I'm trying to compare strings in a macro and the data isn't always entered consistently. The difference comes down to the amount of leading white space (ie " test" vs. "test" vs. " test")
For my macro the three strings in the example should be equivalent. However I can't use Replace, as any spaces in the middle of the string (ex. "test one two three") should be retained. I had thought that was what Trim was supposed to do (as well as removing all trailing spaces). But when I use Trim on the strings, I don't see a difference, and I'm definitely left with white space at the front of the string.
So A) What does Trim really do in VBA? B) Is there a built in function for what I'm trying to do, or will I just need to write a function?
Thanks!
So as Gary's Student aluded to, the character wasn't 32. It was in fact 160. Now me being the simple man I am, white space is white space. So in line with that view I created the following function that will remove ALL Unicode characters that don't actual display to the human eye (i.e. non-special character, non-alphanumeric). That function is below:
Function TrueTrim(v As String) As String
Dim out As String
Dim bad As String
bad = "||127||129||141||143||144||160||173||" 'Characters that don't output something
'the human eye can see based on http://www.gtwiki.org/mwiki/?title=VB_Chr_Values
out = v
'Chop off the first character so long as it's white space
If v <> "" Then
Do While AscW(Left(out, 1)) < 33 Or InStr(1, bad, "||" & AscW(Left(out, 1)) & "||") <> 0 'Left(out, 1) = " " Or Left(out, 1) = Chr(9) Or Left(out, 1) = Chr(160)
out = Right(out, Len(out) - 1)
Loop
'Chop off the last character so long as it's white space
Do While AscW(Right(out, 1)) < 33 Or InStr(1, bad, "||" & AscW(Right(out, 1)) & "||") <> 0 'Right(out, 1) = " " Or Right(out, 1) = Chr(9) Or Right(out, 1) = Chr(160)
out = Left(out, Len(out) - 1)
Loop
End If 'else out = "" and there's no processing to be done
'Capture result for return
TrueTrim = out
End Function
TRIM() will remove all leading spaces
Sub demo()
Dim s As String
s = " test "
s2 = Trim(s)
msg = ""
For i = 1 To Len(s2)
msg = msg & i & vbTab & Mid(s2, i, 1) & vbCrLf
Next i
MsgBox msg
End Sub
It is possible your data has characters that are not visible, but are not spaces either.
Without seeing your code it is hard to know, but you could also use the Application.WorksheetFunction.Clean() method in conjunction with the Trim() method which removes non-printable characters.
MSDN Reference page for WorksheetFunction.Clean()
Why don't you try using the Instr function instead? Something like this
Function Comp2Strings(str1 As String, str2 As String) As Boolean
If InStr(str1, str2) <> 0 Or InStr(str2, str1) <> 0 Then
Comp2Strings = True
Else
Comp2Strings = False
End If
End Function
Basically you are checking if string1 contains string2 or string2 contains string1. This will always work, and you dont have to trim the data.
VBA's Trim function is limited to dealing with spaces. It will remove spaces at the start and end of your string.
In order to deal with things like newlines and tabs, I've always imported the Microsoft VBScript RegEx library and used it to replace whitespace characters.
In your VBA window, go to Tools, References, the find Microsoft VBScript Regular Expressions 5.5. Check it and hit OK.
Then you can create a fairly simple function to trim all white space, not just spaces.
Private Function TrimEx(stringToClean As String)
Dim re As New RegExp
' Matches any whitespace at start of string
re.Pattern = "^\s*"
stringToClean = re.Replace(stringToClean, "")
' Matches any whitespace at end of string
re.Pattern = "\s*$"
stringToClean = re.Replace(stringToClean, "")
TrimEx = stringToClean
End Function
Non-printables divide different lines of a Web page. I replaced them with X, Y and Z respectively.
Debug.Print Trim(Mid("X test ", 2)) ' first place counts as 2 in VBA
Debug.Print Trim(Mid("XY test ", 3)) ' second place counts as 3 in VBA
Debug.Print Trim(Mid("X Y Z test ", 2)) ' more rounds needed :)
Programmers prefer large text as may neatly be chopped with built in tools (inSTR, Mid, Left, and others). Use of text from several children (i.e taking .textContent versus .innerText) may result several non-printables to cope with, yet DOM and REGEX are not for beginners. Addressing sub-elements for inner text precisely (child elements one-by-one !) may help evading non-printable characters.
