I created the following function, but have not been able to finish. I want to return the first 2 characters of each word in the string. Here is what I have so far:
Function SelectWords(ByVal text As String, ByVal maxWords As Integer) As String
If String.IsNullOrEmpty(text) Then Return String.Empty
If maxWords <= 0 Then Return String.Empty
Dim words As String() = text.Split(" "c)
Return String ''I am stuck here
End Function
You did not describe the purpose of maxwords, nor what to do with a. The loop part:
Dim words = str.Split(" "c)
Dim ret As New StringBuilder ' in case it is a long string
For Each w As String In words
If w.Length > 1 Then
ret.Append(w.Substring(0, 2))
Else
' decide if you want 1
End If
Next
return ret.toString
The code you have doesn't do anything that you describing .. Try this function instead.
Function SelectWords(ByVal text As String, ByVal maxWords As Integer) As String
Dim collection As MatchCollection = Regex.Matches(text, "(\w{2})\w*\b")
Dim output As New System.Text.StringBuilder
Dim counter As Integer = 0
For Each M As Match In collection
output.Append(M.Groups(1).Value)
counter += 1
If counter = maxWords Then
Exit For
End If
Next
Return output.ToString
End Function
Related
I would need a little help. If in the Textbox - TxtStringNum1.Text - we have the Digit 3. and TxtIntDraws.Lines (1) - contains the following set: 13,20,21,23,47,49,50,51,63,64,66,70
It shows me that there is Exists a Digit of 3, but it is actually the Digits of 13. It fails to look for it as a whole.
Private Sub ScanareLinia1()
Dim textsrtring As String = TxtStringNum1.Text
Dim words As String() = textsrtring.Split(New Char() {" "c})
' Split string based on space
Dim found As Boolean = False
' Use For Each loop over words
Dim word As Integer
For Each word In words
For i As Integer = 0 To TxtIntDraws.Lines.Count - 1
If TxtIntDraws.Lines(1).Contains(word) Then
TxtResultStr1.Text = word
End If
Next
Next
ScanareLinia2()
End Sub
I have a good comparison code, but I do not know how to use it for the code above.
Private Sub CompareNumbers()
'Pentru funcția de Check-In (For Match Exactly Value Number in List)
'First Textbox that is to be used for compare
Dim textBox1Numbers As List(Of Integer) = GetNumbersFromTextLine(TxtStringNum1.Text)
'Second Textbox that is to be used for compare
Dim textBox2Numbers As List(Of Integer) = GetNumbersFromTextLine(TxtbValBeforeCompar.Text)
'Union List of Common Numbers (this uses a lambda expression, it can be done using two For Each loops instead.)
Dim commonNumbers As List(Of Integer) = textBox1Numbers.Where(Function(num) textBox2Numbers.Contains(num)).ToList()
'This is purely for testing to see if it worked you can.
Dim sb As StringBuilder = New StringBuilder()
For Each foundNum As Integer In commonNumbers
sb.Append(foundNum.ToString()).Append(TextBox25.Text)
TxtbValAfterCompar.Text = (sb.ToString())
Next
End Sub
Private Function GetNumbersFromTextLine(ByVal sTextLine As String) As List(Of Integer)
'Pentru funcția de Check-In (For Match Exactly Value Number in List)
Dim numberList As List(Of Integer) = New List(Of Integer)()
Dim sSplitNumbers As String() = sTextLine.Split(TextBox8.Text)
For Each sNumber As String In sSplitNumbers
If IsNumeric(sNumber) Then
Dim iNum As Integer = CInt(sNumber)
TxtbValAfterCompar.Text = iNum
If Not numberList.Contains(iNum) Then
TxtbValAfterCompar.Text = ("")
numberList.Add(iNum)
End If
Else
End If
Next
Return numberList
End Function
You can use the following condition using String.Split and Array.IndexOf:
If Array.IndexOf(TxtIntDraws.Lines(1).Split(","c), CStr(word)) > -1 Then
TxtResultStr1.Text = word
End If
So the following Array.IndexOf(CStr("13,14,15,16,17").Split(","c), "13") > -1 is True and Array.IndexOf(CStr("13,14,15,16,17").Split(","c), "3") > -1 is False.
