I need to get a certain character from a long string line, that occurs more than once. This is what the string looks like:
<Press T><Press Left><Press A><Press C><Press P><Press U><Press G><Press P><Press NumPad7><Press NumPad7><Press A>
I need to loop through each set of <>'s and get the info that is after each occurrence of the word Press. So in this case I would need the info T, Left, A, C, P, etc
I think this pure String-method approach is the most efficient, but it requires the format to be strict:
Dim text = "<Press T><Press Left><Press A><Press C><Press P><Press U><Press G><Press P><Press NumPad7><Press NumPad7><Press A>"
Dim allKeys As New List(Of String)
Dim pattern = "Press "
Dim index = text.IndexOf(pattern)
While index >= 0
index += pattern.Length
Dim endIndex = text.IndexOf(">", index)
If endIndex >= 0 Then
Dim nextKey = text.Substring(index, endIndex - index)
allKeys.Add(nextKey)
index = text.IndexOf(pattern, endIndex + 1)
Else
Exit While
End If
End While
Console.Write(String.Join(", ", allKeys))
Output: T, Left, A, C, P, U, G, P, NumPad7, NumPad7, A
Here is the Regex that returns all matches, you find the "key" that was pressed in the second group:
pattern = "<Press ([^>]+)>"
Dim regex = New Regex( pattern, RegexOptions.Compiled And RegexOptions.IgnoreCase)
For Each match As Match In regex.Matches(text)
Console.WriteLine(match.Groups(1))
Next
Assuming it's the same format all the time you can do this in one line. Regex would be better if there's a chance the format would be different (e.g. random number of whitespaces etc..)
Dim myString As String = "<Press T><Press Left><Press A><Press C><Press P><Press U><Press G><Press P><Press NumPad7><Press NumPad7><Press A>"
Dim character As String() = myString.Split(New String() {"<Press ", ">"}, StringSplitOptions.RemoveEmptyEntries)
RegEx, short for regular expressions, is an easy way to parse strings. This site Provides good information on using RegEx in .NET. It's how I learned. The site also provides good info on RegEx in general if you are not familiar.
Edit: RegEx expressions can be complicated to create. A great tool to help you out with that is Expresso. It'll help you create and test very complicated expressions with a minimal of fuss.
Dim s As String = "<Press T><Press Left><Press A><Press C><Press P><Press U><Press G><Press P><Press NumPad7><Press NumPad7><Press A>"
Dim ss() As String = s.Replace("<Press ", "").Split(">"c)
For i as integer = 0 to ss.count - 2
Debug.Print(ss(i))
Next
Output:
T
Left
A
C
P
U
G
P
NumPad7
NumPad7
A
Note that the array ss is one longer than the number of key presses due to the final ">" being treated as another separator by .split, you could always remove the final ">"
Related
I'm a programing student, so I've started with vb.net as my first language and I need some help.
I need to know how I delete excess white spaces between words in a sentence, only using these string functions: Trim, instr, char, mid, val and len.
I made a part of the code but it doesn't work, Thanks.
enter image description here
Knocked up a quick routine for you.
Public Function RemoveMyExcessSpaces(str As String) As String
Dim r As String = ""
If str IsNot Nothing AndAlso Len(str) > 0 Then
Dim spacefound As Boolean = False
For i As Integer = 1 To Len(str)
If Mid(str, i, 1) = " " Then
If Not spacefound Then
spacefound = True
End If
Else
If spacefound Then
spacefound = False
r += " "
End If
r += Mid(str, i, 1)
End If
Next
End If
Return r
End Function
I think it meets your criteria.
Hope that helps.
Unless using those VB6 methods is a requirement, here's a one-line solution:
TextBox2.Text = String.Join(" ", TextBox1.Text.Split(New Char() {" "c}, StringSplitOptions.RemoveEmptyEntries))
Online test: http://ideone.com/gBbi55
String.Split() splits a string on a specific character or substring (in this case a space) and creates an array of the string parts in-between. I.e: "Hello There" -> {"Hello", "There"}
StringSplitOptions.RemoveEmptyEntries removes any empty strings from the resulting split array. Double spaces will create empty strings when split, thus you'll get rid of them using this option.
String.Join() will create a string from an array and separate each array entry with the specified string (in this case a single space).
There is a very simple answer to this question, there is a string method that allows you to remove those "White Spaces" within a string.
Dim text_with_white_spaces as string = "Hey There!"
Dim text_without_white_spaces as string = text_with_white_spaces.Replace(" ", "")
'text_without_white_spaces should be equal to "HeyThere!"
Hope it helped!
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
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
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.