Related
I have the following Text sample:
Ins-Si_079_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_May
I want to get the number 079, So what I need is the first instance of digits of length 3. There are certain times the 3 digits are at the end, but they usually found with the first 2 underscores. I only want the digits with length three (079) and not 19, 1920, or 2554 which are different lengths.
Sometimes it can look like this with no underscore:
1920 O-B CLI 353 Tar Traf
Or like this with the 3 digit number at the end:
Ins-Si_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_079
There are also times where what I need is 2 digits but when it's 2 digits its always at the end like this:
FY1920-Or-OLV-B-45
How would I get what I need in all cases?
You can split the listed items and check for 3 digits via Like:
Function Get3Digits(s As String) As String
Dim tmp, elem
tmp = Split(Replace(Replace(s, "-", " "), "_", " "), " ")
For Each elem In tmp
If elem Like "###" Then Get3Digits = elem: Exit Function
Next
If Get3Digits = vbNullString Then Get3Digits = IIf(Right(s, 2) Like "##", Right(s, 2), "")
End Function
Edited due to comment:
I would execute a 2 digit search when there are no 3 didget numbers before the end part and the last 2 digits are 2. if 3 digits are fount at end then get 3 but if not then get 2. there are times when last is a number but only one number. I would only want to get last if there are 2 or 3 numbers. The - would not be relevant to the 2 digets. if nothing is found that is desired then would return " ".
If VBA is not a must you could try:
=TEXT(INDEX(FILTERXML("<t><s>"&SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"_"," "),"-"," ")," ","</s><s>")&"</s></t>","//s[.*0=0][string-length()=3 or (position()=last() and string-length()=2)]"),1),"000")
It worked for your sample data.
Edit: Some explaination.
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"_"," "),"-"," ")," ","</s><s>") - The key part to transform all three potential delimiters (hyphen, underscore and space) to valid XML node end- and startconstruct.
The above concatenated using ampersand into a valid XML construct (adding a parent node <t>).
FILTERXML can be used to now 'split' the string into an array.
//s[.*0=0][string-length()=3 or last() and string-length()=2] - The 2nd parameter of FILTERXML which should be valid XPATH syntax. It reads:
//s 'Select all <s> nodes with
following conditions:
[.*0=0] 'Check if an <s> node times zero
returns zero (to check if a node
is numeric. '
[string-length()=3 or (position()=last() and string-length()=2)] 'Check if a node is 3 characters
long OR if it's the last node and
only 2 characters long.
INDEX(.....,1) - I mentioned in the comments that usually this is not needed, but since ExcelO365 might spill the returned array, we may as well implemented to prevent spilling errors for those who use the newest Excel version. Now we just retrieving the very first element of whatever array FILTERXML returns.
TEXT(....,"000") - Excel will try delete leading zeros of a numeric value so we use TEXT() to turn it into a string value of three digits.
Now, if no element can be found, this will return an error however a simple IFERROR could fix this.
Try this function, please:
Function ExtractThreeDigitsNumber(x As String) As String
Dim El As Variant, arr As Variant, strFound As String
If InStr(x, "_") > 0 Then
arr = Split(x, "_")
Elseif InStr(x, "-") > 0 Then
arr = Split(x, "-")
Else
arr = Split(x, " ")
End If
For Each El In arr
If IsNumeric(El) And Len(El) = 3 Then strFound = El: Exit For
Next
If strFound = "" Then
If IsNumeric(Right(x, 2)) Then ExtractThreeDigitsNumber = Right(x, 2)
Else
ExtractThreeDigitsNumber = strFound
End If
End Function
It can be called in this way:
Sub testExtractThreDig()
Dim x As String
x = "Ins-Si_079_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_May"
Debug.Print ExtractThreeDigitsNumber(x)
End Sub
I have cells that contain various information.
In these cells, there are multiple Uppercase phrases.