I am developing a program where you can input a sentence and then search for a word. The program will then tell you at which positions this word occurs. I have written some code but do not know how to continue.
Module Module1
Sub Main()
Dim Sentence As String
Dim SentenceLength As Integer
Dim L As Integer = 0
Dim LotsofText As String = Console.ReadLine
Console.WriteLine("Enter your word ") : Sentence = Console.ReadLine
For L = 1 To LotsofText.Length
If (Mid(LotsofText, L, 1)) = " " Then
End If
L = L + 1
Dim TextCounter As Integer = 0
Dim MainWord As String = Sentence
Dim CountChar As String = " "
Do While InStr(MainWord, CountChar) > 0
MainWord = Mid(MainWord, 1 + InStr(MainWord, CountChar), Len(MainWord))
TextCounter = TextCounter + 1
'Text = TextCounter + 2
' Console.WriteLine(Text)
Loop
Console.WriteLine(TextCounter)
Console.Write("Press Enter to Exit")
Console.ReadLine()
End Sub
End Module
Transform this piece of code from C# to Visual Basic. match.Index will indicate the position of the given word.
var rx = new Regex("your");
foreach (Match match in rx.Matches("This is your text! This is your text!"))
{
int i = match.Index;
}
To find only words and not sub-strings (for example to ignore "cat" in "catty"):
Dim LotsofText = "catty cat"
Dim Sentence = "cat"
Dim pattern = "\b" & Regex.Escape(Sentence) & "\b"
Dim matches = Regex.Matches(LotsofText, pattern)
For Each m As Match In matches
Debug.Print(m.Index & "") ' 6
Next
If you want to find sub-strings too, you can remove the "\b" parts.
If you add this function to your code:
Public Function GetIndexes(ByVal SearchWithinThis As String, ByVal SearchForThis As String) As List(Of Integer)
Dim Result As New List(Of Integer)
Dim i As Integer = SearchWithinThis.IndexOf(SearchForThis)
While (i <> -1)
Result.Add(i)
i = SearchWithinThis.IndexOf(SearchForThis, i + 1)
End While
Return Result
End Function
And call the function in your code:
Dim Indexes as list(of Integer) = GetIndexes(LotsofText, Sentence)
Now GetIndexes will find all indexes of the word you are searching for within the sentence and put them in the list Indexes.
I would like all possible combinations from two string arrays.
Both arrays must have same length.
Result must keep order
For example :
dim lStr1() as string = {"One", "Two", "Three"}
dim lStr2() as string = {"EditOne", "EditTwo", "EditThree"}
dim res() as string = myAwesomeFunction(lStr1, lStr2)
// res :
One Two Three
One Two EditThree
One EditTwo Three
One EditTwo EditThree
EditOne Two Three
EditOne Two EditThree
EditOne EditTwo Three
EditOne EditTwo EditThree
It's like the binary composition of 2 arrays of strings.
Here's another solution. Since only 2 arrays are involved, we can bit-fiddle to get all of the "combinations". The & " " is just to format the output to match the example.
Private Function myAwesomeFunction(Array1() As String, Array2() As String) As String()
If Array1.Length <> Array2.Length Then
Throw New ArgumentException("Array lengths must be equal.")
End If
Dim combos(CInt(2 ^ Array1.Length) - 1) As String
For i As Integer = 0 To combos.Count - 1
For j = 0 To Array1.Length - 1
If (i And (1 << j)) > 0 Then
combos(i) += Array2(j) & " "
Else
combos(i) += Array1(j) & " "
End If
Next
Next
Return combos
End Function
The following code will produce the array in your example. It should work for any pair of input arrays. The function checks that the input arrays are of the same length.
The GetPermutations function is taken from a more general class I use for generating permutations of numbers. It returns arrays of total Integers between 0 and choose - 1, and being an Iterator function, it returns the next array each time it is called.
