I have a user-defined function in an Excel VBA module. Let's say it's:
Function MyFunc( val1 as Double, val2 as int, val3 as int ) as Double
MyFunc = val1 * val2 + val3
End Function
I also have another Function MyEval:
Function MyEval( formula as string )
MyEval = Evaluate( formula )
End Function
In a worksheet, the cell B1 contains the following string:
"blah blah blah "&MyFunc(VAL,2,0)&" foobar "&MyFunc(VAL,6,5)&" do re mi"
Note that the quotes are part of the string, not just denoting that it is a string. In second cell, I have the following:
=MyEval(SUBSTITUTE(B1,"VAL",8))
In general this works. In the above case I would get, as expected, the string:
blah blah blah 16 foobar 53 do re mi
In some cases, however, I would get #VALUE!. When I examined those examples that failed, I would find that removing different parts would allow the string to evaluate, but that no specific part caused it to fail, i.e., just the first half works, just the second half works, and just the middle part works.
Thinking that it might be a length issue (strings are up to 200 characters), I wrote a new evaluation function, where I break the string into parts on the concatenate symbol (&), evaluate each part separately, and then concatenate the results together:
Function MyEval2(formula As String)
If Len(formula) = 0 Then
Eval2 = ""
Else
Dim parts() As String
parts = Split(formula, "&")
Dim result As String
result = ""
Dim index As Integer
Dim part As String
For index = LBound(parts) To UBound(parts)
part = parts(index)
If (Left(part, 1) = """") And (Right(part, 1) = """") Then
part = Left(part, Len(part) - 1)
part = Right(part, Len(part) - 1)
result = result & part
Else
Dim val As Variant
val = Evaluate(part)
result = result & val
End If
Next index
Eval2 = result
End If
End Function
In this case, sometimes when the variable 'part' was, for example, the string
MyFunc(8,6,5)
the line
val = Evaluate(part)
resulted in 'val' being the string
MyFunc(8,6,5)
instead of the value 53.
Note: I have extensively re-written this to make it clearer.
Related
I'm using this query in vb.net
Raw_data = Alltext_line.Substring(Alltext_line.IndexOf("R|1"))
and I want to increase R|1 to R|2, R|3 and so on using for loop.
I tried it many ways but getting error
string to double is invalid
any help will be appreciated
You must first extract the number from the string. If the text part ("R") is always separated from the number part by a "|", you can easily separated the two with Split:
Dim Alltext_line = "R|1"
Dim parts = Alltext_line.Split("|"c)
parts is a string array. If this results in two parts, the string has the expected shape and we can try to convert the second part to a number, increase it and then re-create the string using the increased number
Dim n As Integer
If parts.Length = 2 AndAlso Integer.TryParse(parts(1), n) Then
Alltext_line = parts(0) & "|" & (n + 1)
End If
Note that the c in "|"c denotes a Char constant in VB.
An alternate solution that takes advantage of the String type defined as an Array of Chars.
I'm using string.Concat() to patch together the resulting IEnumerable(Of Char) and CInt() to convert the string to an Integer and sum 1 to its value.
Raw_data = "R|151"
Dim Result As String = Raw_data.Substring(0, 2) & (CInt(String.Concat(Raw_data.Skip(2))) + 1).ToString
This, of course, supposes that the source string is directly convertible to an Integer type.
If a value check is instead required, you can use Integer.TryParse() to perform the validation:
Dim ValuePart As String = Raw_data.Substring(2)
Dim Value As Integer = 0
If Integer.TryParse(ValuePart, Value) Then
Raw_data = Raw_data.Substring(0, 2) & (Value + 1).ToString
End If
If the left part can be variable (in size or content), the answer provided by Olivier Jacot-Descombes is covering this scenario already.
Sub IncrVal()
Dim s = "R|1"
For x% = 1 To 10
s = Regex.Replace(s, "[0-9]+", Function(m) Integer.Parse(m.Value) + 1)
Next
End Sub
What I want to do is replace all 'A' in a string with "Bb". but it will only loop with the original string not on the new string.
for example:
AAA
BbAA
BbBbA
and it stops there because the original string only has a length of 3. it reads only up to the 3rd index and not the rest.
