This is my first time using VBA and I'm having trouble finding a solution using FOR loops to count how many times a specific letter (entered by user) is in the string (entered by user).
Below I have what I have so I've found other solutions but they don't seem to utilize to FOR loop.
If anyone has any suggestions that'd be awesome.
Sub Week3()
Dim userInput As String
Dim letterSearched As String
Dim counter As Long
Dim occurances As Long
userInput = InputBox("type letters")
letterSearched = InputBox("type letter to be searched")
occurances = 0
For counter = 1 To Len(userInput)
If (InStr(counter, userInput, letterSearched)) > 0 Then
occurances = occurances + 1
Else
occurances = occurances
End If
Next
MsgBox (occurances)
End Sub
No need to use a loop. Is this what you are trying?
Option Explicit
Sub Sample()
Dim Mytext As String
Dim SearchText As String
Mytext = "Sample Text"
SearchText = "e"
MsgBox Len(Mytext) - Len(Replace(Mytext, SearchText, ""))
End Sub
EDIT
Yea I saw that solution but I'm supposed to implement a FOR loop – John Orsa 4 mins ago
Is this what you are trying?
Do not use InStr. Use Mid
For counter = 1 To Len(userInput)
If Mid(userInput, counter, 1) = letterSearched Then occurances = occurances + 1
Next
Note: If you want this not to be case sensitive then try this
For counter = 1 To Len(userInput)
If Mid(UCase(userInput), counter, 1) = UCase(letterSearched) Then occurances = occurances + 1
Next
Related
I'm trying to take a body of text and add line breaks around 80 characters on each line. The issue I'm having is on the last line it's adding an extra line break than would be desired. For instance this string should not have a line break on the second to last line:
Alice was beginning to get very tired of sitting by her sister on the bank, and
of having nothing to do: once or twice she had peeped into the book her sister
was reading, but it had no pictures or conversations in it, and what is the use
of a book, thought Alice without pictures or
conversations?
should look like this (note "conversations" has been moved up):
Alice was beginning to get very tired of sitting by her sister on the bank, and
of having nothing to do: once or twice she had peeped into the book her sister
was reading, but it had no pictures or conversations in it, and what is the use
of a book, thought Alice without pictures or conversations?
Here's the code:
Sub StringChop()
Dim OrigString As String
Dim NewString As String
Dim counter As Long
Dim length As Long
Dim LastSpace As Long
Dim LineBreak As Long
Dim TempString As String
Dim TempNum As Long
OrigString = "Alice was beginning to get very tired of sitting by her sister on the bank, and of having nothing to do: once or twice she had peeped into the book her sister was reading, but it had no pictures or conversations in it, and what is the use of a book, thought Alice without pictures or conversations?"
length = Len(OrigString)
counter = 1
Do While counter < length
'Extract next 80 characters from last position
TempString = Mid(OrigString, counter, 80)
'Determine last space in string
LastSpace = InStrRev(TempString, " ")
'Determine first line break in string
LineBreak = InStr(TempString, vbNewLine)
'If line break exists in sentence...
