I tried almost all the methods (CLEAN,TRIM,SUBSTITUTE) trying to remove the character hiding in the beginning and the end of a text. In my case, I downloaded the bill of material report from oracle ERP and found that the item codes are a victim of hidden characters.
After so many findings, I was able to trace which character is hidden and found out that it's a question mark'?' (via VBA code in another thread) both at the front and the end. You can take this item code: 11301-21
If you paste the above into your excel and see its length =LEN(), you can understand my problem much better.
I need a good solution for this problem. Therefore please help!
Thank you very much in advance.
Thanks to Gary's Student, because his answer inspired me.
Also, I used this answer for this code.
This function will clean every single char of your data, so it should work for you. You need 2 functions: 1 to clean the Unicode chars, and other one to clean your item codes_
Public Function CLEAN_ITEM_CODE(ByRef ThisCell As Range) As String
If ThisCell.Count > 1 Or ThisCell.Count < 1 Then
CLEAN_ITEM_CODE = "Only single cells allowed"
Exit Function
End If
Dim ZZ As Byte
For ZZ = 1 To Len(ThisCell.Value) Step 1
CLEAN_ITEM_CODE = CLEAN_ITEM_CODE & GetStrippedText(Mid(ThisCell.Value, ZZ, 1))
Next ZZ
End Function
Private Function GetStrippedText(txt As String) As String
If txt = "–" Then
GetStrippedText = "–"
Else
Dim regEx As Object
Set regEx = CreateObject("vbscript.regexp")
regEx.Pattern = "[^\u0000-\u007F]"
GetStrippedText = regEx.Replace(txt, "")
End If
End Function
And this is what i get using it as formula in Excel. Note the difference in the Len of strings:
Hope this helps
You have characters that look like a space character, but are not. They are UniCode 8236 & 8237.
Just replace them with a space character (ASCII 32).
EDIT#1:
Based on the string in your post, the following VBA macro will replace UniCode characters 8236 amd 8237 with simple space characters:
Sub Kleanup()
Dim N1 As Long, N2 As Long
Dim Bad1 As String, Bad2 As String
N1 = 8237
Bad1 = ChrW(N1)
N2 = 8236
Bad2 = ChrW(N2)
Cells.Replace what:=Bad1, replacement:=" ", lookat:=xlPart
Cells.Replace what:=Bad2, replacement:=" ", lookat:=xlPart
End Sub
Related
I have a string that contains digits inside. For example, "adf20j83n,m3jh2k9". Is there a direct way to count the number of digits inside the string. As in my example, it should give me "7" as an output.
Also, I have tried RegExp but it's not working in VBScript in QTP.
Btw, I'm not looking for loops and stuff like that. Just a direct way, or a suggestion to make this RegExp work in QTP.
You'll probably need to create the COM object via its ProgId:
Dim re
Set re = CreateObject("VBScript.RegExp")
re.Pattern = "\d"
re.Global = True
MsgBox "Digits: " & re.Execute("adf20j83n,m3jh2k9").Count
Output:
Digits: 7
I know it's techically what you said but it's a good way nonetheless
Try using this function:
Public Sub CountNumeric(ByVal input As String)
Dim numericCount As Integer = 0
For Each c As Char In input
If Char.IsDigit(c) Then numericCount += 1
Next
MessageBox.Show(String.Format("Number of numerics : {0}", numericCount)
End Sub
EDIT:
You could also try this:
Dim charColl As MatchCollection = Regex.Matches(input , "^\d+$")
Console.WriteLine(charColl.Count.ToString())
I am trying to convert letters to numbers.
I have a sub which ensures only numbers are put into the textbox.
My questions is will the following code work. I have a textbox(for numbers) and combobbox(for letters)
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Mid(1, cmbEngletter.Text, 1))
MsgBox "stringposition"
EngNumber = (txtManuNo.Text * 10) + stringposition
My only question above would be will the multiplication work with a .text. I believe it won't because it is a string. Please advise then on how to deal with a situation.
You can use CLng() to convert a string to a Long variable
CLng() will throw an error though if it doesn't like the contents of the string (for example if it contains a non-numeric character), so only use it when you are certain your string will only contain numbers
More forgiving is it to use Val() to convert a string into a numeric variable (a Double by default)
I also suggest you look into the following functions:
Asc() : returns the ASCII value of a character
Chr$() : 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 am looking for a way to count the number of lines in a text file, excluding the CRLF which will be the very last line per say.
Any chance there is a simple code example for this?
Try this (though I found it searching the very title of your question in Google):
IO.File.ReadAllLines("C:\Users\Dan\Desktop\test.txt").Length
If you are worried about empty lines at the end, then loop from all lines until you find one with contents, and delete all of them until the end. This is easy, since:
IO.File.ReadAllLines("C:\Users\Dan\Desktop\test.txt")
Returns a vector of Strings.
Hope this helps.
Try with this:
TextBox1.Text = ""
With OpenFileDialog1
.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Desktop)
.Filter = "Text Files (*.txt)|*.txt"
If .ShowDialog <> DialogResult.OK Then Exit Sub
End With
Dim intLines As Integer = 0
Dim sr As New IO.StreamReader(OpenFileDialog1.FileName)
Do While sr.Peek() >= 0
TextBox1.Text += sr.ReadLine() & ControlChars.CrLf
intLines += 1
Loop
sr.Close()
MessageBox.Show(intLines, Me.Text)
basically you need to calibrate the md5#sum and then check to see if the HTML is linked to the CSS and make sure you have turned on javascript
Hey all, i am trying to replace large spaces between text with just one. My output looks like this right now:
5964215">
This is just the first example of the spaces
5964478">
This would be the 2nd example of showing how many spaces this thing has in each sentence.
5964494">
That comes from a textbox that has multi-line to true. Here is what it looks like when it doesn't have multi-line to true.
http://www.june3rdsoftware.com/forums/vb6.jpg
I can not seem to get the spaces to go away! BTW, this text is from a webpage if that makes any difference.
David
According to the suggestion of MvanGeest, here is some VB code to replace blocks of white spaces:
Sub test()
Dim x As String, y As String
x = "abcd defg 1233"
Dim re As New RegExp
re.Pattern = "\s+"
re.Global = True
y = re.Replace(x, " ")
Debug.Print y
End Sub
To make this work, you will have to add a reference to "Microsoft VBScript Regular Expresssions" to your project.
Assuming no regex support, why not set up a simple state machine that will set the state=1 when a space is found and set state=0 once a non-space is encountered. You can move char by char when state=0 (thus copying over only 1 space per series of spaces).
Also assuming no regex, you could try something like
str = "long text with spaces "
i = LenB(str)
str = Replace(str, " ", " ")
Do While LenB(str) <> i
i = LenB(str)
str = Replace(str, " ", " ")
Loop
Of course this code could be optimized for long space sequences but it might be all you need as well
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))`)