FromIp contains "192.168.1.1". I want to get the last number, but I can't figure out what's wrong here:
Dim str As String
str = FromIP.Text.Substring(FromIP.Text.LastIndexOf("."), FromIP.Text.Length).ToString()
MessageBox.Show(FromIP.Text.Length)
Eduardo has given the correct way of getting the substring - my answer here will just explain why the existing one fails.
String.Substring(int, int) takes a starting position and a count. You're basically saying, "Go from position 9 for 10 characters". The documentation explicitly states it will throw:
ArgumentOutOfRangeException [if]
startIndex plus length indicates a
position not within this instance.
-or-
startIndex or length is less than
zero.
Tested code:
Dim FromIp As String = "192.168.1.1"
Dim str As String
str = FromIp.Substring(FromIp.LastIndexOf(".") + 1).ToString()
MessageBox.Show(str)
You must add 1 to LastIndexOf to skip the dot
There no need put the lenght of the Substring when you want all the rest of the string
But this refactored code will work better:
Dim FromIp As String = "192.168.1.1"
Dim IpPart As String() = FromIp.Split(".")
MessageBox.Show(IpPart(3))
FromIP.Text.LastIndexOf(".") + 1
instead of
FromIP.Text.LastIndexOf(".")
and
FromIP.TextLength-FromIP.Text.LastIndexOf(".")
is the last parameter instead of
FromIP.TextLength
As far as I remember, Substring takes Start,Stop as parameters.
So that would be: txt.Substring(IndexOf, txt.Length - IndexOf) or just with one parameter: Substring(IndexOf + 1)
Related
substring = "frTrig_worldcup"
main string = "0.10/BMM/fTrig_Mast/BOYU/frTrig_worldcup"
The substring value will change. "frTrig_xxxx" but "frTrig_" will be constant
I need to get the substring which is in the LAST PART of the main string. The issue is the "frTrig_" comes in mid of the string also
Please help me to solve this.
We can try using String#split here:
Dim input As String
Dim parts() As String
input = "0.10/BMM/BOYU/fTrig_MastrrRea0.01/frTrig_worldcup"
parts = input.Split("/")
MsgBox(parts(parts.Length-1))
I have a string ABC(N9KGRTLMN9(0J)M3.
I want to return the character after GRTLM which is N. Thanks.
Look at the System.Text.RegularExpressions namespace, and create a RegEx object with this expression:
GRTLM(.)
Then you will be able to check the Matches for the expression to find your character. Depending on what you know about that string, you may be able to narrow things even further. For example:
GRTLM([A-Za-z])
or
GRTLM([A-Z])
If you don't want to use regular expressions (for any reason), here's an alternative:
Private Function ReturnCharAfter(Source As String, after As String) As Char
Dim i As Integer = Source.IndexOf(after)
If i < 0 Then Return Nothing
Return Source(i + after.Length)
End Function
usage:
Dim N As Char = ReturnCharAfter("ABC(N9KGRTLMN9(0J)M3.", "GRTLM")
You could use String.Split() to get the N
Dim input = "ABC(N9KGRTLMN9(0J)M3"
Dim s = "GRTLM"
Dim n = input.Split({s}, StringSplitOptions.RemoveEmptyEntries)(1)(0)
It splits the string into substrings using GRTLM as a delimiter, then returns the first character of the second array item.
Or to get the index of N
Dim i = input.Split({s}, StringSplitOptions.RemoveEmptyEntries)(0).Length + s.Length
It splits the string and returns the length of the first array item plus the length of the delimiter string.
But perhaps the simplest way to do it is using String.IndexOf()
Dim n = input(input.IndexOf(s) + s.Length)
Dim i = input.IndexOf(s) + s.Length
i have two example
"hcg.com.ph?C402-10A-2012-06132017-22"
"hcg.com.?C3032-1B-2012-06132017-1"
output should be
hcg.com.ph?C402-10A-2012-06132017
hcg.com.?C3032-1B-2012-06132017
but i got
hcg.com.ph?C402 and hcg.com.?C3032
Dim FinalSplt() As String
Dim ItemBaseCode As String
FinalSplt = value.ToString.Split("-")
ItemBaseCode = FinalSplt(0)
How to split in the last dash?
Here is some code that uses Substring and LastIndexOf.
'test input
Dim si() As String = {"hcg.com.ph?C402-10A-2012-06132017-22", "hcg.com.?C3032-1B-2012-06132017-1"}
'use to verify
Dim sv() As String = {"hcg.com.ph?C402-10A-2012-06132017", "hcg.com.?C3032-1B-2012-06132017"}
For x As Integer = 0 To si.Length - 1
Dim s As String = si(x).Substring(0, si(x).LastIndexOf("-"c))
'verify
If s = sv(x) Then
Stop 'verified
End If
Next
Ok, without actually writing code I can see you need to split the string more efficiently.
Firstly, strip the quotes.
Secondly, split the string based on the ? mark.
Take the second string in the array, and split that based on the - mark.
Now you have an array of all the portions, join this array with all except the last element.
Join the new string with the original first part.
Add the quotes back if needed.
I have the following string expression:
str="1+2+3+4"
Now from this string I want the value of number (10). How can I get this value from this expression?
For that to work, you'd have to parse all the numbers, something like this
Dim str As String= "1+2+3+4"
Dim numbers() As String = str.Split('+')
Dim result As Integer = 0
For Each number As String In numbers
result += Integer.Parse(number)
Next
Here is an alternative solution that uses Linq. This solution excludes nothing it can not parse...
YOURSTRING.Split("+").ToList.Where(Function(sr) Not String.IsNullOrEmpty(sr) AndAlso Integer.TryParse(sr, New Integer)).Sum(Function(s As String) Integer.Parse(s))
Example Output
1+2+3+4 = 10
1+2+3+4++12+1 = 23 (Notice the typo (4++12), it still works even it there's a mistake)
I have a string, for example:
Dim str as string = xxxxxxxxxxxxxxxxxxxx£xxx£xxxx**£**xxxxxxxxxx
I want to remove £ surrounded from * which is always at a certain position (11th for instance) from the end. The whole string is a long one, always change in size and cannot be counted from the start. I cannot use Replace as well, there may be same characters at other positions that I do not wish to remove.
Solution:
Dim rst As String = str.Remove(str.Length - 11, 1)
Edit: Whoops, I dunno what I was thinking on that first part.
The correct version of the first part would be:
str = str.Substring(0, str.Len -13) + str.Substring(str.Len-11);
There also may be an overload for the String.Delete function that allows you to use a negative number to represent the number of characters from the end of the string -- I know that the C# equivalent does.
If its always going to be the 11th character from the end you can do this...
Dim strTargetString As String = "xxxYxxxxxxxxxx"
Dim strTargetString2 As String = "xxxxxxxYxxxxxxxxxx"
Dim strResult As String = Mid(strTargetString, 1, (Len(strTargetString) - 11)) & Microsoft.VisualBasic.Right(strTargetString, 10)
Dim strResult2 As String = Mid(strTargetString2, 1, (Len(strTargetString2) - 11)) & Microsoft.VisualBasic.Right(strTargetString, 10)
Note that String.SubString is a more modern approach than Mid, but I use it out of preference and example.
This is fairly straightforward with a regular expression replacement operation using look-ahead:
Dim str as String = "xxxxxxxxxxxxxxxxxxxx£xxx£xxxx£xxxxxxxxxx"
Dim str2 as String = Regex.Replace(str, "£(?=.{10}$)", String.Empty)
This will target a single character followed by any ten characters then the end of the string and replace it with the String.Empty value (or just "" if you'd prefer).