In order to match your example, I returned an array of String where each element is a single string consisting of each of the selected strings separated by spaces. You may find it more useful to return a List(Of String()) or even a List(Of List(Of String))
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim lStr1() As String = {"One", "Two", "Three"}
Dim lStr2() As String = {"EditOne", "EditTwo", "EditThree"}
Dim res() As String = myAwesomeFunction(lStr1, lStr2)
End Sub
Function MyAwesomeFunction(lStr1() As String, lStr2() As String) As String()
Dim combos As New List(Of String)
If lStr1.Length <> lStr2.Length Then Throw New ArgumentException("Arrays must have the same length")
For Each combo() As Integer In GetPermutations(lStr1.Length, 2)
Dim elem As New List(Of String)
For i As Integer = 0 To combo.Length - 1
elem.Add(If(combo(i) = 0, lStr1(i), lStr2(i)))
Next
combos.Add(String.Join(" ", elem))
Next
Return combos.ToArray
End Function
Public Iterator Function GetPermutations(choose As Integer, total As Integer) As IEnumerable(Of Integer())
Dim totals() As Integer = Enumerable.Repeat(Of Integer)(total, choose).ToArray
Dim value(choose - 1) As Integer
Do
Yield value
For index As Integer = choose - 1 To 0 Step -1
value(index) += 1
If value(index) < totals(index) Then Continue Do
value(index) = 0
Next
Exit Do
Loop
End Function
Beginner here, bear with me, I apologize in advance for any mistakes.
It's some homework i'm having a bit of trouble going about.
Overall goal: outputting the specific amount of characters in a string using a loop statement. Example being, user wants to find how many "I" is in "Why did the chicken cross the road?", the answer should be 2.
1) The form/gui has 1 MultiLine textbox and 1 button titled "Search"
2) User enters/copys/pastes text into the Textbox clicks "Search" button
3) Search button opens an InputBox where the user will type in what character(s) they want to search for in the Textbox then presses "Ok"
4) (where I really need help) Using a Loop Statement, The program searches and counts the amount of times the text entered into the Inputbox, appears in the text inside the MultiLine Textbox, then, displays the amount of times the character showed up in a "messagebox.show"
All I have so far
Private Sub Search_btn_Click(sender As System.Object, e As System.EventArgs) Handles Search_btn.Click
Dim counterInt As Integer = 0
Dim charInputStr As String
charInputStr = CStr(InputBox("Enter Search Characters", "Search"))
I would use String.IndexOf(string, int) method to get that. Simple example of concept:
Dim input As String = "Test input string for Test and Testing the Test"
Dim search As String = "Test"
Dim count As Integer = -1
Dim index As Integer = 0
Do
index = input.IndexOf(search, index) + 1
count += 1
Loop While index > 0
count is initialized with -1 because of do-while loop usage - it will be set to 0 even if there is no pattern occurrence in input string.
Try this Code
Dim input As String = "Test input string for Test and Testing the Test"
Dim search() As String = {"Te"}
MsgBox(input.Split(input.Split(search, StringSplitOptions.None), StringSplitOptions.RemoveEmptyEntries).Count)
Concept: Increment the count until the input containing the particular search string. If it contains the search string then replace the first occurance with string.empty (In String.contains() , the search starts from its first index, that is 0)
Dim input As String = "Test input string for Test and Testing the Test"
Dim search As String = "T"
Dim count As Integer = 0
While input.Contains(search) : input = New Regex(search).Replace(input, String.Empty, 1) : count += 1 : End While
MsgBox(count)
Edit:
Another solution:
Dim Input As String = "Test input string for Test and Testing the Test"
Dim Search As String = "Test"
MsgBox((Input.Length - Input.Replace(Search, String.Empty).Length) / Search.Length)
try this code.... untested but i know my vb :)
Function lol(ByVal YourLine As String, ByVal YourSearch As String)
Dim num As Integer = 0
Dim y = YourLine.ToCharArray
Dim z = y.Count
Dim x = 0
Do
Dim l = y(x)
If l = YourSearch Then
num = num + 1
End If
x = x + 1
Loop While x < z
Return num
End Function
Its a function that uses its own counter... for every character in the string it will check if that character is one that you have set (YourSearch) and then it will return the number of items that it found. so in your case it would return 2 because there are two i's in your line.
