What i want to do is, after X amount of characters, split the string on last whitespace, and then after the next X amount of characters repeat.
Till now i have something like this , using regex but i get error on "Function"
Dim input = "This is a long sentence with more than 18 letters."
Dim output = Regex.Split(input, "(.{1,18})(?:\s|$)").Where(Function(x) x.Length > 0).ToList()
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
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
I am trying to set the letters after a # symbol to a variable.
For example, x = #BAL
I want to set y = BAL
Or x = #NE
I want y = NE
I am using VBA.
Split() in my opinion is the easiest way to do it:
Dim myStr As String
myStr = "#BAL"
If InStr(, myStr, "#") > 0 Then '<-- Check for your string to not throw error
MsgBox Split(myStr, "#")(1)
End If
As wisely pointed out by Scott Craner, you should check to ensure the string contains the value, which he checks in this comment by doing: y = Split(x,"#")(ubound(Split(x,"#")). Another way you can do it is using InStr(): If InStr(, x, "#") > 0 Then...
The (1) will take everything after the first instance of the character you are looking for. If you were to have used (0), then this would have taken everything before the #.
Similar but different example:
Dim myStr As String
myStr = "#BAL#TEST"
MsgBox Split(myStr, "#")(2)
The message box would have returned TEST because you used (2), and this was the second instance of your # character.
Then you can even split them into an array:
Dim myStr As String, splitArr() As String
myStr = "#BAL#TEST"
splitArr = Split(myStr, "#") '< -- don't append the collection number this time
MsgBox SplitArr(1) '< -- This would return "BAL"
MsgBox SplitArr(2) '< -- This would return "TEST"
If you are looking for additional reading, here is more from the MSDN:
Split Function
Description Returns a zero-based, one-dimensional array containing a specified number of substrings. SyntaxSplit( expression [ ,delimiter [ ,limit [ ,compare ]]] ) The Split function syntax has thesenamed arguments:
expression
Required. String expression containing substrings and delimiters. If expression is a zero-length string(""), Split returns an empty array, that is, an array with no elements and no data.
delimiter
Optional. String character used to identify substring limits. If omitted, the space character (" ") is assumed to be the delimiter. If delimiter is a zero-length string, a single-element array containing the entire expression string is returned.
limit
Optional. Number of substrings to be returned; -1 indicates that all substrings are returned.
compare
Optional. Numeric value indicating the kind of comparison to use when evaluating substrings. See Settings section for values.
You can do the following to get the substring after the # symbol.
x = "#BAL"
y = Right(x,len(x)-InStr(x,"#"))
Where x can be any string, with characters before or after the # symbol.
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.
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.