The string is "What is 8 multiplied by 2?"
How can I get the first 1st number which is "8" and the second number which is "2"?
What I want is to get the first number and use the mathematical operation asked to the second number.
Dim str As String = Label1.Text
Dim i As Integer
For i = 1 To Len(str)
If IsNumeric(Mid(str, i, 1)) Then
Label2.Text = Mid(str, i, 1)
End If
Next
I tried using this but I only get the 2nd number which is "2"
Dim str As String = "What is 8666 multiplied by 2444"
Dim numbers As New List(Of Integer)
For Each word As String In str.Split(" ")
If IsNumeric(word) Then
numbers .Add(CInt(word))
End If
Next
MsgBox("first:" & numbers (0) & ",second:" & numbers (1))
Related
I need to save Multiple(about 20-25) serial number of the specimen in my application. Sometimes serial number will be alphanumeric but will be sequential. I need a way out to increment alphanumeric serial numbers based on the first serial number entered.
My main problem is alphabet position and alphabet count keeps changing. Example : 10MG2015 20562MG0 MGX02526 etc etc
I tried but mine works when Alphabet are in starting position and when there are known number of alphabets. Here is my try
Dim intValue as integer
Dim serialno as string
Dim serialno1 as string
For i =0 to 20
Serialno1 = serialno.Substring(3)
Int32.TryParse(Serialno1, intValue)
intValue = intValue + 1
checkedox1.items.add(serialno.Substring(0,3) + intValue.ToString("D3"))
NEXT
Any help is highly appreciated. Thanks in advance
edit 1
Clarity : I want to increment alphanumeric string. Example : If first entered one is 10MG2015 then I should increment to 10MG2016, 10MG2017, 10MG2018, 10MG2019 and so on... For 20562MG0 it will be 20562MG1, 20562MG2 20562MG3 and so on...
Function FindSequenceNumber(SerialNumber As String) As Integer
'Look for at least four digits in a row, and capture all the digits
Dim sequenceExpr As New Regex("([0-9]{4,11})")
Dim result As Integer = -1
Dim m As Match = sequenceExpr.Match(SerialNumber)
If m.Success AndAlso Integer.TryParse(m.Groups(1).Value, result) Then
Return result
Else
'Throw exception, return -1, etc
End If
End Function
See it here:
https://dotnetfiddle.net/gO2nue
Note: the integer type doesn't preserve leading zeros. You may find it better to return a tuple with either the length of the original string, so you can pad zeros to the left if needed to match the original formatting.
Or maybe this:
Function IncrementSerial(SerialNumber As String) As String
'Look for at least four digits in a row, and capture all the digits
Dim sequenceExpr As New Regex("([0-9]{4,11})")
Dim m As Match = sequenceExpr.Match(SerialNumber)
If Not m.Success Then Throw New Exception("No sequence number found")
Dim c = m.Groups(1).Captures(0)
Dim seq = (Integer.Parse(c.Value) + 1).ToString()
If seq.Length < c.Value.Length Then
seq = seq.PadLeft(c.Value.Length, "0"c)
End If
Dim result As String = ""
If c.Index > 0 Then result & = SerialNumber.Substring(0, c.Index)
result &= seq
If c.Index + seq.Length < SerialNumber.Length Then result &= SerialNumber.SubString(c.Index + seq.Length)
Return result
End Function
I want to make the user input a 3 digit number (100 - 999), then take out the 1st digit, 2nd and 3rd and say what unit it belongs to
e.g.
567 will make
5 hundreds
6 tens
7 ones
I've tried mid and right but mid shows me all my units and right only works for the ones
Im stuck on the tens
Module Module1
Dim Num As Integer
Sub Main()
Console.WriteLine("Enter a number between 100 and 999")
Num = CStr(Console.ReadLine)
Console.Write(Left(Num, 1))
Console.Write(" Hundreds")
Console.WriteLine(Mid(Num, 1))
Console.Write(" Tens")
Console.WriteLine(Right(Num, 1))
Console.Write(" One's")
Console.ReadKey()
End Sub
End Module
Use string.SubString.
For example:
Dim strNumber As String = Num.ToString()
Dim hundreds As String = strNumber.SubString(0, 1) 'shows the hundreds.
Dim tens As String = strNumber.SubString(1, 1) 'shows the tens.
Dim ones As String = strNumber.SubString(2, 1) 'the ones.
As somebody commented on your question:
In your case since you only want one digit to return you can access the strings char array and return the char at position X like so:
Dim strNumber As String = Num.ToString()
Dim hundreds As String = strNumber(0)
Dim tens As String = strNumber(1)
Dim ones As String = strNumber(2)
I'm working to create an Excel macro using VBA to convert bit strings to numbers. They are not binary numbers, each '1' stands for it's own number.
e.g: 1100000000000000000010001
from the left, the first bit represents "1", the second bit represents "2", third bit represents "0", and so on. The total quantity of bits in each string is 25.
I want VBA to convert it and show results like so: 1, 2, 21, 25.
I tried using Text to Columns but was not successful.
Try something like this:
Sub Execute()
Dim buff() As String
Dim i As Integer, total As Double
buff = Split(StrConv(<theString>, vbUnicode), Chr$(0))
total = 0
For i = 0 To UBound(buff)
Debug.Print (buff(i))
'total = total + buff(i) * ??