I would like to be able to split the contents of the cell by adding the CHAR(13) + CHAR(10) Carriage return - linefeed combination
to the start of each new Uppercase phrase.
The only consistency is that the multiple Uppercase phrases begin after a period (.) and before open parenthesis "("
Example:
- Add CRLF to start of PERSUADER
- Add CRLF to start of RIVER JEWEL
- Add CRLF to start of TAHITIAN DANCER
- Add CRLF to start of AMBLEVE
- Add CRLF to start of GINA'S HOPE
NOTE:
There are multiple periods (.) in the text.
I have highlighted the text in red for a visual purpose only (normal text/font during import).
I am OK with either formula, UDF or VBA sub.
TEXT
PERSUADER (1) won by a margin first up at Kyneton. Bit of authority about her performance there and with the stable finding form it's easy to see her going right on with that. Ran really well when placed at Caulfield second-up last prep and that rates well against these. RIVER JEWEL (2) has been racing well at big odds. I have to like the form lines that she brings back in class now. Shapes as a key danger. TAHITIAN DANCER (5) will run well. She was okay without a lot of room at Flemington last time. AMBLEVE (13) is winning and can measure up while GINA'S HOPE (11) wasn't too far from River Jewel at Flemington and ties in as a hope off that form line.
I was able to extract with this function - but not able to manipulate the data in the cell
This is my code so far:
Function UpperCaseWords(ByVal S As String) As String
Dim X As Long, Words() As String
Const OkayPunctuation As String = ",."";:'&,-?!"
For X = 1 To Len(OkayPunctuation)
S = Replace(S, Mid(OkayPunctuation, X, 1), " ")
Next
Words = Split(WorksheetFunction.Trim(S))
For X = 0 To UBound(Words)
If Words(X) Like "*[!A-Z]*" Then Words(X) = ""
Next
UpperCaseWords = Trim(Join(Words))
End Function
Your description is not the same as your examples.
None of your examples start after a dot.
Most start after a dot-space except
PERSUADER starts at the start of the string
GINA'S HOPE starts after a space
I incorporated those rules into a regular expression, but, since your upper case words can include punctuation, for brevity I just looked for
- words that excluded lower case letters and digits
- words at least three characters long
If that is not sufficient in your real data, the regex can easily be made more specific:
Option Explicit
Function upperCaseWords(S As String) As String
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.MultiLine = True
.Pattern = "^|\s(\b[^a-z0-9]+\b\s*\()"
upperCaseWords = .Replace(S, vbCrLf & "$1")
End With
End Function
as per your wording
The only consistency is that the multiple Uppercase phrases begin
after a period (.) and before open parenthesis "("
this should do:
Function UpperCaseWords(ByVal s As String) As String
Dim w As Variant
Dim s1 As String
For Each w In Split(s, ". ")
If InStr(w, "(") Then w = Chr(13) + Chr(10) & w
s1 = s1 & w
Next
UpperCaseWords = s1
End Function
Since the OP accepted the formula solution, and here is a formula answer .
Assume data put in A1
In B1, enter formula and copied across until blank :
=TRIM(RIGHT(SUBSTITUTE(TRIM(MID(SUBSTITUTE(SUBSTITUTE(" (. "&$A1," while ",". ")," (",REPT(" ",700)),COLUMN(A1)*700,700))&" ",". ",REPT(" ",300)),300))
I have text in Excel like this:
120
124569 abasd 12345
There are sapces both to the left and to the right side.
I copy this from Excel and paste as text. When I check this, it shows like this when I click on button.
Code:
abArray= abArray & "," & gridview1.Rows(i).Cells(2).Text
For k = 3 To 17
bArray= abArray& "," & Val(gridview1.Rows(i).Cells(k).Text)
Next
In abArray this shows as:
0, abasd ,12345,0,0,0,0,0
I want to remove/trim spaces both from left and right.
I have tried abArray.Trim() but this still show spaces.
If you want to remove all the spaces out of the end result consider String.Replace:
Returns a new string in which all occurrences of a specified Unicode character or String in the current string are replaced with another specified Unicode character or String.