Dim txt As String
txt = output_text.Text
Dim a As String = a_equi.Text
Dim index As Integer = txt.Length - 1
Dim output As String = ""
For i = 0 To index
If (txt(i) = TextBox1.Text) Then
output = txt.Remove(i, 1).Insert(i, a)
txt = output
TextBox2.Text += txt + Environment.NewLine
End If
Next
End Sub
I think this leaves us looking for a String.ReplaceFirst function. Since there isn't one, we can just write that function. Then the code that calls it becomes much more readable because it's quickly apparent what it's doing (from the name of the function.)
Public Function ReplaceFirst(searched As String, target As String, replacement As String) As String
'This input validation is just for completeness.
'It's not strictly necessary.
'If the searched string is "null", throw an exception.
If (searched Is Nothing) Then Throw New ArgumentNullException("searched")
'If the target string is "null", throw an exception.
If (target Is Nothing) Then Throw New ArgumentNullException("target")
'If the searched string doesn't contain the target string at all
'then just return it - were done.
Dim foundIndex As Integer = searched.IndexOf(target)
If (foundIndex = -1) Then Return searched
'Build a new string that replaces the target with the replacement.
Return String.Concat(searched.Substring(0, foundIndex), replacement, _
searched.Substring(foundIndex + target.Length, searched.Length - (foundIndex + target.Length)))
End Function
Notice how when you read the code below, you don't even have to spend a moment trying to figure out what it's doing. It's readable. While the input string contains "A", replace the first "A" with "Bb".
Dim input as string = "AAA"
While input.IndexOf("A") > -1
input = input.ReplaceFirst(input,"A","Bb")
'If you need to capture individual values of "input" as it changes
'add them to a list.
End While
You could optimize or completely replace the function. What matters is that your code is readable, someone can tell what it's doing, and the ReplaceFirst function is testable.
Then, let's say you wanted another function that gave you all of the "versions" of your input string as the target string is replaced:
Public Function GetIterativeReplacements(searched As String, target As String, replacement As String) As List(of string)
Dim output As New List(Of String)
While searched.IndexOf(target) > -1
searched = ReplaceFirst(searched, target, replacement)
output.Add(searched)
End While
Return output
End Function
If you call
dim output as List(of string) = GetIterativeReplacments("AAAA","A","Bb")
It's going to return a list of strings containing
BbAAA, BbBbAA, BbBbBbA, BbBbBbBb
It's almost always good to keep methods short. If they start to get too long, just break them into smaller methods with clear names. That way you're not trying to read and follow and test one big, long function. That's difficult whether or not you're a new programmer. The trick isn't being able to create long, complex functions that we understand because we wrote them - it's creating small, simpler functions that anyone can understand.
Check your comments for a better solution, but for future reference you should use a while loop instead of a for loop if your condition will be changing and you're wanting to take that change into account.
I've made a simple example below to help you understand. If you tried the same with a for loop, you'd only get "one" "two" and "three" printed because the for loop doesn't 'see' that vals was changed
Dim vals As New List(Of String)
vals.Add("one")
vals.Add("two")
vals.Add("three")
Dim i As Integer = 0
While i < vals.Count
Console.WriteLine(vals(i))
If vals(i) = "two" Then
vals.Add("four")
vals.Add("five")
End If
i += 1
End While
If you do want to replace one by one instead of using the Replace function, you could use a while loop to look for the index of your search character/string, and then replace/insert at that index.
Sub Main()
Dim a As String = String.Empty
Dim b As String = String.Empty
Dim c As String = String.Empty
Dim d As Int32 = -1
Console.Write("Whole string: ")
a = Console.ReadLine()
Console.Write("Replace: ")
b = Console.ReadLine()
Console.Write("Replace with: ")
c = Console.ReadLine()
d = a.IndexOf(b)
While d > -1
a = a.Remove(d, b.Length)
a = a.Insert(d, c)
d = a.LastIndexOf(b)
End While
Console.WriteLine("Finished string: " & a)
Console.ReadLine()
End Sub
Output would look like this:
Whole string: This is A string for replAcing chArActers.