Hope this helps!
EDIT:
This only works if you are searching for individual Characters not words
You can try with something like this:
Dim sText As String = TextBox1.Text
Dim searchChars() As Char = New Char() {"i"c, "a"c, "x"c}
Dim index, iCont As Integer
index = sText.IndexOfAny(searchChars)
While index >= 0 Then
iCont += 1
index = sText.IndexOfAny(searchChars, index + 1)
End While
Messagebox.Show("Characters found " & iCont & " times in text")
If you want to search for words and the times each one is appearing try this:
Dim text As String = TextBox1.Text
Dim wordsToSearch() As String = New String() {"Hello", "World", "foo"}
Dim words As New List(Of String)()
Dim findings As Dictionary(Of String, List(Of Integer))
'Dividing into words
words.AddRange(text.Split(New String() {" ", Environment.NewLine()}, StringSplitOptions.RemoveEmptyEntries))
findings = SearchWords(words, wordsToSearch)
Console.WriteLine("Number of 'foo': " & findings("foo").Count)
With this function used:
Private Function SearchWords(ByVal allWords As List(Of String), ByVal wordsToSearch() As String) As Dictionary(Of String, List(Of Integer))
Dim dResult As New Dictionary(Of String, List(Of Integer))()
Dim i As Integer = 0
For Each s As String In wordsToSearch
dResult.Add(s, New List(Of Integer))
While i >= 0 AndAlso i < allWords.Count
i = allWords.IndexOf(s, i)
If i >= 0 Then dResult(s).Add(i)
i += 1
End While
Next
Return dResult
End Function
You will have not only the number of occurances, but the index positions in the file, grouped easily in a Dictionary.
well i know that there are a lot of these threads but im new to vb.net yet i cant edit the sources given to make what i really want
so i want a function that will generate random strings which will contain from 15-32 characters each and each of them will have the following chars ( not all at the same string but some of them ) :
A-Z
a-z
0-9
here is my code so far
Functon RandomString()
Dim s As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
Dim r As New Random
Dim sb As New StringBuilder
For i As Integer = 1 To 8
Dim idx As Integer = r.Next(0, 35)
sb.Append(s.Substring(idx, 1))
Next
return sb.ToString()
End Function
Change the string to include the a-z characters:
Dim s As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
Change the loop to create a random number of characters:
Dim cnt As Integer = r.Next(15, 33)
For i As Integer = 1 To cnt
Note that the upper boundary in the Next method is exclusive, so Next(15, 33) gives you a value that can range from 15 to 32.
Use the length of the string to pick a character from it:
Dim idx As Integer = r.Next(0, s.Length)
As you are going to create random strings, and not a single random string, you should not create the random number generator inside the function. If you call the function twice too close in time, you would end up with the same random string, as the random generator is seeded using the system clock. So, you should send the random generator in to the function:
Function RandomString(r As Random)
So, all in all:
Function RandomString(r As Random)
Dim s As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
Dim sb As New StringBuilder
Dim cnt As Integer = r.Next(15, 33)
For i As Integer = 1 To cnt
Dim idx As Integer = r.Next(0, s.Length)
sb.Append(s.Substring(idx, 1))
Next
return sb.ToString()
End Function
Usage example:
Dim r As New Random
Dim strings As New List<string>()
For i As Integer = 1 To 10
strings.Add(RandomString(r))
Next
Try something like this:-
stringToReturn&= Guid.NewGuid.ToString().replace("-","")
You can also check this:-
Sub Main()
Dim KeyGen As RandomKeyGenerator
Dim NumKeys As Integer
Dim i_Keys As Integer
Dim RandomKey As String
''' MODIFY THIS TO GET MORE KEYS - LAITH - 27/07/2005 22:48:30 -
NumKeys = 20
KeyGen = New RandomKeyGenerator
KeyGen.KeyLetters = "abcdefghijklmnopqrstuvwxyz"
KeyGen.KeyNumbers = "0123456789"
KeyGen.KeyChars = 12
For i_Keys = 1 To NumKeys
RandomKey = KeyGen.Generate()
Console.WriteLine(RandomKey)
Next
Console.WriteLine("Press any key to exit...")