Example use:
Dim s As String = "0, abasd ,12345,0,0,0,0,0"
s = s.Replace(" ", "")
This would output:
0,abasd,12345,0,0,0,0,0
It may also be worth using a StringBuilder to join all your values together as this is good practice when looping as you are. At this point you could use String.Trim. This would preserve any spaces that are within your value. In order words it would only remove the spaces from the beginning and the end of the value.
Example use:
Dim sb As New StringBuilder
For k = 0 To 17
sb.Append(String.Format("{0},", gridview1.Rows(i).Cells(k).Text.Trim()))
Next
Dim endResult As String = sb.ToString().TrimEnd(","c)
endResult would output:
0,abasd,12345,0,0,0,0,0
You will have to import System.Text in order to make use of the StringBuilder class.
Use the VB.NET Trim function to remove leading and trailing spaces, change this one line of code:
abArray= abArray& "," & Val(Trim(gridview1.Rows(i).Cells(k).Text))
abArray.Trim() does not work because you did not give the Trim function anything to trim.
Try it like this
abArray = abArray & "," & gridview1.Rows(i).Cells(2).Text.Trim
For k = 3 To 17
abArray= abArray& "," & Val(gridview1.Rows(i).Cells(k).Text.Trim)
Next
I have a table where one of the fields has 13 leading spaces (no visible characters in them). I tried TRIM() and REPLACE([Field1], " ", "") but neither one worked. Could anyone venture a guess as to what's going on and how to fix this?
This is not an answer but it will allow you to see what the chr value is for the string. Just call it like MsgBox WhatAreTheAscValues([Field1]) and you will see what value the characters that make up the string and allow you to adjust your code accordingly.
Function WhatAreTheAscValues(str As String)
Dim i As Integer
Dim answer As String
answer = "The Chr(x) values in this string are listed below" & vbCrLf
For i = 1 To Len(str)
answer = answer + CStr(Asc(Mid(str, i, 1))) & ", "
Next i
WhatAreTheAscValues = Left(answer, Len(answer) - 2)
End Function
I am currently encountering a problem which doesn't seem that hard to fix but, yet, I can't find a clean way of doing it on my own.
I am using the "Replace" function to change some expressions in a sentence typed by an user. For example, if the user types "va", I want it to be turned into "V. A." instead so it will match more easily with my database for further operations.
Here is my simple code to do it :
sMain.Range("J3").Replace "VA", "V. A."
It works well.
Problem is, it's not only spotting "VA" as an individual expression, but also as a part of words.
So if my user types "Vatican", it's gonna turn it into : "V. A.tican"... which of course I don't want.
Do you know how to easily specify my code to make it ONLY consider replacing the whole words matching the expression? (I have dozens of lines of these replacement so ideally, it would be better to act directly on the "replace" functions - if possible).
Thanks in advance !
Do this:
sMain.Range("J3").Replace " VA ", "V. A."
then handle the cases where the original string starts or ends with VA
also, handle all cases of separators which could be (for example) tab, space or comma.
To do that:
const nSep As Integer = 3
Dim sep(nSep) As String
sep(1) = " "
sep(2) = vbTab
sep(3) = ","
for i=1 to nSep
for j=1 to nSep
sMain.Range("J3").Replace sep(i) & "VA" & sep(j), "V. A."
next
next
Can split it up and check each word. I have put it into a function for easy of use and flexibility.
Function ReplaceWordOnly(sText As String, sFind As String, sReplace As String) As String
On Error Resume Next
Dim aText As Variant, oText As Variant, i As Long
aText = Split(sText, " ")
For i = 0 To UBound(aText)
oText = aText(i)
' Check if starting with sFind
If LCase(Left(oText, 2)) = LCase(sFind) Then
Select Case Asc(Mid(oText, 3, 1))
Case 65 To 90, 97 To 122
' obmit if third character is alphabet (checked by ascii code)
Case Else
aText(i) = Replace(oText, sFind, sReplace, 1, -1, vbTextCompare)
End Select
End If
Next
ReplaceWordOnly = Join(aText, " ")
End Function
Example output:
?ReplaceWordOnly("there is a vatican in vA.","Va","V. A.")
there is a vatican in V. A..