Can someone show me how to make a number occurrences application in Visual Basic 2008? The program is supposed to prompt the user for a number and then it counts the occurrences of each number (0-9) in the user number.
For example, a number of 122378 would print out:
0=0, 1=1, 2=2, 3=1, 4=0, 5=0, 6=0, 7=1, 8=1, 9=0,
It says to treat the number like a string, and the answer should be displayed in a listbox.
There are other ways to do it but so you understand - do this:
'prepare placeholder
dim dict as New Dictionary(of String, Integer)
dict.Add("0", 0)
dict.Add("1", 0)
dict.Add("2", 0)
dict.Add("3", 0)
dict.Add("4", 0)
dict.Add("5", 0)
dict.Add("6", 0)
dict.Add("7", 0)
dict.Add("8", 0)
dict.Add("9", 0)
' Assuming you ask user to type into text box txtSubmittedNumber
' Parse your numbers and collect counts
For Each c as Char in txtSubmittedNumber.Text.ToCharArray()
Dim key as String = c.ToString()
dict(key) = dict(key) + 1
Next
' Assuming you fill results into listbox lstNumbers, just add your strings to display
For each kvp as KeyValuePair(of String, Integer) in dict
lstNumbers.Items.Add(kvp.Key & "=" & kvp.Value.ToString())
Next
This is program for beginners. There is Linq of course but it is not going to be as clear what is going on because it substitutes loops, group by, etc.
Related
Dim LineNo as Integer
LineNo = CStr(channel) 'This can have a value of 1 to 100
If LineNo = 1 then
Text1.Text = "Line one selected"
Elseif LineNo = 2 then
Text2.Text = "Line one selected"
'Etc etc
End if
I need to replace the number "1" in Text1.Text and every other TextBox with the value of LineNo? For example:
Text{LineNo}.Text
So I would not have to do a repeated "If" and have a smaller one line code like this:
Text{LineNo}.Text = "Line " & LineNo & " selected"
How would I do this?
Look into a Control array of text boxes. You could have txtLine(), for example, indexed by the channel number.
LineNo = CStr(channel)
txtLine(channel).Text = "Line " & LineNo & " selected"
To create the array, set the Index property of each of the text boxes to an increasing integer, starting at 0.
If you have a finite and relatively small number, you can use a property. I've used this approach with up to 30+ elements in a pinch. Super simple, easy pattern to recognize and replicate in other places. A bit if a pain if the number of elements changes in the future, but extensible nevertheless.
It uses the Choose statement, which takes an index N and returns the Nth element (1-based), hence, the check makes sure that N is > 0 and <= MAX (which you would configure).
Public Property Get TextBox txt(ByVal N As Long)
Const MAX As Long = 10
If N <= 0 || N > MAX Then Exit Property ' Will return a "Nothing". You could return the bound element if you prefer
set txt = Choose(Text1, Text2, Text3, Text4, Text5, Text6, Text7, Text8, Text9, Text10)
End Property
Then, you can simply reference them with the Property, much like an alias:
txt(1).Text = "Line 1 text"
txt(2).Text = "Line 2 text"
If you have an arbitrary number, then you are likely using a control array already, which is simpler because it can be referenced by Index already, so you can directly reference it.
If neither of these work for you and you have a very large number of controls, you can scan the Controls collection in a similar Property, attempting to match ctrl.Name with the pattern of your choice (e.g., matching the first 4 characters to the string "Text", thus matching Text1, Text2, etc, for an unlimited number). Theoretically, this should be future-proofed, but that's just theoretical, because anything can happen. What it does do for you is to encapsulate the lookup in a way that "pretends" to be a control array. Same syntax, just you control the value.
Public Property Get TextBox txt(ByVal N As Long)
Dim I As Long
For I = 0 To Controls.Count - 1 ' Controls is zero-based
' Perform whatever check you need to. Obviously, if you have a "Label" named
' "Text38", the assignment will throw an error (`TextBox = Label` doesn't work).
If Left(Controls(I).Name, 4) = "Text" Then
set txt = Controls(I)
End If
Next
' If you want the Property to never return null, you could uncomment the following line (preventing dereference errors if you insist on the `.Text` for setting/getting the value):
' If txt Is Nothing Then Set txt = Text1
End Property
Use the same way as above: txt(n).Text = "..."
I am trying to add a random set of numbers to the end of a string. I'm still learning the basics of VBS but this has really tricked me and I can't seem to find anything online.
I've tried:
string2 = "hello" + (Rnd() * Len(VALID_TEXT)) + 1
And:
x = rnd*10
string2 = "hello" + x
What am I doing wrong?
All random number generators rely on an underlying algorithm, usually fed by what’s called a seed number. You can use the Randomize statement to create a new seed number to ensure your random numbers don’t follow a predictable pattern.
To get the random numbers, using rnd alone is not sufficient as you will keep on getting the same random number again and again. You have to use randomize to achieve the task as shown below:
Dim strTest:strTest = "Hello"
Dim intNoOfDigitsToAppend:intNoOfDigitsToAppend = 5
Randomize
Msgbox "String before appending: " & strTest
strTest = fn_appendRandomNumbers(strTest,intNoOfDigitsToAppend)
Msgbox "String before appending: " & strTest
function fn_appendRandomNumbers(strToAppend,intNoOfRandomDigits)
Dim i
for i=1 to intNoOfRandomDigits
strToAppend= strToAppend & int(rnd*10) 'rnd gives a random number between 0 and 1, something like 0.8765341. Then, we multiply it by 10 so that the number comes in the range of 0 to 9. In this case, it becomes 8.765341. After that, we use the int method to truncate the decimal part so that we are only left with the Integer part. In this case, 765341 is truncated and we are left with only the integer 8
next
fn_appendRandomNumbers = strToAppend
end function
Reference 1
Reference 2
& is the string concatenation character. + is an old compatability concat character and will error if you mix text and numbers. Use + for maths only.
