builder.Append(c)
End If
Next
return builder.ToString()
End Function