My problem is to get a string as described above ! In my case
1234;0020001212;6565656AEBCD698798832
The string,I want is "0020001212" , My known words are ";".
I tried Mid() but my value is not stable , variable !
so, please help me to find this answer. Thank y
Explode it:
theLine = "1234;0020001212;6565656AEBCD698798832"
Dim arr() As String
arr = Split(theLine, ";")
Debug.Print arr(0)
Debug.Print arr(1)
Debug.Print arr(2)
For
1234
0020001212
6565656AEBCD698798832
Use the Split function and use ";" as the delimiter to convert the string into an array of strings.
Dim arr() As String
arr = Split("a;b;c", ";")
Then you can access the second element of the array to get your value.
Use split function, using ";" as separator
Related
Using this SO question / answer as a starting point: Splitting a single word into an array of the consituent letters
I have this simple bit of code to take a word and split the word into single letters:
<%
Dim word1, i
word1 = "particle"
For i = 1 To Len(word1)
Response.Write "<p>" & Mid(word1, i, 1) & "</p>"
Next
%>
I would like to know how to take a word (variable length, rather than a word that is 8 characters long as in the example above), and randomly rearrange the letters of the word - so that e.g. particle could be e.g.:
alpreict
lircpaet
ctelaipr
teapclir
raeitclp
This is an example of what I'd like to achieve: https://onlinerandomtools.com/shuffle-letters
However, I realise that is easier said than done.
I wondered if anyone has any advice about how it might be possible to achieve this using Classic ASP please?
Thanks
Here's one way to do it:
Function ShuffleText(p_sText)
Dim iLength
Dim iIndex
Dim iCounter
Dim sLetter
Dim sText
Dim sShuffledText
' Copy text passed as parameter
sText = p_sText
' Get text length
iLength = Len(sText)
For iCounter = iLength To 1 Step -1
' Get random index
iIndex = 1 + Int(Rnd * (iCounter))
' Get character at that index
sLetter = Mid(sText, iIndex, 1)
' Remove character from string
sText = Left(sText, iIndex - 1) & Mid(sText, iIndex + 1)
' Add character to shuffled string
sShuffledText = sShuffledText & sLetter
Next
' Return shuffled text
ShuffleText = sShuffledText
End Function
This code selects a random character in the string, removes it and adds it to a shuffled string. It repeats this process until it has gone through all characters.
There are probably more efficient ways to do this by randomizing an array of numbers first and using those numbers as iIndex, without manipulating the sText string.
I need help on how to remove spaces/emtpy in data without compromising spaces on other data. Here's my sample data.
12345," ","abcde fgh",2017-06-06,09:00,AM," ", US
expected output:
12345,,"abcde fgh",2017-06-06,09:00,AM,, US
since " " should be considered as null.
I tried the Trim() function but it did not work. I also tried Regex pattern but still no use.
Here's my sample function.
Private Sub Transform(delimiter As String)
Dim sFullPath As String
Dim strBuff As String
Dim re As RegExp
Dim matches As Object
Dim m As Variant
If delimiter <> "," Then
strBuff = Replace(strBuff, delimiter, ",")
Else
With re
.Pattern = "(?!\B""[^""]*)" & delimiter & "(?![^""]*""\B)"
.IgnoreCase = False
.Global = True
End With
Set matches = re.Execute(strBuff)
For Each m In matches
strBuff = re.Replace(strBuff, ",")
Next
Set re = Nothing
Set matches = Nothing
End If
End Sub
I think you're on the right track. Try using this for your regular expression. The two double quotes in a row are how a single double quote is included in a string literal. Some people prefer to use Chr(34) to include double quotes inside a string.