Replace: A
Replace with: Bb
Finished string: This is Bb string for replBbcing chBbrBbcters.
I was going to write a while loop to answer your question, but realized (with assistance from others) that you could just .replace(x,y)
Output.Text = Input.Text.Replace("A", "Bb")
'Input = N A T O
'Output = N Bb T O
Edit: There is probably a better alternative, but i quickly jotted this loop down, hope it helps.
You've said your new and don't fully understand while loops. So if you don't understand functions either or how to pass arguments to them, I'd suggest looking that up too.
This is your Event, It can be a Button click or Textbox text change.
'Cut & Paste into an Event (Change textboxes to whatever you have input/output)
Dim Input As String = textbox1.Text
Do While Input.Contains("A")
Input = ChangeString(Input, "A", "Bb")
' Do whatever you like with each return of ChangeString() here
Loop
textbox2.Text = Input
This is your Function, with 3 Arguments and a Return Value that can be called in your code
' Cut & Paste into Code somewhere (not inside another sub/Function)
Private Function ChangeString(Input As String, LookFor As Char, ReplaceWith As String)
Dim Output As String = Nothing
Dim cFlag As Boolean = False
For i As Integer = 0 To Input.Length - 1
Dim c As Char = Input(i)
If (c = LookFor) AndAlso (cFlag = False) Then
Output += ReplaceWith
cFlag = True
Else
Output += c
End If
Next
Console.WriteLine("Output: " & Output)
Return Output
End Function
I am building a tool for Autodesk Inventor that works with an expression string.
=Pipe Ø<Pipe_OD> x <Pipe_t> - lg. <Pipe_length>mm
The text between <...> can be almost anything it's what the user made as input so these values are not fixed. And the number of <...> in the string can vary from 0 to 5.
What I would like to have as result for this string is following:
A converted string where the values between the <...> are replaced by number with an ascending value.
=Pipe Ø<1> x <2> - lg. <3>mm
And a string() where the values that are replaced by the numbers (above) are stored in.
I found a method that can work for strings with 1 <...> in the string but now that the quantity is variable I'm clueless about how I can do this.
Link to method
Edit: New answer with regular expressions (much better, will still have problems with additional < and > though)
Dim s As String = "abc <pipe_val1> 123 <pipe_val2> &*( <pipe_val3>k"
Dim myValues As New List(Of String)
Dim matches As MatchCollection = Regex.Matches(s, "<(.|\n)*?>", RegexOptions.IgnoreCase)
Dim totalDiff As Integer = 0
Dim idx As Integer = 0
For Each ma As Match In matches
Dim realIndex = ma.Index - totalDiff
s = s.Remove(realIndex, ma.Length).Insert(realIndex, "<" & idx.ToString & ">")
idx += 1
totalDiff += ma.Length - (idx.ToString.Count + 2)
myValues.Add(ma.Value.Trim({"<"c, ">"c}))
Next
"abc <pipe_val1> 123 <pipe_val2> &*( <pipe_val3>" will be changed to "abc <0> 123 <1> &*( <2>" and myValues will contain "pipe_val1", "pipe_val2" and "pipe_val3"
Suppose I need to add the ASCII version of each character in the word "hello" to "hi" so that the result would be something like this: (h+h = )(e+i = )(l+h = )(l+i = )(o+h = ) etc how would I go about looping the "hi" string?
I have already managed to loop the "hello" string, but not quite sure how to do the second without getting (h+h = )(h+i = )(e+h = )(e+i = ) etc.
Thanks!
You can use the Mod opreator to make the index start over. Example:
Dim str1 as String = "hello"
Dim str2 as String = "hi"
' This gets the length of the longest string
Dim longest = Math.Max(str1.Length, str2.Length)
' This loops though all characters
' The Mod operator makes the index wrap over for the shorter string
For i As Integer = 0 To longest - 1
Console.Write(str1(i Mod str1.Length))
Console.WriteLine(str2(i Mod str2.Length))
Next
Output:
hh
ei
lh
li
oh
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