I need to separate following strings into Name and Number: e.g.
evil333 into evil and 333
bili454 into bili and 454
elvis04 into elvis and 04
Split(String, "#") ' don't work here because numbers are unknown
similarly
Mid(String, 1, String - #) ' don't work because Numbers length is unknown
so what should be the best way to start? Just want to keep it simple as possible
Update:
For further info follow - https://youtu.be/zjF7oLLgtms
Two more ways for solving this:
Sub test()
Dim sInputString As String
Dim i As Integer
Dim lFirstNumberPos As Long
sInputString = "evil333"
'loop through text in input string
'if value IsNumeric (digit), stop looping
For i = 1 To Len(sInputString)
If IsNumeric(Mid(sInputString, i, 1)) Then
lFirstNumberPos = i
Exit For
End If
Next i
Dim Name As String
Dim Number As String
'return result
Name = Left$(sInputString, lFirstNumberPos - 1)
Number = Mid$(sInputString, lFirstNumberPos)
End Sub
Or another method:
Sub test2()
'if you are going to have too long string it would maybe better to use "instr" method
Dim sInputString As String
Dim lFirstNumberPos As Long
Dim i As Integer
sInputString = "evil333"
Dim lLoopedNumber as Long
LoopedNumber = 0
lFirstNumberPos = Len(sInputString) + 1
'loop through digits 0-9 and stop when any of the digits will be found
For i = 0 To 9
LoopedNumber = InStr(1, sInputString, cstr(i), vbTextCompare)
If LoopedNumber > 0 Then
lFirstNumberPos = Application.Min(LoopedNumber,lFirstNumberPos)
End If
Next i
Dim Name As String
Dim Number As String
'return result
Name = Left$(sInputString, lFirstNumberPos - 1)
Number = Mid$(sInputString, lFirstNumberPos)
End Sub
You should regular expressions (regex) to match the two parts of your strings. The following regex describes how to match the two parts:
/([a-z]+)([0-9]+)/
Their use in VBA is thorougly explained in Portland Runner's answer to How to use Regular Expressions (Regex) in Microsoft Excel both in-cell and loops
Have some problems with my function. In database i could have diffrent numbers. For instance below: ( i know it looks strange )
12 312323.3
013.43.9
3.23.14353.55 WHATEVER 345.193
728937.3
87.3 ojojo 23.434blabla 24.424.7
What i need to do is increase number after LAST DOT so just make + 1.
The problem is its not working when it comes after dot more than one digit then.
here is my current code:
Dim inputValue as String = "34.234234.6.12"
'--Get Last char from string and add 1 to it
Dim lastChar As String = CInt(CStr(inputValue.Last)) + 1
'--Remove last char and add lastChar
Dim nextCombinNummer As String = lastValue.Nummer.Substring(0, lastValue.Nummer.Length - 1) & lastChar
Return nextCombinNummer
I think the problem is lastValue.Last + 1 as it will take only one digit, and also when i remove by substring last digit but only 2 will be removed.
Can you help me out with this? How to always take number after last dot from string and then increase that number by 1 and return new entire number?
EDIT:
I think i am able to get and increase the number but still dont know how to remove and put it at the end:
Think that's ok:
Dim inputValue as String = "34.234234.6.12"
Dim number As String = inputValue .Substring(inputValue .LastIndexOf("."c) + 1)
Dim numberIncreased as integer = CInt(number) + 1
'How to do this correctly? :
Dim nextCombinNummer As String = lastValue.Nummer.Substring(0, lastValue.Nummer.Length - 1) & numberIncreased
An easy solution is to cast as Integer the last part of the string, add one, then recompose your string :
'Original Value
Dim val As String = "123.456.789"
'We take only the last part and add one
Dim nb = Integer.Parse(val.Substring(val.LastIndexOf(".") + 1)) + 1
'We recompose the string
Dim FinalVal As String = val.Substring(0, val.LastIndexOf(".") + 1) & nb.ToString()
I'd use following which uses String.Split, Int32.TryParse and String.Join:
Dim numbers As New List(Of String) From {"12.312323.3", "013.43.9", "3.231435355345.193", "728937.3", "87.323.43424.424.7"}
for i As Int32 = 0 To numbers.Count -1
Dim num = numbers(i)
Dim token = num.Split("."c)
dim lastNum = token.Last() ' or token(token.Length-1)
Dim n As Int32
If int32.TryParse(lastNum, n)
n += 1
token(token.Length-1) = n.ToString()
End If
numbers(i) = string.Join(".", token)
Next
I have a code like this :
Dim mincatval As String
Dim strarr() As String = dr1(0).ToString().Split(New Char() {"-"c})
Dim i As String
i = (Integer.Parse(strarr(0)) + 1)
mincatval = i
my dr(1) value is L1 i want to add 1,so i want the out put L2,but i am getting error like this :Input string was not in a correct format.
