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
Related
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 am importing longer form text into a Unity program. I need one word of the longer text to be displayed on each line...
Thanks
The problem with working with large blocks of text in Word is that operations like Find and Replace can only be performed with Find text strings of 255 characters or less without causing an error. Once you import your text and assign it to a string variable, you can use Len() to determine the length of the string and then use Left() Mid() and Right() to breakup the larger string into shorter chunks of 250 characters each. Here's some code I wrote for just a find and replace situation:
With Selection.Find
y = Len(Selection.Text)
Select Case y
Case Is <= 250
x = 1
.Text = stFound
.Execute Replace:=wdReplaceAll
Case Is <= 500
Dim stFound2 As String
x = 2
z = Len(stFound) - 250
stFound1 = Left(stFound, 250)
stFound2 = Right(stFound, z)
Case Is <= 750
Dim stFound2 As String
Dim stFound3 As String
x = 3
stFound1 = Left(stFound, 250)
stFound2 = Mid(stFound, 251, 249)
stFound3 = Right(stFound, Len(stFound) - 500)
End Select
End With
I then used a For Next loop to run a Find and Replace on each string.
In your situation, it's going to be important to not break up the strings in the middle of a word. To do this you can use the InStr() function to find the position of spaces within your string and then break up the text according to where the spaces are. I wouldn't try using the Split() function on the raw text as depending on the size of the string you could run into a Subscript Out of Range error.
Once the text is chunked down into useable pieces, use the Split() function to send each word to an array and then run the following code to put each word on it's own line or paragraph:
Dim stTxt as String
dim stWord as String
dim stArr() as String
dim x as long
stTxt = 'One of your text strings
stArr() = Split(stTxt)
For x = LBound(stArr()) to UBound(stArr())
stWord = stArr(x) & "^p"
Selection.Typetext stWord
Next
After a little more research, I determined that the 255 character limit to text strings only affects some functions, not all. So I took a 17,335 character (including spaces) Word document and ran Split() on it to create an Array. There were no errors and the resulting array had a UBound of 2690.
So the next question is what kind of text is being imported into Word and what size is it. Is it just a list of words separated by spaces, or another delimiter? Does it contain any punctuation? If it's just a list of words separated by spaces or another delimiter such as a comma or semicolon, the Split() function will sort the words into an Array, at least up to 17,000 characters. More testing would be required for a larger text block. If the text contains punctuation, you would have to process the text to remove the unwanted punctuation which can be done with a Wildcard Find and Replace as long as the Find string is <= 255 characters. But if all you have are words and spaces or some other delimiter, using Split() to separate each word into an array element would work and then just run code as in the second half of my previous example:
For x = LBound(stArr()) to UBound(stArr())
stWord = stArr(x) & "^p"
Selection.Typetext stWord
Next
what I want to do is insert a desired character like a "space" into a desired string like "123456789" at a specific point. Example: insert a space at the position 5 in the string 123456789 = 1234 56789. Here is my code:
Dim str As String = sum2.Text '123456789
Dim insStr As String = " " 'space
Dim strRes As String = str.Insert(5, insStr) '5th position
the code looks fine and i dont get any errors when i use it or run it but it will not add the space at the 5th position so i need some help!
You must not that the startIndex in String.Insert(Integer, String) is zero based. That means if your intention is to insert a space in the 5th position, you will have to adjust that by -1:
Dim insertPosition = 5 ' Assuming this came from the user who says put it in position 5
Dim inStr = " "
Dim strRes = str.Insert(insertPosition - 1, inStr) ' assuming your str already had a value.
That will insert the space between 4 and 5 and will produce
1234 56789
I saw in one your comments that you might want to insert spaces in positions 1,5,7. In that case, you will have to do it in reverse, starting with the largest position, to the smallest. This is of cause assuming that you wanted
_1234_6_89
I have used underscores to represent spaces so that you can see it better.
Before working with that, ensure that your string has enough characters to be indexed by your index otherwise you'll get a ArgumentOutOfRangeException.
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*$"
stringToClean = re.Replace(stringToClean, "")
TrimEx = stringToClean
End Function
Non-printables divide different lines of a Web page. I replaced them with X, Y and Z respectively.
Debug.Print Trim(Mid("X test ", 2)) ' first place counts as 2 in VBA
Debug.Print Trim(Mid("XY test ", 3)) ' second place counts as 3 in VBA
Debug.Print Trim(Mid("X Y Z test ", 2)) ' more rounds needed :)
Programmers prefer large text as may neatly be chopped with built in tools (inSTR, Mid, Left, and others). Use of text from several children (i.e taking .textContent versus .innerText) may result several non-printables to cope with, yet DOM and REGEX are not for beginners. Addressing sub-elements for inner text precisely (child elements one-by-one !) may help evading non-printable characters.
How do trim off characters in a string, by how much you want?
For example, say your string is "Tony", but you wanted to display "ny" by trimming of the first two characters, how can this be done?
Sub Main()
Dim s As String
Dim Result As String
s = "Tony"
Result = LTrim(s)
msgbox(Result)
I have this so far using the LTrim function, so how do you specify by how much you want to cut to just display "ny" in the MessageBox?
You don't want LTrim. You want Right:
Result = Right(s, Len(s) - 2);
This will take all but the two left-most characters of s.
You could use the additional string functions to do the same thing,
for example:
X$ = RIGHT$(V$, 2) ' get the ending 2 chars of string
X$ = LEFT$(V$, 2) ' get the leading 2 chars of string
X$ = MID$(V$, 2, 2) ' get 2 chars from the inside of string
Well... If I was trying to clip off the beginning of a string, I would use two functions: StrReverse, and Remove.
I would first reverse the string, then use the remove function to cut off what is now the end, Then flip the remaining string back to it's original state using the reverse function again.
The code would look something like this:
Dim s As String = "Anthony"
Dim index As Integer = 2
Debug.Print(StrReverse(StrReverse(s).Remove(2)))
The output of this would be "ny" and the length will correspond to the index.