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.
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
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
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.