'only count characters up to line break, and set counter to that amount
Select Case LastSpace 'What to do if there are spaces in sentence
Case Is > 0 'There are spaces in sentence
Select Case LineBreak 'What to do if there are line breaks in sentence
Case Is = 0
'From last counter position,
NewString = NewString & Mid(OrigString, counter, LastSpace) & vbNewLine
counter = counter + LastSpace
Case Is <> 0
NewString = NewString & Mid(OrigString, counter, LineBreak)
counter = counter + LineBreak
End Select
Case Is = 0 'There are no more spaces left in remaining sentence
NewString = NewString & Mid(OrigString, counter)
counter = length
End Select
Loop
Debug.Print NewString
End Sub
Word wrapping is an interesting problem. I wrote the following code once as an experiment. You might find it helpful:
Option Explicit
'Implements a dynamic programming approach to word wrap
'assumes fixed-width font
'a word is defined to be a white-space delimited string which contains no
'whitespace
'the cost of a line is the square of the number of blank spaces at the end
'of a line
Const INFINITY As Long = 1000000
Dim optimalCost As Long
Function Cost(words As Variant, i As Long, j As Long, L As Long) As Long
'words is a 0-based array of strings, assumed to have no white spaces
'i, j are indices in range 0,...,n, where n is UBOUND(words)+1
'L is the maximum length of a line
'Cost returns the cost of a line which begins with words(i) and ends with
'words(j-1). It returns INFINITY if the line is too short to hold the words
'or if j <= i
Dim k As Long
Dim sum As Long
If j <= i Or Len(words(i)) > L Then
Cost = INFINITY
Exit Function
End If
sum = Len(words(i))
k = i + 1
Do While k < j And sum <= L
sum = sum + 1 + Len(words(k)) 'for space
k = k + 1
Loop
If sum > L Then
Cost = INFINITY
Else
Cost = (L - sum) ^ 2
End If
End Function
Function WordWrap(words As Variant, L As Long) As String
'returns string consisting of words with spaces and
'line breaks inserted at the appropriate places
Dim v() As Long, d() As Long
Dim n As Long
Dim i As Long, j As Long
Dim candidate As Long
n = UBound(words) + 1
ReDim v(0 To n)
ReDim d(0 To n)
v(0) = 0
d(0) = -1
For j = 1 To n
v(j) = INFINITY 'until something better is found
i = j - 1
Do
candidate = v(i) + Cost(words, i, j, L)
If candidate < v(j) Then
v(j) = candidate
d(j) = i
End If
i = i - 1
Loop While i >= 0 And candidate < INFINITY
If v(j) = INFINITY Then
MsgBox "Some words are too long for the given length"
Exit Function
End If
Next j
optimalCost = v(n)
'at this stage, optimal path has been found
'just need to follow d() backwards, inserting line breaks
i = d(n) 'beginning of current line
WordWrap = words(n - 1)
j = n - 2
Do While i >= 0
Do While j >= i
WordWrap = words(j) & " " & WordWrap
j = j - 1
Loop
If i > 0 Then WordWrap = vbCrLf & WordWrap
i = d(i)
Loop
End Function
The above function expects an array of words. You would have to split a string before using it as input:
Sub test()
Dim OrigString As String
OrigString = "Alice was beginning to get very tired of sitting by her sister on the bank, and of having nothing to do: once or twice she had peeped into the book her sister was reading, but it had no pictures or conversations in it, and what is the use of a book, thought Alice without pictures or conversations?"
Debug.Print WordWrap(Split(OrigString), 80)
End Sub
Output:
Alice was beginning to get very tired of sitting by her sister on the bank,
and of having nothing to do: once or twice she had peeped into the book
her sister was reading, but it had no pictures or conversations in it, and
what is the use of a book, thought Alice without pictures or conversations?
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
I need a code that could give me a list of unique combinations from a set of elements in an array, something like this:
Say myArray contains [A B C]
So, the output must be:
A
B
C
A B
A C
B C
A B C
or
A B C
B C
A C
A B
A
B
C
either output is OK for me (Starts with 1 combination, followed by 2 combinations and ends with all combination OR vice versa).
The position of the letters are not critical and the order of letters within the same combination type is also not critical.
I'd found a suggestion by 'Dick Kusleika' in a thread: Creating a list of all possible unique combinations from an array (using VBA) but when I tried, it did not present me with the arrangement that I wanted.
I'd also found a suggestion by 'pgc01' in a thread: http://www.mrexcel.com/forum/excel-questions/435865-excel-visual-basic-applications-combinations-permutations.html and it gave me the arrangement that I wanted however, the combinations was not being populated in an array but it was being populated in excel cells instead, using looping for each combination.
So, I wanted the arrangement of combinations to be like what 'pgc01' suggested and being populated in an array as what 'Dick Kusleika' presented.