Every day I am receiving values as given below, since to automate I want fetch last part of the string from a text. For example, from the following
a) 999999000045090
b) 9990090105
c) 9999010000
d) 990000660000
from the above I need to fetch actual values in the right as given here
a) 45090
b) 90105
c) 10000
d) 660000
since the length is varying and not fixed I need help to resolve
You can use a regular expression:
Dim inputString, regularExpression, outputString
inputString = "999999000045090"
Set regularExpression = New Regexp
regularExpression.Pattern = "^9+0+"
outputString = regularExpression.Replace(inputString, "")
^ will match the start of the string, 9+ will match 1 or more nines and 0+ will match 1 or more zeroes.
Which means that, in this example, 9999990000 will be replaced by an empty string, and that outputString will be 45090.
Disclaimer: not tested. I am not familiar with VBScript.
I am trying to convert letters to numbers.
I have a sub which ensures only numbers are put into the textbox.
My questions is will the following code work. I have a textbox(for numbers) and combobbox(for letters)
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Mid(1, cmbEngletter.Text, 1))
MsgBox "stringposition"
EngNumber = (txtManuNo.Text * 10) + stringposition
My only question above would be will the multiplication work with a .text. I believe it won't because it is a string. Please advise then on how to deal with a situation.
You can use CLng() to convert a string to a Long variable
CLng() will throw an error though if it doesn't like the contents of the string (for example if it contains a non-numeric character), so only use it when you are certain your string will only contain numbers
More forgiving is it to use Val() to convert a string into a numeric variable (a Double by default)
I also suggest you look into the following functions:
Asc() : returns the ASCII value of a character
Chr$() : coverts an ASCII value into a character
Left$() : returns the first characters of a string
CStr() : convert a number into a string
I think in your code you mean to show the contents of your variable stringposition instead of the word "stringposition", so you should remove the ""
I do wonder though what you are trying to accomplish with your code, but applying the above to your code gives:
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Left$(cmbEngletter.Text, 1))
MsgBox CStr(stringposition)
EngNumber = (Val(txtManuNo.Text) * 10) + stringposition
I used Val() because I am not certain your txtManuNo will contain only numbers
To ensure an user can only enter numbers you can use the following code:
Private Sub txtManuNo_KeyPress(KeyAscii As Integer)
Select Case KeyAscii
Case vbKeyBack
'allowe backspace
Case vbKey0 To vbKey9
'allow numbers
Case Else
'refuse any other input
KeyAscii = 0
End Select
End Sub
An user can still input non-numeric charcters with other methods though, like copy-paste via mouse actions, but it is a quick and easy first filter
I'm a total noob as far as visual basic goes.
In class I need to do the following: "Write a program that requests a sentence from the user and then records the number of times each letter of the alphabet occurs."
How would I go about counting the number of occurrences in the string?
Loop the chars and add to the list
Dim s As String = "tafata"
Dim count As New Dictionary(Of Char, Integer)()
For i As Integer = 0 To s.Length - 1
count(s(i)) = (If(count.ContainsKey(s(i)), count(s(i)) + 1, 1))
Next
Havent worked with linq that mutch, but i think you can try
Dim s As String = "tafata"
Dim t = From c In s _
Group c By c Into Group _
Select Group
As Dave said, the easiest solution would be to have an array of length 26 (for English), and loop through each character in the string, incrementing the correct array element. You can use the ASCII value of each character to identify which letter it is, then translate the ASCII number of the letter into the corresponding index number:
'Dimension array to 26 elements
Dim LetterCount(0 to 25) As Long
'Temporary index number
Dim tmpIdx As Long
'Temporary character
Dim tmpChar as String
'String to check
Dim checkStr As String
checkStr = "How many of each letter is in me?"
'Change all letters to lower case (since the upper case
'of each letter has a different ASCII value than its
'lower case)
checkStr = LCase(checkStr)
'Loop through each character
For n = 1 to Len(checkStr)
'Get current character
tmpChar = Mid(checkStr, n, 1)
'Is the character a letter?
If (Asc(tmpChar) >= Asc("a")) And (Asc(tmpChar) <= Asc("z")) Then
'Calcoolate index number from letter's ASCII number
tmpIdx = Asc(tmpChar) - Asc("a")
'Increase letter's count
LetterCount(tmpIdx) = LetterCount(tmpIdx) + 1
End If
Next n
'Now print results
For n = 0 to 25
Print Chr(Asc("a") + n) & " - " & CStr(LetterCount(n))
Next n
The quick and dirty way:
Public Function CountInStr(ByVal value As String, ByVal find As String, Optional compare As VbCompareMethod = VbCompareMethod.vbBinaryCompare) As Long
CountInStr = (LenB(value) - LenB(Replace(value, find, vbNullString, 1&, -1&, compare))) \ LenB(find)
End Function
I would count and remove every instance of the first character, and repeat until there are no characters left. Then display the count for each character detected. Much better than looping through once for every possible character, unless you know the possible range of characters is smaller than the length of the sentence.
Imports System.Linq
Dim CharCounts = From c In "The quick brown fox jumped over the lazy dog." _
Group c By c Into Count() _
Select c, Count
from itertools import groupby
sentence = raw_input("Your sentence:")
for char,group in groupby(sorted(sentence.lower())):
print(char+": "+`len(list(group))`)