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.
Related
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
I have a string called indicators, that the original developer of this application used to store single characters to indicate certain components of a model. I need to change the 7th character in the string, which I tried to do with the following code:
indicators[6] = "R"
The problem, I discovered quickly, was that the string is not always 7 characters long. For example, I have one set of values with U 2, that I need to convert to U 2 R (adding an additional space after the 2). Is there an easy way to force character count with Ruby?
use String.ljust(integer, padstr=' ')
If integer is greater than the length of [the receiver], returns a new String of
length integer with [the return value] left justified and padded with padstr;
otherwise, returns [an unmodified version of the receiver].
indicators = indicators.ljust(7)
indicators[6] = "R"
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 need to write some code in VBScript and have a version number string in a text file that I need to compare against. If I write this code as a test:
option explicit
Dim VersionString
VersionString = "6.2.1"
Dim Version
Version = CDbl (VersionString)
Version = Version * 100
I get an error on the CDbl line:
Microsoft VBScript runtime error: Type mismatch: 'CDbl'
How should I read and compare this string value?
"6.2.1" is not a Double formatted as a String. So CDbl() can't convert it. Your options are:
treat versions as strings; ok if you only need to compare for equality, bad if you need "6.9.1" to be smaller that "6.10.2"
Split() the string on "." and deal with the parts (perhaps converted to Integer/Long) separately; you'll need to write a comparison function for such arrays
Remove the "."s and CLng the resulting string; will break for versions like "6.10.2"
Split() the string on "*" and multiply + add the 'digits' to get one (integer) version number (6 * 100 + 2 * 10 + 1 * 1 = 621 for your sample); may be more complex for versions like "15.00.30729.01"
The conversion to a double isn't working because there are two decimal points in your string. To convert the string, you will have to remove one or both of them.
For this, you can use the Replace function. The syntax for Replace is
Replace(string, find, replacewith [, start [, count [, compare]]])
where string is the string to search, find is the substring to find, replacewith is the substring to replace find with, start is an optional parameter specifying the index to start searching at, count is an optional parameter specifying how many replaces to make, and compare is an optional parameter that is either 0 (vbBinaryCompare) to perform a binary comparison, or 1 (vbTextCompare) to perform a textual comparison
' Remove all decimals
Version = CDbl(Replace(VersionString, ".", "")
' Remove only the first decimal
Version = CDbl(Replace(VersionString, ".", "", 1, 1)
' Remove only the second decimal
Version = CDbl(Replace(VersionString, ".", "", 3, 1)
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))`)