I have a Note as stringBuilder with word and date: reval 41/50/50
I want to manipulate it, so I will have: reval 05/05/14.
(The date is only when I have the word reval before)
My function is:
Sub correctDateShowing(ByRef Note As StringBuilder)
Dim index, i, j As Integer
For index = 0 To Note.Length - 2
If (Note(index).ToString = "r" And Note(index + 1).ToString = "e") Then
For i = 6 To Note.Length - 1 'start from 6,because I want only the date to be reversed
'Here I am Stuck!!
Next
End If
Next
End Sub
I try to do some replacing with a tmp variable but it didn't work.
I will be glad to get some help.
Thanks All!!!
Sub CorrectDateShowing(ByRef Note As StringBuilder)
Dim str As String = Note.ToString()
Dim arr As String() = str.Split(" "c)
arr(1) = StrReverse(arr(1))
Note = New StringBuilder(arr(0) & " " & arr(1))
End Sub
Split the text into two parts, reverse the second part (the date) and then reconnect them.
Try this:
Dim tempCharArray As char[]
Dim dateStartIndex, dateLength As int
'' Here you need to calculate the date start index and date length (i guess the date length is always 8)
Note.CopyTo(dateStartIndex, tempCharArray, 0, dateLength)
Note.Replace(new String(tempCharArray), new String(Array.Reverse(tempCharArray)), dateStartIndex, dateLength)
Related
I want all of my strings to be formatted consistently. It is grabbing values from the database and sometimes there are not spaces after colons, but there should be. My goal is to add a space after a colon (:) if there isn't one already. I would prefer to do this without regex, but I am open to any solutions! Thanks so much.
Edit: Apologies I read the title as VBA, have updated now.
Here is a nice simple solution for you.
Replace everything with a Colon and a space with just a Colon meaning all Colons no longer have a space regardless of weather they did or not initially.
Then Replace all Colons with a Colon and a Space:
Dim value1 As String = "Hello:World"
value1 = value1.Replace(": ", ":").Replace(":", ": ")
I can't say it's the most elegant. And its performance is not optimized, though using String.IndexOf should in general be faster than looping over every character. But being a brute force solution, it sure doesn't use regex. ;)
Besides regex, I'm sure there's a cute one-line solution using LINQ, but it's probably hard to read and maintain. Someone is welcome to post that for comparison.
Option Strict On
Module Module1
Sub Main()
Console.WriteLine(EnsureSpaceAfterColon("first: second:asdf third::"))
'prints:
'"first: second: asdf third: : "
End Sub
Public Function EnsureSpaceAfterColon(input As String) As String
Dim colon As Char = CChar(":")
Dim space As Char = CChar(" ")
Dim returnString As String = String.Copy(input) 'leave the original alone
Dim index As Integer = input.IndexOf(colon)
While index > -1 'String.IndexOf returns -1 if the index is not found
'if the index is the last index, there is no space, so add it
'or else if the Char at the next index is not a space, make it so
If index = returnString.Length - 1 OrElse returnString.Chars(index + 1) <> space Then
returnString = returnString.Insert(index + 1, space)
End If
'get the next index
index = returnString.IndexOf(colon, index + 1)
End While
Return returnString
End Function
End Module
Public Module Module1
Public Sub Main()
dim nospace as string = EnsureSpace("xxx:")
dim space as string = EnsureSpace("xxx: ")
Console.WriteLine("|" + nospace + "|")
Console.WriteLine("|" + space + "|")
End Sub
private function EnsureSpace(val as string)
dim temp as string = val.trim
return temp.padright(temp.length + 1)
end function
' OR
private function EnsureSpace1(val as string)
return val.substring(0, val.LastIndexOf(":") + 1) + " "
end function
End Module
|xxx: |
|xxx: |
