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.
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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!
Related
I'm trying to convert WText into its ASCII code and put it into a TextBox; Numencrypt. But I don't want to convert the spaces into ASCII code.
How do I replace the spaces with null?
Current code:
Dim withSpace As String = Numencrypt.Text
For h = 1 To lenText
wASC = wASC & CStr(Asc(Mid$(WText, h, 1)))
Next h
Numencrypt.Text = wASC
Numencrypt2.Text = Numencrypt2.Replace(Numencrypt.Text, " ", "")
By the way, the TextBox Numencrypt2 is the WText without a space inside it.
Without knowing whether or not you want the null character or empty string I did the following in a console app so I don't have your variables. I also used a string builder to make the string concatenation more performant.
Dim withSpaces = "This has some spaces in it!"
withSpaces = withSpaces.Replace(" "c, ControlChars.NullChar)
Dim wASC As New StringBuilder
For h = 1 To withSpaces.Length
wASC.Append($"{AscW(Mid(withSpaces, h, 1))} ") ' Added a space so you can see the boundaries ascii code boundaries.
Next
Dim theResult = wASC.ToString()
Console.WriteLine(theResult)
You will find that if you use ControlChars.NewLine as I have, the place you had spaces will be represented by a zero. That position is completely ignored if you use Replace(" ", "")
'How to reverse "This is Friday" to "Friday is this" in vb.net in Easiest Way
Dim str As String = txtremarks.Text
Dim arr As New List(Of Char)
arr.AddRange(str.ToCharArray)
arr.Reverse()
Dim a As String = ""
For Each l As Char In arr
a &= l
Next
' I saw on a few forums that to use SPLIT function. Please help
Yes you can use split. You can also use join and the reverse method:
Dim test = "This is Friday"
Dim reversetest = String.Join(" ", test.Split().Reverse)
First you'll want to split your sentence into individual words. This is where you'd use the String.Split method.
Once you have an array containing your individual words, you can reverse that array. Perhaps using Linq's Enumerable.Reverse extension method.
Finally, you can put the words back together into a string. The String.Join method allows you to join the elements of a string array back into a single string.
I'm not a VB programmer, but something like this should work:
Dim str As String = "this is friday"
Dim split As String() = str.Split(" ")
Dim result as String = String.Join(" ", split.Reverse())
Here's a way to do it in 1 line:
Dim reverse As String = "This is friday".Split().Reverse().Aggregate(Function(left, right) String.Join(" ", left, right))
Do note that this has a horrible performance overhead.
Yes, you can Split your String via " " (space) and insert results into array.
Next, read array from the end to start.
Good luck!
Try this...
Dim txt As String = "This is friday"
Dim txtarray() As String = Split(txt.Trim(), " ")
Dim result As String = ""
For x = txtarray.GetUpperBound(0) To 0 Step -1
result += txtarray(x) & " "
Next x
MsgBox(result.Trim())
I am trying to take a string in a rich text box and replace them with a different string.
Now how this should work is that if two same characters are entered into the text box
e.g tt the "tt" will be replaced with "Ǿt" , it adds back one of the t's to the replaced string. Only the most recently entered string is manipulated if two same characters are entered .
I read the LAST string that is in the RichTextBox by using this method
Dim laststring As String = RichTextBox1.Text.Split(" ").Last
'hitting space bar breaks the operation so if i enter t t there will be no replacement
this is the replacement method which I use , it works correctly .
if laststring = "tt"
RichTextBox1 .Text = RichTextBox1 .Text.Replace("tt", "Ǿt")
This method is inefficient because i need to check id there are double letters for all letters and if i was to use this method it would tavke up a lot of code .
how can I accomplish this using a shorter method??
You need to put the if then section in a loop.
Dim holdstring As String
Dim doubleinstance() As String = {"bb", "tt", "uu"} ' array
Dim curstring As String = RichTextBox1.Text.Split(" ").Last
For Each item As String In doubleinstance
If RichTextBox1.Text.EndsWith(item) Then
holdstring = RichTextBox1.Text.Split(" ").Last.Length - 1 ' change to subtract 1 character from doubleinstance
RichTextBox1.Text = RichTextBox1.Text.Replace(curstring, "Ǿt" & holdstring)
MsgBox(curstring)
End If
Next item
Here's a bit of code to get you in the right direction...
