So I have an array of indexes of characters in a string that I wish to insert a character before, how do i easily insert a character before each index? So for example:
"The big brown fox ... "
the positions
array = 4,9
the character to insert ','
the result: "The, big, brown fox ..."
Is there a method that provides such an easy utility?
String.insert(originalStr, index, stringToInsert) for example???
Update
The example I provided is just an example implementation. I also may want to do the following:
orginalText = "some text with characters like ; : } <"
in which I may want to insert "\" with the result being:
result = "some text with characters like \; : } \<"
This is hacky and a bit rushed but try this:
Dim sString: sString = "the something something"
Dim position: position = 1
Dim character: character = "F"
if position = 0 then
sString = character + Left(Mid(sString, 1), Len(sString) + 1)
else
sString = Left(sString, position) + character + Left(Mid(sString, position), Len(sString) - position + 1)
end if
Assuming that the indexes are sorted, loop backwards and insert each character.
For lngPos = UBound(alngPositions) to 0 step -1
strText = Left(strText, alngPositions(lngPos) - 1) + "," + Mid(strText, alngPositions(lngPos))
Next
Note that with your example data it will of course produce the string "The, big ,brown fox ... ". The indexes are not pre-added to match the position in the resulting string, are they?
Edit:
An alternative that would be faster for large strings, is to split up the string at the index positions into an array, then join the strings with commas in between:
Dim astrSubstrings(UBound(alngPositions) + 1)
lngLeft = 1
For lngPos = 0 to UBound(alngPositions)
astrSubstrings(lngPos) = Mid(strText, lngLeft, alngPositions(lngPos) - lngLeft)
lngLeft = alngPositions(lngPos)
Next
astrSubstrings(UBound(alngPositions) + 1) = Mid(strText, lngLeft)
strText = Join(astrSubstrings, ",")
I'm not a classic ASP user but you can use substring to get the part of the string up to the index where you have to insert the character, substring the other part of the string and take these two parts and build a new string doing part1 & "," & part2.
Hope it helps.
You should be able to use the split function based on the space between the words - this will return an array of words. You then put a comma after each item in the array and you can get to the requried string that you are looking for. Example here http://www.w3schools.com/VBscript/func_split.asp
It's been a while, but Mid(str, start, [end]) would be the way to go.
Related
i have problem to split string into newline in vb.net.
right now i can make it to split by a single space.i want split new line after 3 space.
Dim s As String = "SOMETHING BIGGER THAN YOUR DREAM"
Dim words As String() = s.Split(New Char() {" "c})
For Each word As String In words
Console.WriteLine(word)
Next
output :
SOMETHING
BIGGER
THAN
YOUR
DREAM
Desire output :
SOMETHING BIGGER THAN
YOUR DREAM
Another alternative added to existing efficient answers might to be:
Dim separator As Char = CChar(" ")
Dim sArr As String() = "SOMETHING BIGGER THAN YOUR DREAM".Split(separator)
Dim indexOfSplit As Integer = 3
Dim sFinal As String = Join(sArr.Take(indexOfSplit).ToArray, separator) & vbNewLine &
Join(sArr.Skip(indexOfSplit).ToArray, separator)
Console.WriteLine(sFinal)
You can split your input string, then loop the array of parts generated and add them to a StringBuilder object.
When you have read a number of parts that is multiple of a defined value, (wordsPerLine, here), you append vbNewLine to the current part.
When the loop completes, print the content of the StringBuilder to the Console:
Dim input As String = "SOMETHING BIGGER THAN YOUR DREAM, NOT MORE THAN YOUR ACCOUNT BALANCE"
Dim wordsPerLine As Integer = 3
Dim wordsCounter As Integer = 1
Dim sb As StringBuilder = New StringBuilder()
For Each word As String In input.Split()
sb.Append(word & If(wordsCounter Mod wordsPerLine = 0, vbNewLine, " "))
wordsCounter += 1
Next
Console.WriteLine(sb.ToString())
Prints:
SOMETHING BIGGER THAN
YOUR DREAM, NOT
MORE THAN YOUR
ACCOUNT BALANCE
Instead of using split, you might capture 3 words in a capturing group and match the trailing whitespace chars.
In the replacement use the group followed by a newline.
Pattern
(\S+(?:\s+\S+){2})\s*
That will match:
( Capture group 1
\S+ Match 1+ non whitespace chars
(?:\s+\S+){2} Repeat 2 times matching 1+ whitespace chars and 1+ non whitespace chars
) Close group 1
\s* Match trailing whitespace chars
.NET Regex demo | VB.NET demo
Example code
Dim s As String = "SOMETHING BIGGER THAN YOUR DREAM"
Dim output As String = Regex.Replace(s, "(\S+(?:\s+\S+){2})\s*", "$1" + Environment.NewLine)
Console.WriteLine(output)
Output
SOMETHING BIGGER THAN
YOUR DREAM
String.Join has an overload that will help you.
