I am trying to create an application which will determine whether a string entered by user is a palindrome or not.
Is it possible to do without StrReverse, possibly with for next loop. That's what i have done so far.
Working one, with StrReverse:
Dim userInput As String = Me.txtbx1.Text.Trim.Replace(" ", "")
Dim toBeComparedWith As String = StrReverse(userInput)
Select Case String.Compare(userInput, toBeComparedWith, True)
Case 0
Me.lbl2.Text = "The following string is a palindrom"
Case Else
Me.lbl2.Text = "The following string is not a palindrom"
End Select
Not working one:
Dim input As String = TextBox1.Text.Trim.Replace(" ", "")
Dim pallindromeChecker As String = input
Dim output As String
For counter As Integer = input To pallindromeChecker Step -1
output = pallindromeChecker
Next counter
output = pallindromeChecker
If output = input Then
Me.Label1.Text = "output"
Else
Me.Label1.Text = "hi"
End If
While using string reversal works, it is suboptimal because you're iterating over the string at least 2 full times (as string reversal creates a copy of a string because strings are immutable in .NET) (plus extra iterations for your Trim and Replace calls).
However consider the essential properties of a palindrome: the first half of a string is equal to the second half of the string in reverse.
The optimal algorithm for checking a palindrome needs only iterate through half of the input string - by comparing value[n] with value[length-n] for n = 0 to length/2.
In VB.NET:
Public Shared Function IsPalindrome(value As String) As Boolean
' Input validation.
If value Is Nothing Then Throw New ArgumentNullException("value")
value = value.Replace(" ", "") // Note String.Replace(String,String) runs in O(n) time and if replacement is necessary then O(n) space.
' Shortcut case if the input string is empty.
If value.Length = 0 Then Return False ' or True, depends on your preference
' Only need to iterate until half of the string length.
' Note that integer division results in a truncated value, e.g. (5 / 2 = 2)...
'... so this ignores the middle character if the string is an odd-number of characters long.
Dim max As Integer = value.Length - 1
For i As Integer = 0 To value.Length / 2
If value(i) <> value(max-i) Then
' Shortcut: we can abort on the first mismatched character we encounter, no need to check further.
Return False
End If
Next i
' All "opposite" characters are equal, so return True.
Return True
End Function
How can i check for a character after other text within a listbox?
e.g
Listbox contents:
Key1: V
Key2: F
Key3: S
Key4: H
How do I find what comes after Key1-4:?
Key1-4 will always be the same however what comes after that will be user defined.
I figured out how to save checkboxes as theres only 2 values to choose from, although user defined textboxes is what im struggling with. (I have searched for solutions but none seemed to work for me)
Usage:
Form1_Load
If ListBox1.Items.Contains("Key1: " & UsersKey) Then
TextBox1.Text = UsersKey
End If
Which textbox1.text would then contain V / whatever the user defined.
I did try something that kind of worked:
Form1_Load
Dim UsersKey as string = "V"
If ListBox1.Items.Contains("Key1: " & UsersKey) Then
TextBox1.Text = UsersKey
End If
but i'm not sure how to add additional letters / numbers to "V", then output that specific number/letter to the textbox. (I have special characters blocked)
Reasoning I need this is because I have created a custom save settings which saves on exit and loads with form1 as the built in save settings doesn't have much customization.
e.g Can't choose save path, when filename is changed a new user.config is generated along with old settings lost.
Look at regular expressions for this.
Using the keys from your sample:
Dim keys As String = "VFSH"
Dim exp As New RegEx("Key[1-4]: ([" & keys& "])")
For Each item As String in ListBox1.Items
Dim result = exp.Match(item)
If result.Success Then
TextBox1.Text = result.Groups(1).Value
End If
Next
It's not clear to me how your ListBoxes work. If you might find, for example, "Key 2:" inside ListBox1 that you need to ignore, you will want to change the [1-4] part of the expression to be more specific.
Additionally, if you're just trying to exclude unicode or punctuation, you could also go with ranges:
Dim keys As String = "A-Za-z0-9"
If you are supporting a broader set of characters, there are some you must be careful with: ], \, ^, and - can all have special meanings inside of a regular expression character class.
You have multiple keys, I assume you have multiple textboxes to display the results?