Anyone can help? Appreciate it.
Start from here:
Sub TestRoutine()
Dim inputt() As String, i As Long
Dim outputt As Variant
inputt = Split("A B C", " ")
outputt = Split(ListSubsets(inputt), vbCrLf)
For i = LBound(outputt) + 2 To UBound(outputt)
MsgBox i & vbTab & outputt(i)
Next i
End Sub
Function ListSubsets(Items As Variant) As String
Dim CodeVector() As Long
Dim i As Long
Dim lower As Long, upper As Long
Dim SubList As String
Dim NewSub As String
Dim done As Boolean
Dim OddStep As Boolean
OddStep = True
lower = LBound(Items)
upper = UBound(Items)
ReDim CodeVector(lower To upper) 'it starts all 0
Do Until done
'Add a new subset according to current contents
'of CodeVector
NewSub = ""
For i = lower To upper
If CodeVector(i) = 1 Then
If NewSub = "" Then
NewSub = Items(i)
Else
NewSub = NewSub & " " & Items(i)
End If
End If
Next i
If NewSub = "" Then NewSub = "{}" 'empty set
SubList = SubList & vbCrLf & NewSub
'now update code vector
If OddStep Then
'just flip first bit
CodeVector(lower) = 1 - CodeVector(lower)
Else
'first locate first 1
i = lower
Do While CodeVector(i) <> 1
i = i + 1
Loop
'done if i = upper:
If i = upper Then
done = True
Else
'if not done then flip the *next* bit:
i = i + 1
CodeVector(i) = 1 - CodeVector(i)
End If
End If
OddStep = Not OddStep 'toggles between even and odd steps
Loop
ListSubsets = SubList
End Function
Note we discard the first two elements of the output array.
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 am having a sheet which contains range of values like "5670&&2","1281&&-3&-5&&7",... etc. in Column A.
Kindly help me to extract the output in VBA in following way:
For E.g 5670&&2 I require A1 cell contains 5670,B1 cell contains &&,C1 cell contains 2.
For E.g 1281&&-3&-5&&7,I would require that A1 cell contains 1281,B1 cell contains &&-,C1 cell contains 3,D1 cell contains &-,E1 cell contains 5,F1 cell contains && and G1 cell contains 7.
Pls help in the same .
Thanks.,
Here i have tried to write code to separate numbers from non-numbers. Numbers and non-numbers are copied to different columns, like Excel Text-To-Columns. Code is a little crazy, if u need i will provide comments. As input the ActiveSheet.UsedRange.Columns(1).Cells is used.
Option Explicit
Sub SeparateNumbers()
Dim targetRange As Range
Dim cellRange As Range
Dim charIndex As Integer
Dim oneChar As String
Dim nextChar As String
Dim start As Integer
Dim copiedCharsCount As Integer
Dim cellValue As String
Dim columnIndex As Integer
Set targetRange = ActiveSheet.UsedRange.Columns(1).Cells
For Each cellRange In targetRange
columnIndex = cellRange.Column
start = 1
copiedCharsCount = 0
cellValue = cellRange.Value
If (VBA.Strings.Len(cellValue) <= 1) Then GoTo nextCell
For charIndex = 2 To Len(cellValue)
oneChar = VBA.Strings.Mid(cellValue, charIndex - 1, 1)
nextChar = VBA.Strings.Mid(cellValue, charIndex, 1)
If VBA.IsNumeric(oneChar) And VBA.IsNumeric(nextChar) Then GoTo nextCharLabel
If Not VBA.IsNumeric(oneChar) And Not VBA.IsNumeric(nextChar) Then GoTo nextCharLabel
cellRange.Offset(0, columnIndex).Value = VBA.Strings.Mid(cellValue, start, charIndex - start)
columnIndex = columnIndex + 1
copiedCharsCount = copiedCharsCount + (charIndex - start)
start = charIndex
nextCharLabel:
If charIndex = Len(cellValue) Then
cellRange.Offset(0, columnIndex).Value = VBA.Strings.Right(cellValue, charIndex - copiedCharsCount)
End If
Next charIndex
nextCell:
Next cellRange
End Sub
Here is one more code. As a side product, function TextSplitToNumbersAndOther can be used independently as a formula to achieve the same effect.
