Okay this one might be a little tougher. I'm using VB that looks like this:
string = Replace(string.ToLower, chr(63), "A")
But I also want chr(63) = "B" as well, like this:
string = Replace(string.ToLower, chr(63), "B")
My problem is that when chr(63) is at the end of a string I need it to be B, and when it's not the end I need it to be A. I suppose that I can use an if/then/else statement. Is there a way to do this?
Example:
XXXXXchr(63)XXXXX = A
but
XXXXXXXXXXchr(63) = B
Thanks!
pseudo:
if (string[string.Length] == chr(63))
{
string[string.Length] = B
}
string = Replace(string.ToLower, chr(63), "A")
string = Replace(string.ToLower, chr(63), "A", 1, Len(string) - 1)
If Right(string, 1) = chr(63) then
Mid$(string, Len(string), 1) = 'B'
End if
Update: in response to comment:
VB String Functions
VB String Array Functions - Split, Join, Filter (very useful)
I haven't used Visual Basic since version 6, but it should be something like this:
If Robert.EndsWith(chr(63)) Then
Robert = Left(Robert, Robert.Length - 1) + "B"
End If
Then do the usual replacement with A.
This ought to do it
Dim s As String
Dim char63 As String = Convert.ToChar(63).ToString
If s.EndsWith(char63) Then
s = s.Substring(0, s.Length - 1) & "B"
End If
s = s.Replace(char63, "A")
Related
I have this string abcd, and I want to replace the a with [a|b] and the b with [c|d]
I try many ways to do it, like
Dim varString As String = "abcd"
varString = varString.Replace("a", "[a|b]")
varString = varString.Replace("b", "[c|d]")
The result I get is
[a|[c|d]][c|d]cd
Instead I want it like this
[a|b][c|d]cd
The problem is every time I use the replace function it backs to change the values I already replaced before so I replaced a with [a|b] but then when I do my second command to replace the b it changes the b in [a|b] that I just changed and I don't want this.
I tried to use StringBuilder but it gives the same result.
Please advise me,
I solved the problem by making an array in this way
Dim NewCommand As String = "abcd"
For i = 0 To LikeCommand.Length - 1
If LikeCommand(i) = "a" Then
NewCommand += "[a|b]"
ElseIf LikeCommand(i) = "b" Then
NewCommand += "[c|d]"
Else
NewCommand += LikeCommand(i)
End If
Next
LikeCommand = NewCommand
Or just switch the logic up. But obviously I'm thinking you're using a basic example for a more complex question.
dim varString as string = "abcd"
varString = varString.Replace("b" ,"[c|d]")
varString = varString.Replace("a" ,"[a|b]")
That would get you the desired results.
I'm trying to check whether the main string contains the entire substring, even if there are interruptions.
For example:
main string = 12ab34cd,
substring = 1234d
should return a positive, since 1234d is entirely contained in my main string, even though there are extra characters.
Since InStr doesn't take wildcards, I wrote my own VBA using the mid function, which works well if there are extra characters at the start/end, but not with extra characters in the middle.
In the above example, the function I wrote
works if the main string is ab1234dc,
but not if it's 12ab34cd.
Is there a way to accomplish what I'm trying to do using VBA?
Note Both of the methods below are case sensitive. To make them case insensitive, you can either use Ucase (or Lcase) to create phrases with the same case, or you can prefix the routine with the Option Compare Text statement.
Although this can be done with regular expressions, here's a method using Mid and Instr
Option Explicit
Function ssFind(findStr, mainStr) As Boolean
Dim I As Long, J As Long
I = 1: J = 1
Do Until I > Len(findStr)
J = InStr(J, mainStr, Mid(findStr, I, 1))
If J = 0 Then
ssFind = False
Exit Function
End If
I = I + 1: J = J + 1
Loop
ssFind = True
End Function
Actually, you can shorten the code further using Like:
Option Explicit
Function ssFind(findStr, mainStr) As Boolean
Dim I As Long
Dim S As String
For I = 1 To Len(findStr)
S = S & "*" & Mid(findStr, I, 1)
Next I
S = S & "*"
ssFind = mainStr Like S
End Function
Assuming you have 3 columns "SUBSTR","MAIN" and "CHECK" and your "Substring" data range is named "SUBSTR"
Sub check_char()
Dim c As Range
For Each c In Range("SUBSTR")
a = 1
test = ""
For i = 1 To Len(c.Offset(0, 1))
If Mid(c.Offset(0, 1), i, 1) = Mid(c, a, 1) Then
test = test & Mid(c.Offset(0, 1), i, 1)
a = a + 1
End If
Next i
If test = c Then
c.Offset(0, 2) = "MATCH"
Else
c.Offset(0, 2) = "NO MATCH"
End If
Next
End Sub
I have a script that takes the contents of a cell, and puts the first 2 characters of the cell into a string array. I need to later compare that string array to a string, but I can't seem to get that to work. Here's what I have:
For i = 2 To 600
colStr = Sheets("smartlist").Cells(i, "A").Value
If colStr <> "" Then
ReDim charArray(Len(colStr) - 1)
For j = 1 To Len(colStr)
charArray(j - 1) = Mid$(colStr, j, 1)
Next
strArray = LCase(charArray(0)) & LCase(charArray(1))
If CStr(Join(strArray)) = CStr(Join(pwArray)) Then
Now, I've tried:
If charArray = "ab"
If Join(charArray) = "ab"
If CStr(Join(charArray)) = "ab"
I'm pretty lost at this point. Any suggestions would be welcome!