Supposing that strarr(0) is the word "L1" and you want it to become "L2" then you need to isolate the numeric part from the text part and then rebuild the string taking the first part of strarr and the incremented value
Dim mincatval As String
Dim strarr() As String = dr1(0).ToString().Split(New Char() {"-"c})
Dim i As String
Dim itShouldBeAnumber = strarr(0).Substring(1)
if Int32.TryParse(itShouldBeAnumber, i) Then
mincatval = strarr(0).Substring(0,1) & (i + 1)
else
MessageBox.Show("Not a valid number from position 1 of " & strarr(0))
End if
Of course this solution assumes that your string is Always composed of an initial letter followed by a numeric value that could be interpreted as an integer
I have a string (for example: "Hello there. My name is John. I work very hard. Hello there!") and I am trying to find the number of occurrences of the string "hello there". So far, this is the code I have:
Dim input as String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase as String = "hello there"
Dim Occurrences As Integer = 0
If input.toLower.Contains(phrase) = True Then
Occurrences = input.Split(phrase).Length
'REM: Do stuff
End If
Unfortunately, what this line of code seems to do is split the string every time it sees the first letter of phrase, in this case, h. So instead of the result Occurrences = 2 that I would hope for, I actually get a much larger number. I know that counting the number of splits in a string is a horrible way to go about doing this, even if I did get the correct answer, so could someone please help me out and provide some assistance?
Yet another idea:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "Hello there"
Dim Occurrences As Integer = (input.Length - input.Replace(phrase, String.Empty).Length) / phrase.Length
You just need to make sure that phrase.Length > 0.
the best way to do it is this:
Public Function countString(ByVal inputString As String, ByVal stringToBeSearchedInsideTheInputString as String) As Integer
Return System.Text.RegularExpressions.Regex.Split(inputString, stringToBeSearchedInsideTheInputString).Length -1
End Function
str="Thisissumlivinginsumgjhvgsum in the sum bcoz sum ot ih sum"
b= LCase(str)
array1=Split(b,"sum")
l=Ubound(array1)
msgbox l
the output gives u the no. of occurences of a string within another one.
You can create a Do Until loop that stops once an integer variable equals the length of the string you're checking. If the phrase exists, increment your occurences and add the length of the phrase plus the position in which it is found to the cursor variable. If the phrase can not be found, you are done searching (no more results), so set it to the length of the target string. To not count the same occurance more than once, check only from the cursor to the length of the target string in the Loop (strCheckThisString).
Dim input As String = "hello there. this is a test. hello there hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer = 0
Dim intCursor As Integer = 0
Do Until intCursor >= input.Length
Dim strCheckThisString As String = Mid(LCase(input), intCursor + 1, (Len(input) - intCursor))
Dim intPlaceOfPhrase As Integer = InStr(strCheckThisString, phrase)
If intPlaceOfPhrase > 0 Then
Occurrences += 1
intCursor += (intPlaceOfPhrase + Len(phrase) - 1)
Else
intCursor = input.Length
End If
Loop
You just have to change the input of the split function into a string array and then delare the StringSplitOptions.
Try out this line of code:
Occurrences = input.Split({phrase}, StringSplitOptions.None).Length
I haven't checked this, but I'm thinking you'll also have to account for the fact that occurrences would be too high due to the fact that you're splitting using your string and not actually counting how many times it is in the string, so I think Occurrences = Occurrences - 1
Hope this helps
You could create a recursive function using IndexOf. Passing the string to be searched and the string to locate, each recursion increments a Counter and sets the StartIndex to +1 the last found index, until the search string is no longer found. Function will require optional parameters Starting Position and Counter passed by reference:
Function InStrCount(ByVal SourceString As String, _
ByVal SearchString As String, _
Optional ByRef StartPos As Integer = 0, _
Optional ByRef Count As Integer = 0) As Integer
If SourceString.IndexOf(SearchString, StartPos) > -1 Then
Count += 1
InStrCount(SourceString, _
SearchString, _
SourceString.IndexOf(SearchString, StartPos) + 1, _
Count)
End If
Return Count
End Function
Call function by passing string to search and string to locate and, optionally, start position:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer
Occurrances = InStrCount(input.ToLower, phrase.ToLower)
Note the use of .ToLower, which is used to ignore case in your comparison. Do not include this directive if you do wish comparison to be case specific.
One more solution based on InStr(i, str, substr) function (searching substr in str starting from i position, more info about InStr()):
Function findOccurancesCount(baseString, subString)
occurancesCount = 0
i = 1
Do
foundPosition = InStr(i, baseString, subString) 'searching from i position
If foundPosition > 0 Then 'substring is found at foundPosition index
occurancesCount = occurancesCount + 1 'count this occurance
i = foundPosition + 1 'searching from i+1 on the next cycle
End If
Loop While foundPosition <> 0
findOccurancesCount = occurancesCount
End Function
As soon as there is no substring found (InStr returns 0, instead of found substring position in base string), searching is over and occurances count is returned.