To prevent accidental firing of the macro in a wrong sheet or a wrong column and overwriting neighbouring columns with scrap, named range "Start_point" should be defined by a user. Below this range in the same column, all data will be processed till the first blank row.
Spreadsheet example: http://www.bumpclub.ee/~jyri_r/Excel/Extracting_symbols_into_columns.xls
Option Explicit
Sub ExtractSymbolsIntoColumns()
Dim rng As Range
Dim row_processed As Integer
Dim string_to_split As String
Dim columns_needed As Long
Dim counter As Long
row_processed = 1
counter = 0
Set rng = Range("Start_point")
While rng.Offset(row_processed, 0).Value <> ""
string_to_split = rng.Offset(row_processed, 0).Value
columns_needed = TextSplitToNumbersAndOther(string_to_split)
For counter = 1 To columns_needed
rng.Offset(row_processed, counter).Value = _
TextSplitToNumbersAndOther(string_to_split, counter)
Next
row_processed = row_processed + 1
Wend
End Sub
Function TextSplitToNumbersAndOther(InputText As String, _
Optional SplitPieceNumber As Long) As Variant
Dim piece_from_split(100) As Variant
Dim char_from_input As String
Dim word_count As Long
Dim counter As Long
Dim char_type(100) As Variant
InputText = Trim(InputText)
If Not IsNull(InputText) Then
word_count = 1
piece_from_split(word_count) = ""
For counter = 1 To Len(InputText)
char_from_input = CharFromTextPosition(InputText, counter)
char_type(counter) = CharTypeAsNumber(char_from_input)
If counter = 1 Then
piece_from_split(word_count) = char_from_input
Else
If (char_type(counter - 1) = char_type(counter)) Then
piece_from_split(word_count) = piece_from_split(word_count) & char_from_input
'Merge for the same type
Else
word_count = word_count + 1
piece_from_split(word_count) = char_from_input
End If
End If
Next
End If
If SplitPieceNumber = 0 Then
TextSplitToNumbersAndOther = word_count
Else
If SplitPieceNumber > word_count Then
TextSplitToNumbersAndOther = ""
Else
TextSplitToNumbersAndOther = piece_from_split(SplitPieceNumber)
End If
End If
End Function
Function CharTypeAsNumber(InputChar As String, Optional PositionInString As Long) As Long
If PositionInString = 0 Then PositionInString = 1
If Not IsNull(InputChar) Then
InputChar = Mid(InputChar, PositionInString, 1)
Select Case InputChar
Case 0 To 9
CharTypeAsNumber = 1
Case "a" To "z"
CharTypeAsNumber = 2
Case "A" To "Z"
CharTypeAsNumber = 3
Case Else
CharTypeAsNumber = 4
End Select
Else
CharTypeAsNumber = 0
End If
End Function
Function CharFromTextPosition(InputString As String, TextPosition As Long) As String
CharFromTextPosition = Mid(InputString, TextPosition, 1)
End Function
You can write a UDF (user defined function) to achieve the objective.
Your two example are in an order (ascending) to filter out into adjacent columns in Excel (A, B, C, D...)
So is it correct to assume logically, that you will never have scenarios where you will have to break the string into non-adjacent columns? e.g. 1234 goes to A, && goes to C, 3 goes to D... resulting in A, C, D.
Asumption 2: That your splitted-string is not going to need columns more than Excel can provide.
Steps you may try:
1. Check your string is not empty
2. Split it by the characters other than numerics
3. At the start and end of each non-numeric character you may proceed to the next adjacent column.
search help: Split a string into multiple columns in Excel - VBA