Next i
End Sub
Consider:
Public Function BitPicker(sIn As String) As String
For i = 1 To Len(sIn)
If Mid(sIn, i, 1) = 1 Then
BitPicker = BitPicker & i & ","
End If
Next
BitPicker = Mid(BitPicker, 1, Len(BitPicker) - 1)
End Function
Another non-VBA solution, based on the OP' initial approach and with a layout designed to facilitate multiple 'conversions' (ie copy formulae down to suit):
Does this have to be VBA? Give a data setup like this:
The formula in cell B4 and copied down to B33 is:
=IF(ROWS(B$3:B3)>LEN($B$1)-LEN(SUBSTITUTE($B$1,"1","")),"",FIND("#",SUBSTITUTE($B$1,"1","#",ROWS(B$3:B3))))
The formula cells are formatted as General and the the Bit String cell (B1) is formatted as Text.
Try this:
Function ConvertMyRange(Rng As Range) As String
Dim MyString As String
MyString = Rng.Text
Dim OutPutString As String
For i = 1 To Len(MyString)
If Mid(MyString, i, 1) = "1" Then OutPutString = OutPutString & ", " & i
Next i
' Get rid of first ", " that was added in the loop
If Len(OutPutString) > 0 Then
OutPutString = Mid(OutPutString, 2)
End If
ConvertMyRange = OutPutString
End Function
For your input, the output is 1, 2, 21, 25
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 want to display the first and last characters of any given string entered into a textbox. The strings can be of any length as the user wants (as long as it is one word) I would like to be able to do something like this... "william = will and iam" or "Celtic = Cel and tic"
I understand I would have to split or divide the string. How would I go about doing this? Any help is appreciated, thanks.
EDIT:
Thanks for your help once again guys, this is how the code ended up!
Dim strInput = txtString.Text
Dim halflength = strInput.Length / 2
Dim firsthalf = strInput.Substring(0, halflength)
Dim secondhalf = strInput.Substring(halflength)
Dim strResults = firsthalf
Dim secondResult = secondhalf
MessageBox.Show(firsthalf)
MessageBox.Show(secondhalf)
MessageBox.Show("First half of string contains... " & " " & strResults.Length.ToString & " characters", "Character Count")
MessageBox.Show("Second half of string contains... " & " " & secondResult.Length.ToString & " characters", "Character Count")
EDIT:
Also meant to mention my current incorrect code.
Dim strInput As String
Dim strLength As String
Dim strResults As String
strInput = txtString.Text
strLength = strInput.Length / 2
strResults = txtString.Text
MessageBox.Show(strInput.Length.ToString, "Length of characters")
MessageBox.Show(strLength.ToString)
MessageBox.Show(strResults.Substring(0, 3))
String.Substring and String.Length should give you everything you need to get started on this.
Seeing your existing code will make this easier. Let's walk through what we have now.
Let's assume we have just a plain, simple string like this instead of a textbox for the sake of making things easier:
Dim txtString = "Hello World"
Now, in order to split the length of the string in half; we need to get the length. The `Length property will give is that, and then divide it by two.
Dim halfLength = txtString.Length \ 2
This will perform integer division; so any remaining decimal is truncated.
Now we know where the middle of the string is. We can now use String.Substring to carve out a peice of the string by index. Substring takes two parameters, the index where to start the string, and number of characters to take. There is a second overload that takes the index to start at and consumes till the end of the string. Indexes are zero based. So for example, if we wanted to start at the beginning of the string, we'd use zero. If we wanted to skip the first character, we'd use one.
For the first half of the string, we don't want to skip any characters, so we'll use zero. The number of characters we want is half length of the string, so we pass in halfLength:
Dim firstHalf = txtString.Substring(0, halfLength)
For the second half, we want to start in the middle of the string, and consume characters till the end, so we'll use the other overload:
Dim secondHalf = txtString.Substring(halfLength)
You now have your string split in half.
The final result looks like this:
Dim txtString = "Hello World"
Dim halfLength = txtString.Length \ 2
Dim firstHalf = txtString.Substring(0, halfLength)
Dim secondHalf = txtString.Substring(halfLength)
Assuming the rules are "each side is half the length with the left side taking precedence", you would use Substring and some simple division:
Dim str As String = "william"
Dim part1 As String = str.Substring(0, CInt(Math.Ceiling(str.Length / 2.0#)))
Dim part2 As String = str.Substring(part1.Length)
part1 & " and " & part2 'will and iam
Here's a demo.
My code displays first half and last half of any number of characters entered.
Declare Variable
Dim strResults As String
Fetch text from textbox
strResults = Textbox1.Text
Display the first half of the text
MessageBox.Show(strResults.Substring(0, strResults.Length / 2), "First Half Characters")
Display the last half of the text
MessageBox.Show(strResults.Substring(strResults.Length / 2), "Last Half Characters")
Full code:
Dim strResults As String
strResults = Textbox1.Text
MessageBox.Show(strResults.Substring(0, strResults.Length / 2), "First Half Characters")
MessageBox.Show(strResults.Substring(strResults.Length / 2), "Last Half Characters")