Console.Read()
End Sub
Using your function as a guide, I modified it to:
Randomize the length (between minChar & maxCharacters)
Randomize the string produced each time (by using the static Random)
Code:
Function RandomString(minCharacters As Integer, maxCharacters As Integer)
Dim s As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
Static r As New Random
Dim chactersInString As Integer = r.Next(minCharacters, maxCharacters)
Dim sb As New StringBuilder
For i As Integer = 1 To chactersInString
Dim idx As Integer = r.Next(0, s.Length)
sb.Append(s.Substring(idx, 1))
Next
Return sb.ToString()
End Function
Try this out:
Private Function RandomString(ByRef Length As String) As String
Dim str As String = Nothing
Dim rnd As New Random
For i As Integer = 0 To Length
Dim chrInt As Integer = 0
Do
chrInt = rnd.Next(30, 122)
If (chrInt >= 48 And chrInt <= 57) Or (chrInt >= 65 And chrInt <= 90) Or (chrInt >= 97 And chrInt <= 122) Then
Exit Do
End If
Loop
str &= Chr(chrInt)
Next
Return str
End Function
You need to change the line For i As Integer = 1 To 8 to For i As Integer = 1 To ? where ? is the number of characters long the string should be. This changes the number of times it repeats the code below so more characters are appended to the string.
Dim idx As Integer = r.Next(0, 35)
sb.Append(s.Substring(idx, 1))
My $.02
Dim prng As New Random
Const minCH As Integer = 15 'minimum chars in random string
Const maxCH As Integer = 35 'maximum chars in random string
'valid chars in random string
Const randCH As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
Private Function RandomString() As String
Dim sb As New System.Text.StringBuilder
For i As Integer = 1 To prng.Next(minCH, maxCH + 1)
sb.Append(randCH.Substring(prng.Next(0, randCH.Length), 1))
Next
Return sb.ToString()
End Function
please note that the
r.Next(0, 35)
tend to hang and show the same result Not sure whay; better to use
CInt(Math.Ceiling(Rnd() * N)) + 1
see it here Random integer in VB.NET
I beefed up Nathan Koop's function for my own needs, and thought I'd share.
I added:
Ability to add Prepended and Appended text to the random string
Ability to choose the casing of the allowed characters (letters)
Ability to choose to include/exclude numbers to the allowed characters
NOTE: If strictly looking for an exact length string while also adding pre/appended strings you'll need to deal with that; I left out any logic to handle that.
Example Usages:
' Straight call for a random string of 20 characters
' All Caps + Numbers
String_Random(20, 20, String.Empty, String.Empty, 1, True)
' Call for a 30 char string with prepended string
' Lowercase, no numbers
String_Random(30, 30, "Hey_Now_", String.Empty, 2, False)
' Call for a 15 char string with appended string
' Case insensitive + Numbers
String_Random(15, 15, String.Empty, "_howdy", 3, True)
.
Public Function String_Random(
intMinLength As Integer,
intMaxLength As Integer,
strPrepend As String,
strAppend As String,
intCase As Integer,
bIncludeDigits As Boolean) As String
' Allowed characters variable
Dim s As String = String.Empty
' Set the variable to user's choice of allowed characters
Select Case intCase
Case 1
' Uppercase
s = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
Case 2
' Lowercase
s = "abcdefghijklmnopqrstuvwxyz"
Case Else
' Case Insensitive + Numbers
s = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
End Select
' Add numbers to the allowed characters if user chose so
If bIncludeDigits = True Then s &= "0123456789"
Static r As New Random
Dim chactersInString As Integer = r.Next(intMinLength, intMaxLength)
Dim sb As New StringBuilder
' Add the prepend string if one was passed
If String.IsNullOrEmpty(strPrepend) = False Then sb.Append(strPrepend)
For i As Integer = 1 To chactersInString
Dim idx As Integer = r.Next(0, s.Length)
sb.Append(s.Substring(idx, 1))
Next
' Add the append string if one was passed
If String.IsNullOrEmpty(strAppend) = False Then sb.Append(strAppend)
Return sb.ToString()
End Function