First parameter is the character to use between elements of your array.
Second parameter is the array you wish to join.
Third parameter is the starting position, for the first line in your desired output this would be the element at index 0.
Fourth parameter is the length to use, for the first line we want three array elements.
Private Sub OPCode()
Dim s As String = "SOMETHING BIGGER THAN YOUR DREAM"
Dim words As String() = s.Split(New Char() {" "c})
Dim line1 As String = String.Join(" ", words, 0, 3)
Console.WriteLine(line1)
Dim line2 As String = String.Join(" ", words, 3, words.Length - 3)
Console.WriteLine(line2)
End Sub
im having a hard time getting a function working. I need to search message.text for each "," found, for each "," found I need to get the number position of where the "," is located in the string. For example: 23232,111,02020332,12 it would return 6/10/19 where the "," are located (index of). My code finds the first index of the first , but then just repeats 6 6 6 6 over, any help would be appreciated thanks.
Heres my code:
For Each i As Char In message.Text
If message.Text.Contains(",") Then
Dim data As String = message.Text
Dim index As Integer = System.Text.RegularExpressions.Regex.Match(data, ",").Index
commas.AppendText(index & " ")
End If
Next
You can try it this way; instantiate a Regex object and increment each time the position from which you start the matching (this possibility is not available with the static method Match).
Dim reg As New System.Text.RegularExpressions.Regex(",")
Dim Index As Integer = reg.Match(data).Index
Do While Index > 0
commas.AppendText(index & " ")
Index = reg.Match(data, Index + 1).Index
Loop
p.s the returned indices are zero-based.
Just use the Regex.Matches method
Dim message As String = "23232,111,02020332,12"
Dim result As String = ""
For Each m As Match In Regex.Matches(message, ",")
result &= m.Index + 1 & " "
Next
I should also add that indexes are 0 based (which is why +1 is added to m.Index). If you later need these values to point to the position of a particular comma, you may be off by 1 and could potentially try to access an index larger than the actual string.
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 am working on a macro for a document-tracking sheet at work. I use a button that prompts the user to enter in the document number and I'd like to specify a default number based on the following numbering convention. The first two characters of the document number are the latter two year digits (15 in this case), then there is a "-" followed by a five digit serialization.
My current code looks at the last-entered document and increments those last 5 characters, but chops off any leading zeroes, which I want to keep. This is an extraction of the code to generate this default number (assuming the variable "prevNCRF" is the previous document name found in the doc):
Sub codeChunkTester()
Dim prevNCRF, defNCRFNum As String
Dim NCRFNumAr() As String
'pretend like we found this in the sheet.
prevNCRF = "15-00100"
'split the string into "15" and "00100" and throw those into an array.
NCRFNumAr() = Split(prevNCRF, "-")
'reconstruct the number by reusing the first part and dash, then converting
'the "00100" to a number with Val(), adding 1, then back to a string with CStr().
defNCRFNum = NCRFNumAr(0) & "-" & CStr(Val(NCRFNumAr(1)) + 1)
'message box shows "15-101" rather than "15-00101" as I had hoped.
MsgBox (defNCRFNum)
End Sub
So can anyone help me preserve those zeroes? I suppose I could include a loop that checks the length of the string and adds a leading zero until there are 5 characters, but perhaps there's a better way...
Converting "00100" to a Double using Val turned it into 100, so CStr(100) returns "100" as it should.
You need to format the string to what you want it to look like:
defNCRFNum = NCRFNumAr(0) & "-" & Format(Val(NCRFNumAr(1)) + 1, "00000")
If you need to parameterize the length of the string, you can use the String function to generate the format string:
Const digits As Integer = 5
Dim formatString As String
formatString = String(digits, "0")
defNCRFNum = NCRFNumAr(0) & "-" & Format(Val(NCRFNumAr(1)) + 1, formatString)
Here is that loop solution I mentioned above. If anyone's got something better, I'm all ears!
prevNCRF = "15-00100"
NCRFNumAr() = Split(prevNCRF, "-")
zeroAdder = CStr(Val(NCRFNumAr(1)) + 1)
'loop: everytime the zeroAdder string is not 5 characters long,
'put a zero in front of it.
Do Until Len(zeroAdder) = 5
zeroAdder = "0" & zeroAdder
Loop
defNCRFNum = NCRFNumAr(0) & "-" & zeroAdder
MsgBox (defNCRFNum)
defNCRFNum = NCRFNumAr(0) & "-" & Format(CStr(Val(NCRFNumAr(1)) + 1), String(Len(NCRFNumAr(1)), "0"))
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.