There are a couple of variations of .Find, but you probably want to look at the .Select method.
With RichTextBox1
.Find("Don")
.SelectedText = "Mr. Awesome"
End With
Here is a way I came up with
Dim holdstring As String
Dim doubleinstance() As String = {"bb", "tt", "uu"} ' array
Dim curstring As String = RichTextBox1.Text.Split(" ").Last
If curstring = doubleinstance(0) And RichTextBox1.Text.EndsWith(doubleinstance(0)) Then
holdstring = RichTextBox1.Text.Split(" ").Last.Length - 1 ' change to subtract 1 character from doubleinstance
RichTextBox1.Text = RichTextBox1.Text.Replace(curstring, "Ǿt" + holdstring)
MsgBox(curstring)
End If
where i have doubleinstance(0) how do i get the if statement to not only check a single index but all of the index from 0 to 2 in this example ?
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 writing a program in Visual Basic 2010 that lists how many times a word of each length occurs in a user-inputted string. Although most of the program is working, I have one problem:
When looping through all of the characters in the string, the program checks whether there is a next character (such that the program does not attempt to loop through characters that do not exist). For example, I use the condition:
If letter = Microsoft.VisualBasic.Right(input, 1) Then
Where letter is the character, input is the string, and Microsoft.VisualBasic.Right(input, 1) extracts the rightmost character from the string. Thus, if letter is the rightmost character, the program will cease to loop through the string.
This is where the problems comes in. Let us say the string is This sentence has five words. The rightmost character is an s, but an s is also the fourth and sixth character. That means that the first and second s will break the loop just as the others will.
My questions is whether there is a way to ensure that only the last s, or whatever character is the last one in the string can break the loop.
There are a few methods you can use for this, one as Neolisk shows; here are a couple of others:
Dim breakChar As Char = "s"
Dim str As String = "This sentence has five words"
str = str.Replace(".", " ")
str = str.Replace(",", " ")
str = str.Replace(vbTab, " ")
' other chars to replace
Dim words() As String = str.ToLower.Split(New Char() {" "}, StringSplitOptions.RemoveEmptyEntries)
For Each word In words
If word.StartsWith(breakChar) Then Exit For
Console.WriteLine("M1 Word: ""{0}"" Length: {1:N0}", word, word.Length)
Next
If you need to loop though chars for whatever reason, you can use something like this:
Dim breakChar As Char = "s"
Dim str As String = "This sentence has five words"
str = str.Replace(".", " ")
str = str.Replace(",", " ")
str = str.Replace(vbTab, " ")
' other chars to replace
'method 2
Dim word As New StringBuilder
Dim words As New List(Of String)
For Each c As Char In str.ToLower.Trim
If c = " "c Then
If word.Length > 0 'support multiple white-spaces (double-space etc.)
Console.WriteLine("M2 Word: ""{0}"" Length: {1:N0}", word.ToString, word.ToString.Length)
words.Add(word.ToString)
word.Clear()
End If
Else
If word.Length = 0 And c = breakChar Then Exit For
word.Append(c)
End If
Next
If word.Length > 0 Then
words.Add(word.ToString)
Console.WriteLine("M2 Word: ""{0}"" Length: {1:N0}", word.ToString, word.ToString.Length)
End If
I wrote these specifically to break on the first letter in a word as you ask, adjust as needed.
VB.NET code to calculate how many times a word of each length occurs in a user-inputted string:
Dim sentence As String = "This sentence has five words"
Dim words() As String = sentence.Split(" ")
Dim v = From word As String In words Group By L = word.Length Into Group Order By L
Line 2 may need to be adjusted to remove punctuation characters, trim extra spaces etc.
In the above example, v(i) contains word length, and v(i).Group.Count contains how many words of this length were encountered. For debugging purposes, you also have v(i).Group, which is an array of String, containing all words belonging to this group.