Three different approaches (all different from the other answers)...pick your poison:
(1) Using a helper Iterator function and String.Join():
Public Function AddSpaceAfterColons(ByVal input As String) As String
Return String.Join("", ColonHelper(input))
End Function
Public Iterator Function ColonHelper(ByVal input As String) As IEnumerable(Of Char)
Dim lastCh As Nullable(Of Char)
For Each ch As Char In input
If lastCh.HasValue AndAlso lastCh.Value = ":"c AndAlso Not (ch = " "c) Then
Yield " "c
End If
Yield ch
lastCh = ch
Next
If lastCh.HasValue AndAlso lastCh.Value = ":"c Then
Yield " "c
End If
End Function
(2) Walking backwards through a StringBuilder:
Public Function AddSpaceAfterColons(ByVal input As String) As String
Static colon As Char = ":"c
Static space As Char = " "c
Dim sb As New System.Text.StringBuilder(input)
For i As Integer = sb.Length - 1 To 0 Step -1
If sb(i) = colon Then
If i = (sb.Length - 1) OrElse sb(i + 1) = space Then
sb.Insert(i + 1, space)
End If
End If
Next
Return sb.ToString
End Function
(3) Using String.Split() and a StringBuilder:
Public Function AddSpaceAfterColons(ByVal input As String) As String
Dim sb As New System.Text.StringBuilder
Dim parts() As String = input.Split(":")
sb.Append(parts(0))
For i As Integer = 1 To parts.Length - 1
sb.Append(":")
If Not parts(i).StartsWith(" ") Then
sb.Append(" ")
End If
sb.Append(parts(i))
Next
Return sb.ToString
End Function
I am having a problem where I just can't seem to get it to split or even display the message. The message variable is predefined in another part of my code and I have debugged to make sure that the value comes through. I am trying to get it so that every 100 characters it goes onto a new line and with every message it also goes onto a new line.
y = y - 13
messagearray.AddRange(Message.Split(ChrW(100)))
Dim k = messagearray.Count - 1
Dim messagefin As String
messagefin = ""
While k > -1
messagefin = messagefin + vbCrLf + messagearray(k)
k = k - 1
End While
k = 0
Label1.Text = Label1.Text & vbCrLf & messagefin
Label1.Location = New Point(5, 398 + y)
You can use regular expression. It will create the array of strings where every string contains 100 characters. If the amount of remained characters is less than 100, it will match all of them.
Dim input = New String("A", 310)
Dim mc = Regex.Matches(input, ".{1,100}")
For Each m As Match In mc
'// Do something
MsgBox(m.Value)
Next
You can use LINQ to do that.
When you do a Select you can get the index of the item by including a second parameter. Then group the characters by that index divided by the line length so, the first character has index 0, and 0 \ 100 = 0, all the way up to the hundredth char which has index 99: 99 \ 100 = 0. The next hundred chars have 100 \ 100 = 1 to 199 \ 100 = 1, and so on (\ is the integer division operator in VB.NET).
Dim message = New String("A"c, 100)
message &= New String("B"c, 100)
message &= New String("C"c, 99)
Dim lineLength = 100
Dim q = message.Select(Function(c, i) New With {.Char = c, .Idx = i}).
GroupBy(Function(a) a.Idx \ lineLength).
Select(Function(b) String.Join("", b.Select(Function(d) d.Char)))
TextBox1.AppendText(vbCrLf & String.Join(vbCrLf, q))
It is easy to see how to change the line length because it is in a variable with a meaningful name, for example I set it to 50 to get the output
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
You can use String.SubString to do that. Like this
Dim Message As String = "your message here"
Dim MessageList As New List (Of String)
For i As Integer = 0 To Message.Length Step 100
If (Message.Length < i + 100) Then
MessageList.Add(Message.SubString (i, Message.Length - i)
Exit For
Else
MessageList.Add(Message.SubString (i, 100))
End If
Next
Dim k = MessageList.Count - 1
...
