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
Related
Function ChewQuery(QueryList As String)
'This function takes in a string and partitions it into a weight tree
Dim ElementList() As String
ElementList = QueryList.Split(";")
Dim LoadedWeight(ElementList.Length) As WeightElement
For x = 0 To ElementList.Length - 1
'Using this method, in which the front portion of the query is repeatedly removed, allows for a simpler query structure, so that we don't need to partition between the pointers and values
LoadedWeight(x).LowPointer = Int(BiteQuery(ElementList(x)))
LoadedWeight(x).HighPointer = Int(BiteQuery(ElementList(x)))
LoadedWeight(x).TraitPointer = Int(BiteQuery(ElementList(x)))
LoadedWeight(x).Num = Int(BiteQuery(ElementList(x)))
ChewQueryValues(ElementList(x), LoadedWeight(x))
Next
Return LoadedWeight
End Function
Function BiteQuery(ByRef QueryList As String)
Dim Marker As Integer
Dim Bite As String
'This function partitions the input string around the first comma
'It returns the section before the comma, and stores the section behind the comma as the new value for the string
Try
Marker = InStr(QueryList, ",")
Bite = Left(QueryList, Marker - 1)
Marker = Len(QueryList) - Marker
QueryList = Right(QueryList, Marker)
Catch
'This is used in the case that a list without a comma is input
Bite = QueryList
QueryList = ""
End Try
Return Bite
End Function
Sub ChewQueryValues(ByRef QueryList As String, ByRef LoadedWeight As WeightElement)
LoadedWeight.Values = {}
While Len(QueryList) > 0
'This While loop is so that an arbitrary number of values can be inserted
'Because BiteQuery takes in functions by reference, each loop reduces the length of the string until it is empty
ReDim Preserve LoadedWeight.Values(LoadedWeight.Values.Length + 1)
LoadedWeight.Values(LoadedWeight.Values.Length - 1).TraitName = BiteQuery(QueryList)
LoadedWeight.Values(LoadedWeight.Values.Length - 1).TraitNum = Int(BiteQuery(QueryList))
LoadedWeight.Values(LoadedWeight.Values.Length - 1).WeightValue = CDec(BiteQuery(QueryList))
End While
End Sub
This set of functions is exhibiting some sort of undefined/random behavior when run in the following case:
FullToken = Strings.Right(Strings.Left(Query, 10), 5) 'This set of functions will extract the 5 rightmost characters of the 10 leftmost characters. This is equivalent to the 5th to 10th characters, which is where the token is stored
QueryText = Strings.Right(Query, Len(Query) - 11)
ReDim Preserve Weight(Weight.Length + 1)
Weight(Weight.Length - 1).Token = FullToken 'This puts the token in the list
Weight(Weight.Length - 1).Weight = ChewQuery(QueryText) 'This puts the weight in the list
WeightList.Text += vbCrLf + FullToken 'This adds the token to the viewable label
This is causing other important parts of the program to fail, which is not desirable. How do I fix this code so that it performs identically on each run of the program?
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
how can i check if a string only contains 4 digit numbers ( or a year )
i tried this
Dim rgx As New Regex("^/d{4}")
Dim number As String = "0000"
Console.WriteLine(rgx.IsMatch(number)) // true
number = "000a"
Console.WriteLine(rgx.IsMatch(number)) // false
number = "000"
Console.WriteLine(rgx.IsMatch(number)) //false
number = "00000"
Console.WriteLine(rgx.IsMatch(number)) // true <<< :(
this returns false when its less than 4 or at characters but not at more than 4
thanks!
I actually wouldn't use a regex for this. The expression is deceptively simple (^\d{4}$), until you realize that you also need to evaluate that numeric value to determine a valid year range... unless you want years like 0013 or 9015. You're most likely going to want the value as an integer in the end, anyway. Given that, the best validation is probably just to actually try to convert it to an integer right off the bat:
Dim numbers() As String = {"0000", "000a", "000", "00000"}
For Each number As String In numbers
Dim n As Integer
If Integer.TryParse(number, n) AndAlso number.Length = 4 Then
'It's a number. Now look at other criteria
End If
Next
Use LINQ to check if All characters IsDigit:
Dim result As Boolean = ((Not number Is Nothing) AndAlso ((number.Length = 4) AndAlso number.All(Function(c) Char.IsDigit(c))))
You should use the .NET string manipulation functions.
Firstly the requirements, the string must:
Contain exactly four characters, no more, no less;
Must consist of a numeric value
However your aim is to validate a Date:
Function isKnownGoodDate(ByVal input As String) As Boolean 'Define the function and its return value.
Try 'Try..Catch statement (error handling). Means an input with a space (for example ` ` won't cause a crash)
If (IsNumeric(input)) Then 'Checks if the input is a number
If (input.Length = 4) Then
Dim MyDate As String = "#01/01/" + input + "#"
If (IsDate(MyDate)) Then
Return True
End If
End If
End If
Catch
Return False
End Try
End Function
You may experience a warning:
Function isKnownGoodDate does not return a value on all code
paths. Are you missing a Return statement?
this can be safely ignored.
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