Edit: added the whole function up until I get the 'Type mismatch'
You could use Join(charArray, "") - without "" it joins the elements with space so the result of your initial try was "a b"
Firstly, you really need to clarify what you're doing. You say that later you need to check what's in the string. If that is the case, then you don't need an array, you simply need another string...
Dim chars As String
chars = Left$(cellToTest, 2)
Later you can test more simply using the InStr function, like so...
Dim loc As Integer
loc = Instr(colStr, chars)
If loc > 0 Then
...
If you want to turn this into a function on your spreadsheet you can use the following in the cells as formulas...
=LEFT(A1, 2)
=SEARCH(A3, A1)
Here's a little screen shot of what I mean...
Hi I have a function that finds the longest common substring between two strings. It works great except it seems to break when it reaches any single quote mark: '
This causes it to not truly find the longest substring sometimes.
Could anyone help me adjust this function so it includes single quotes in the substring? I know it needs to be escaped someplace I'm just not sure where.
Example:
String 1: Hi there this is jeff's dog.
String 2: Hi there this is jeff's dog.
After running the function the longest common substring would be:
Hi there this is jeff
Edit: seems to also happen with "-" as well.
It will not count anything after the single quote as part of the substring.
Here's is the function:
Public Shared Function LongestCommonSubstring(str1 As String, str2 As String, ByRef subStr As String)
Try
subStr = String.Empty
If String.IsNullOrEmpty(str1) OrElse String.IsNullOrEmpty(str2) Then
Return 0
End If
Dim num As Integer(,) = New Integer(str1.Length - 1, str2.Length - 1) {}
Dim maxlen As Integer = 0
Dim lastSubsBegin As Integer = 0
Dim subStrBuilder As New StringBuilder()
For i As Integer = 0 To str1.Length - 1
For j As Integer = 0 To str2.Length - 1
If str1(i) <> str2(j) Then
num(i, j) = 0
Else
If (i = 0) OrElse (j = 0) Then
num(i, j) = 1
Else
num(i, j) = 1 + num(i - 1, j - 1)
End If
If num(i, j) > maxlen Then
maxlen = num(i, j)
Dim thisSubsBegin As Integer = i - num(i, j) + 1
If lastSubsBegin = thisSubsBegin Then
subStrBuilder.Append(str1(i))
Else
lastSubsBegin = thisSubsBegin
subStrBuilder.Length = 0
subStrBuilder.Append(str1.Substring(lastSubsBegin, (i + 1) - lastSubsBegin))
End If
End If
End If
Next
Next
subStr = subStrBuilder.ToString()
Return subStr
Catch e As Exception
Return ""
End Try
End Function
I tried it with dotnetfiddle and there it is working with your Code you posted. Please activate your warnings in your project. You have function with no return value and you return an integer or a string. This is not correct. How are you calling your function?
Here is my example I tested for you:
https://dotnetfiddle.net/mVBDQp
Your code works perfectly like Regex! As far as I can see, there is really nothing wrong with your code.
Here I even tested it under more severe case:
Public Sub Main()
Dim a As String = ""
Dim str1 As String = "Hi there this is jeff''s dog.-do you recognize this?? This__)=+ is m((a-#-&&*-ry$##! <>Hi:;? the[]{}re this|\ is jeff''s dog." 'Try to trick the logic!
Dim str2 As String = "Hi there this is jeff''s dog. ^^^^This__)=+ is m((a-#-&&*-ry$##! <>Hi:;? the[]{}re this|\ is jeff''s dog."
LongestCommonSubstring(str1, str2, a)
Console.WriteLine(a)
Console.ReadKey()
End Sub
Note that I put '-$#^_)=+&|\{}[]?!;:.<> all there. Plus I tried to trick your code by giving early result.
But the result is excellent!
You could probably put more actual samples on the inputs which give you problems. Else, you could possibly describe the environment that you use/deploy your code into. Maybe the problem lies elsewhere and not in the code.
The quickest way to solve this would be to use an escape code and replace all the ' with whatever escape code you use
Hello I am trying to get part of my string out and was wondering if anyone can help so here is what I am working with
Dim tempName, NewName As String
'This is an example of what tempName will Equal
'What I need is the information in between the first -
'and the second one. In this case it would be 120
'I can not do it by number because the dashes are the only thing
'that stays the same.
tempName = "3-120-12-6"
NewName = tempName 'Do not know what String Manipulation to use.
Another few examples would be 6-56.5-12-12 I need the 56.5 or 2-89-12-4 I would need the 89
Thank you for the help.
You can do this too... You can break it up into pieces by detecting the first -. Then get the second...
Dim x as String = Mid(tempName, InStr(tempName, "-") + 1)
NewName = Mid(x, 1, InStr(x , "-") - 1)
Or make something like this to get it into 1 line of code...
NewName = Mid(Mid(tempName, InStr(tempName, "-") + 1), 1, InStr(Mid(tempName, InStr(tempName, "-") + 1), "-") - 1)
Or the quickest approach to this would be to use Split like the code below... This code equates the values between - into an array. Since you want the second value, you'd want the array with the index of 1 since an array starts with 0. If you're only starting, do try to create your own way to manipulate the string, this will help your mind think of new things and not just rely on built-in functions...
Dim tempName As String = "3-120-12-6"
Dim secondname() As String = Split(tempName, "-")
NewName = secondname(1)