Here is what your code produced with a bit of clean up. I ignored the new position of the label.
Private Sub OpCode()
Dim messagearray As New List(Of String) 'I guessed that messagearray was a List(Of T)
messagearray.AddRange(Message.Split(ChrW(100))) 'ChrW(100) is lowercase d
Dim k = messagearray.Count - 1
Dim messagefin As String
messagefin = ""
While k > -1
messagefin = messagefin + vbCrLf + messagearray(k)
k = k - 1
End While
k = 0 'Why reset k? It falls out of scope at End Sub
Label1.Text = Label1.Text & vbCrLf & messagefin
End Sub
I am not sure why you think that splitting a string by lowercase d would have anything to do with getting 100 characters. As you can see the code reversed the order of the list items. It also added a blank line between the existing text in the label (In this case Label1) and the new text.
To accomplish your goal, I first created a List(Of String) to store the chunks. The For loop starts at the beginning of the input string and keeps going to the end increasing by 10 on each iteration.
To avoid an index out of range which would happen at the end. Say, we only had 6 characters left from start index. If we tried to retrieve 10 characters we would have an index out of range.
At the end we join the elements of the string with the separated of new line.
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
BreakInto10CharacterChunks("The quick brown fox jumped over the lazy dogs.")
End Sub
Private Sub BreakInto10CharacterChunks(input As String)
Dim output As New List(Of String)
Dim chunk As String
For StartIndex = 0 To input.Length Step 10
If StartIndex + 10 > input.Length Then
chunk = input.Substring(StartIndex, input.Length - StartIndex)
Else
chunk = input.Substring(StartIndex, 10)
End If
output.Add(chunk)
Next
Label1.Text &= vbCrLf & String.Join(vbCrLf, output)
End Sub
Be sure to look up String.SubString and String.Join to fully understand how these methods work.
https://learn.microsoft.com/en-us/dotnet/api/system.string.substring?view=netframework-4.8
and https://learn.microsoft.com/en-us/dotnet/api/system.string.join?view=netframework-4.8
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
Here is some sample code:
Dim arrValue(3) as Integer
arrValue(0) = 5
arrValue(1) = 4
arrValue(2) = 7
arrValue(3) = 1
How can I display those four values next to each other.
More specifically, given those values how can I make txtValue.Text = 5471
Edit:
The idea I had would be to use some sort of function to append each one to the end using a loop like this:
Dim finalValue
For i As Integer = 3 To 0 Step -1
arrValue(i).appendTo.finalValue
Next
Obviously that code wouldn't work though the premise is sound I don't know the syntax for appending things and I'm sure I wouldn't be able to append an Integer anyway, I would need to convert each individual value to a string first.
Another method is to use String.Join:
Sub Main
Dim arrValue(3) as Integer
arrValue(0) = 5
arrValue(1) = 4
arrValue(2) = 7
arrValue(3) = 1
Dim result As String = String.Join("", arrValue)
Console.WriteLine(result)
End Sub
If I understand your question correctly, you can use StringBuilder to append the values together.
Dim finalValue as StringBuilder
finalValue = new StringBuilder()
For i As Integer = 3 To 0 Step -1
finalValue.Append(arrValue(i))
Next
Then just return the finalValue.ToString()
Convert the integers to strings, and concatenate them:
Dim result as String = ""
For Each value as Integer in arrValue
result += value.ToString()
Next
Note: using += to concatenate strings performs badly if you have many strings. Then you should use a StringBuilder instead:
Dim builder as New StringBuilder()
For Each value as Integer in arrValue
builder.Append(value)
Next
Dim result as String = builder.ToString()
for i = lbound(arrValue) to ubound(arrValue)
ss=ss & arrValue(i)
next i
end for
debug.print ss
Dim value as string = ""
For A As Integer = 1 To Begin.nOfMarks
value += "Mark " & A & ": " & (Begin.Marks(A)) & vbCrLf
Next A