Looking at your original attempt, I have found that this should do the trick as "Split" creates an array.
Occurrences = input.split(phrase).ubound
This is CaSe sensitive, so in your case the phrase should equal "Hello there", as there is no "hello there" in the input
Expanding on Sumit Kumar's simple solution, here it is as a one-line working function:
Public Function fnStrCnt(ByVal str As String, ByVal substr As String) As Integer
fnStrCnt = UBound(Split(LCase(str), substr))
End Function
Demo:
Sub testit()
Dim thePhrase
thePhrase = "Once upon a midnight dreary while a man was in a house in the usa."
If fnStrCnt(thePhrase, " a ") > 1 Then
MsgBox "Found " & fnStrCnt(thePhrase, " a ") & " occurrences."
End If
End Sub 'testit()
I don't know if this is more obvious?
Starting from the beginning of longString check the next characters up to the number characters in phrase, if phrase is not found start looking from the second character etc. If it is found start agin from the current position plus the number of characters in phrase and increment the value of occurences
Module Module1
Sub Main()
Dim longString As String = "Hello there. My name is John. I work very hard. Hello there! Hello therehello there"
Dim phrase As String = "hello There"
Dim occurences As Integer = 0
Dim n As Integer = 0
Do Until n >= longString.Length - (phrase.Length - 1)
If longString.ToLower.Substring(n, phrase.Length).Contains(phrase.ToLower) Then
occurences += 1
n = n + (phrase.Length - 1)
End If
n += 1
Loop
Console.WriteLine(occurences)
End Sub
End Module
I used this in Vbscript, You can convert the same to VB.net as well
Dim str, strToFind
str = "sdfsdf:sdsdgs::"
strToFind = ":"
MsgBox GetNoOfOccurranceOf( strToFind, str)
Function GetNoOfOccurranceOf(ByVal subStringToFind As String, ByVal strReference As String)
Dim iTotalLength, newString, iTotalOccCount
iTotalLength = Len(strReference)
newString = Replace(strReference, subStringToFind, "")
iTotalOccCount = iTotalLength - Len(newString)
GetNoOfOccurranceOf = iTotalOccCount
End Function
I know this thread is really old, but I got another solution too:
Function countOccurencesOf(needle As String, s As String)
Dim count As Integer = 0
For i As Integer = 0 to s.Length - 1
If s.Substring(i).Startswith(needle) Then
count = count + 1
End If
Next
Return count
End Function
I have code below. How do I get strings inside brackets? Thank you.
Dim tmpStr() As String
Dim strSplit() As String
Dim strReal As String
Dim i As Integer
strWord = "hello (string1) there how (string2) are you?"
strSplit = Split(strWord, "(")
strReal = strSplit(LBound(strSplit))
For i = 1 To UBound(strSplit)
tmpStr = Split(strSplit(i), ")")
strReal = strReal & tmpStr(UBound(tmpStr))
Next
Dim src As String = "hello (string1) there how (string2) are you?"
Dim strs As New List(Of String)
Dim start As Integer = 0
Dim [end] As Integer = 0
While start < src.Length
start = src.IndexOf("("c, start)
If start <> -1 Then
[end] = src.IndexOf(")"c, start)
If [end] <> -1 Then
Dim subStr As String = src.Substring(start + 1, [end] - start - 1)
If Not subStr.StartsWith("(") Then strs.Add(src.Substring(start + 1, [end] - start - 1))
End If
Else
Exit While
End If
start += 1 ' Increment start to skip to next (
End While
This should do it.
Dim result = Regex.Matches(src, "\(([^()]*)\)").Cast(Of Match)().Select(Function(x) x.Groups(1))
Would also work.
This is what regular expressions are for. Learn them, love them:
' Imports System.Text.RegularExpressions
Dim matches = Regex.Matches(input, "\(([^)]*)\)").Cast(of Match)()
Dim result = matches.Select(Function (x) x.Groups(1))
Two lines of code instead of more than 10.
In the words of Stephan Lavavej: “Even intricate regular expressions are easier to understand and modify than equivalent code.”
Use String.IndexOf to get the position of the first opening bracket (x).
Use IndexOf again the get the position of the first closing bracket (y).
Use String.Substring to get the text based on the positions from x and y.
Remove beginning of string up to y+1.
Loop as required
That should get you going.
This may also work:
Dim myString As String = "Hello (FooBar) World"
Dim finalString As String = myString.Substring(myString.IndexOf("("), (myString.LastIndexOf(")") - myString.IndexOf("(")) + 1)
Also 2 lines.