Then something like this would work. Loop thru the total number of keys, inside that you loop thru the alphabet. When you find a match, output to the correct textbox:
Dim UsersKey As String
For i As Integer = 1 To 4
For Each c In "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray()
UsersKey = c
If ListBox1.Items.Contains("Key" & i & ": " & UsersKey) Then
Select Case i
Case 1
TextBox1.Text = UsersKey
Case 2
TextBox2.Text = UsersKey
Case 3
TextBox3.Text = UsersKey
Case 4
TextBox4.Text = UsersKey
End Select
Exit For 'match found so exit inner loop
End If
Next
Next
Also, you say your settings are lost when the filename is changed. I assume when the version changes? The Settings has an upgrade method to read from a previous version. If you add an UpgradeSettings boolean option and set it to True and then do this at the start of your app, it will load the settings from a previous version:
If My.Settings.UpgradeSettings = True Then
My.Settings.Upgrade()
My.Settings.Reload()
My.Settings.UpgradeSettings = False
My.Settings.Save()
End If
Updated Answer:
Instead of using a listtbox, read the settings file line by line and output the results to the correct textbox based on the key...something like this:
Dim settingsFile As String = "C:\settings.txt"
If IO.File.Exists(settingsFile) Then
For Each line As String In IO.File.ReadLines(settingsFile)
Dim params() As String = Split(line, ":")
If params.Length = 2 Then
params(0) = params(0).Trim
params(1) = params(1).Trim
Select Case params(0)
Case "Key1"
Textbox1.Text = params(1)
Case "Key2"
Textbox2.Text = params(1)
End Select
End If
Next line
End If
You can associate text box with a key via its Name or Tag property. Lets say you use Name. In this case TextBox2 is associated with key2. TextBox[N] <-> Key[N]
Using this principle the code will look like this [considering that your list item is string]
Sub Test()
If ListBox1.SelectedIndex = -1 Then Return
Dim data[] As String = DirectCast(ListBox1.SelectedItem, string).Split(new char(){":"})
Dim key As String = data(0).Substring(3)
Dim val As String = data(1).Trim()
' you can use one of the known techniques to get control on which your texbox sits.
' I omit this step and assume "Surface1" being a control on which your text boxes sit
DirectCast(
(From ctrl In Surface1.Controls
Where ctrl.Name = "TextBox" & key
Select ctrl).First()), TextBox).Text = val
End Sub
As you can see, using principle I just explained, you have little parsing and what is important, there is no growing Select case if, lets say, you get 20 text boxes. You can add as many text boxes and as many corresponding list items as you wish, the code need not change.
[{"name":"chrisjj","uuid":"d086112c-6e25-31a0-acf0-f95c3ca98784","expiresOn":"2016-02-22 23:04:35 +0000"}]
[{"name":"ben","uuid":"d086112c-7a26-33b5-ucf3-j96c1ca26854","expiresOn":"2015-011-12 22:04:35 +0000"}]
Basically im working on a project for a while now and I am trying to keep the names chrisjj and ben and removing the rest of the text from textbox in visual basic 2012 if you have any idea that would be great help
You may use regex to achieve what you want.
Dim Input As String = RichTextBox1.Text
Dim MC As MatchCollection = Regex.Matches(Input, Regex.Escape("[{""name"":""") & "[chrisjj|ben].*?" & Regex.Escape("]"), RegexOptions.IgnoreCase)
Dim Output As New List(Of String)
For i = 0 To MC.Count - 1
Output.Add(MC(i).Value)
Next
MsgBox(String.Join(vbNewLine, Output.ToArray()))
I think this is what you want. this regex matches [{"name":" then chrisjj or ben and goes on until ] is found.
You can do this:
If InStr(Textbox1.Text, "chrisjj") Then
Textbox1.text = "chrisjj"
else if InStr(Textbox1.Text, "ben") Then
Textbox1.text = "ben"
end if
The InStr Function returns the position of the first occurrence of one string within another.
Also
if TextBox1.Text.Contains("chrisjj") Then
TextBox1.Text = TextBox1.Text = "chrisjj"
ElseIf TextBox2.Text.Contains("ben") Then
TextBox1.Text = TextBox1.Text = ben
end if
The String.Contains Method (String) returns a value indicating whether a specified substring occurs within this string.
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