\B(\s)(?!(?:[^""]*""[^""]*"")*[^""]*$)
Using that expression on your example string
12345," ","abcde fgh",2017-06-06,09:00,AM," ", US
yields
12345,"","abcde fgh",2017-06-06,09:00,AM,"", US
Example function
Private Function Transform(ByVal strLine As String) As String
Dim objRegEx As RegExp
On Error GoTo ErrTransForm
Set objRegEx = New RegExp
With objRegEx
.Pattern = "\B(\s)(?!(?:[^""]*""[^""]*"")*[^""]*$)"
.IgnoreCase = False
.Global = True
Transform = .Replace(strLine, "")
End With
ExitTransForm:
If Not objRegEx Is Nothing Then
Set objRegEx = Nothing
End If
Exit Function
ErrTransForm:
'error handling code here
GoTo ExitTransForm
End Function
And credit where credit is due. I used this answer, Regex Replace Whitespaces between single quotes as the basis for the expression here.
I would add an output string variable and have a conditional statement saying if the input is not empty, add it on to the output string. For example (VB console app format with the user being prompted to enter many inputs):
Dim input As String
Dim output As String
Do
input = console.ReadLine()
If Not input = " " Then
output += input
End If
Loop Until (end condition)
Console.WriteLine(output)
You can throw any inputs you don't want into the conditional to remove them from the output.
Your CSV file isn't correctly formatted.
Double quote shouldn't exists, then open your CSV with Notepad and replace them with a null string.
After this, you now have a real CSV file that you can import whitout problems.
I'm trying to make a program that can take the letters of a string and invert their case. I know that in vb.net there exists the IsUpper() command, but I don't know of such a thing in vb6.
What can I use in place of it?
Thanks!
Something like this should work:
Private Function Invert(strIn As String) As String
Dim strOut As String
Dim strChar As String
Dim intLoop As Integer
For intLoop = 1 To Len(strIn)
strChar = Mid(strIn, intLoop, 1)
If UCase(strChar) = strChar Then
strChar = LCase(strChar)
Else
strChar = UCase(strChar)
End If
strOut = strOut + strChar
Next
Invert = strOut
End Function
This loops through the supplied string, and extracts each character. It then tries to convert it to upper case and checks it against the extracted character. If it's the same then it was already upper case, so it converts it to lower case.
It handles non alpha characters just fine as UCase/LCase ignores those.
I have a string like X5BC8373XXX. Where X = a special character equals a Square.
I also have some special characters like \n but I remove them, but I can't remove the squares...
I'd like to know how to remove it.
I Found this method:
Dim Test As String
Test = Replace(Mscomm1.Input, Chr(160), Chr(64) 'Here I remove some of the special characters like \n
Test = Left$(Test, Len(Test) -2)
Test = Right$(Test, Len(Test) -2)
This method DOES remove those special characters, but it's also removing my first character 5.
I realize that this method just remove 2 characters from the left and the right,
but how could I work around this to remove these special characters ?
Also I saw something with vblF, CtrlF something like this, but I couldn't work with this ;\
You can use regular expressions. If you want to remove everything that's not a number or letter, you can use the code below. If there are other characters you want to keep, regular expressions are highly customizable, but can get a little confusing.
This also has the benefit of doing the whole string at once, instead of character by character.
You'll need to reference Microsoft VBScript Regular Expressions in your project.
Function AlphaNum(OldString As String)
Dim RE As New RegExp
RE.Pattern = "[^A-Za-z0-9]"
RE.Global = True
AlphaNum = RE.Replace(OldString, "")
End Function
Cleaning out non-printable characters is easy enough. One brute-force but easily customizable method might be:
Private Function Printable(ByVal Text As String) As String
Dim I As Long
Dim Char As String
Dim Count As Long
Printable = Text 'Allocate space, same width as original.
For I = 1 To Len(Text)
Char = Mid$(Text, I, 1)
If Char Like "[ -~]" Then
'Char was in the range " " through "~" so keep it.
Count = Count + 1
Mid$(Printable, Count, 1) = Char
End If
Next
Printable = Left$(Printable, Count)
End Function
Private Sub Test()
Dim S As String
S = vbVerticalTab & "ABC" & vbFormFeed & vbBack
Text1.Text = S 'Shows "boxes" or "?" depending on the font.
Text2.Text = Printable(S)
End Sub
This will remove control characters (below CHR(32))
Function CleanString(strBefore As String) As String
CleanString = ""
Dim strAfter As String
Dim intAscii As Integer
Dim strTest As String
Dim dblX As Double
Dim dblLen As Double
intLen = Len(strBefore)
For dblX = 1 To dblLen
strTest = Mid(strBefore, dblX, 1)
If Asc(strTest) < 32 Then
strTest = " "
End If
strAfter = strAfter & strTest
Next dblX
CleanString = strAfter
End Function
I have a space delimited string variable. I would like to store the contents of the variable into an array. Using split, I can store every space delimited value in an array. I would prefer if I could separate the string variable at every 7th space. For example, the text could read:
"hello hello hello hello hello hello hello hi hi hi hi hi hi hi hey hey hey hey hey hey hey"
This isn't the actual content of the string, but a simpler version that is easier to read. I want to separate at the places where the words change, or every 7th space. Any help would be greatly appreciated. My current code looks like this, which splits at every space.
ReDim StatsArray(ArrInc)
StatsArray = Split(Stats)
For i = 0 To UBound(StatsArray())
If i > UBound(StatsArray()) Then
ReDim Preserve StatsArray(i + ArrInc)
End If
' MsgBox StatsArray(i) ' When not commented out, this help me check the array
Next
There isn't any built-in function that will do this for you. A couple of solutions come to mind: (1) Do your split, then iterate your array. Concatenate seven array elements in a string variable. Write the result to a new array. Rinse and repeat. (2) Create an Array. Iterate through the string character by character, pushing each character into a variable and keeping track of the spaces you encounter; when you reach the seventh space add an element to your array, copy the variable to it, and clear the variable. Rinse and repeat.
The first one strikes me as a bit faster, but I could be quite wrong about that.
If I understand what you're trying to do, a string of "1234 12345678 123" would get split into 1234, 1234567, 8, 123. Is this correct?
If so, then you can use Regular Expressions to do the split for you.
Function Split7(S As String)
Dim regex As New RegExp
regex.Pattern = "[^ ]{7}"
regex.Global = True
Split7 = Split(regex.Replace(S, "$& "), " ")
End Function
This will insert a space after every 7th character that is not a space. Then use the split function to get the whole thing into an array.
Another stab at it, though I'd probably optimize the ReDim Preserves, doing them in chunks:
Option Explicit
Private Function SplitN( _
ByRef Source As String, _
ByVal Delim As String, _
ByVal N As Long) As String()
'Delim can be multi-character.
'Always returns at least 1-element result,
'even if Source = "".
Dim SearchPos As Long
Dim PartPos As Long
Dim DelimPos As Long
Dim Parts() As String
Dim DelimCount As Long
Dim PartsCount As Long
SearchPos = 1
PartPos = SearchPos
Do
DelimPos = InStr(SearchPos, Source, Delim)
If DelimPos > 0 Then
DelimCount = DelimCount + 1
If DelimCount = N Then
DelimCount = 0
ReDim Preserve Parts(PartsCount)
Parts(PartsCount) = _
Mid$(Source, PartPos, DelimPos - PartPos)
PartsCount = PartsCount + 1
PartPos = DelimPos + Len(Delim)
End If
SearchPos = DelimPos + Len(Delim)
End If
Loop While DelimPos > 0
ReDim Preserve Parts(PartsCount)
Parts(PartsCount) = Mid$(Source, PartPos)
SplitN = Parts
End Function
Private Sub Form_Load()
Dim S As String
Dim Parts() As String
Dim P As Long
S = "hello hello hello hello hello hello hello " _
& "hi hi hi hi hi hi hi " _
& "hey hey hey hey hey hey hey"
Text1.SelStart = 0
Text1.SelText = S & vbNewLine & vbNewLine
Parts = SplitN(S, " ", 7)
Text1.SelText = "Ubound() = " & UBound(Parts) & vbNewLine
For P = 0 To UBound(Parts)
Text1.SelText = Parts(P) & vbNewLine
Next
End Sub