Perhaps this is a simple solution for most, but I can't get this to work like it should according to syntax.
I have this line of text "Part Number123456Price$50.00"
I want to pull the part number out of it, so I use this function...
str = Mid(str, str.IndexOf("Part Number") + 12, str.IndexOf("Price"))
My results are str = "123456Price$50.0" every time. I know the part number can vary in length so I need a solid solution of pulling this out.
It can be confusing to mix the legacy VB string methods (such as Mid) with the .Net string methods (like IndexOf). The VB methods use 1 as the index of the first character while the .Net methods use 0.
The following code will extract the part number from a string
Dim str As String = "Part Number123456Price$50.00"
Dim iPart As Integer = str.IndexOf("Part Number") + 11
Dim iPrice As Integer = str.IndexOf("Price")
str = str.Substring(iPart, iPrice - iPart).Trim
The Mid() function of Visual Basic is documented as having three arguments: (1) a string, (2) the beginning location in the string, and (3) the number of characters to copy.
So if your string is "Part Number123456Price$50.00" and you want to pull the part number as a series of digits, the "123456" part of the string, using the Mid() function then you need to find the beginning of the part number digit string and to then know the number of digits.
If your string is in the variable str then you can find the offset by something like str.IndexOf("Number") + len("Number") which will provide the offset to after the string "Number".
Next you need to find the number of digits so you would do something like str.IndexOf("Price") to find where the text "Price" begins and then subtract from that offset the offset of where the digits begin.
The result of all of this is you need a bit of code something like the following. I have not tested this source as I am not a VB programmer so it may need a tweak and you might want to put some checks on data validity as well.
Dim TextNumber as String = "Number"
Dim TextPrice as String = "Price"
iOffset = str.IndexOf(TextNumber) + len(TextNumber)
str = Mid(str, iOffset, str.IndexOf(TextPrice) - iOffset)
Alternatively, if Price is always the format $00.00, this will also work.
Dim str as String = "Part Number123456Price$50.00"
str = str.Remove(str.IndexOf("Price"))
Related
In VB.net (Visual Studio 2015) how can I get the nth string (or number) in a comma-separated list?Say I have a comma-separated list of numbers like so:13,1,6,7,2,12,9,3,5,11,4,8,10How can I get, say, the 5th value in this string, in this case 12?I've looked at the Split function, but it converts a string into an array. I guess I could do that and then get the 5th element of that array, but that seems like a lot to go through just to get the 5th element. Is there a more direct way to do this, or am I pretty much limited to the Split function?
In case you are looking for an alternative method, which is more basic, you can try this:
Module Module1
Sub Main()
Dim a As String = "13,1,6,7,2,12,9,3,5,11,4,8,10"
Dim counter As Integer = 5 'the number you want (in this case, 5th one)
Dim movingcounter As Integer = 0 'how many times we have moved
Dim startofnumber, endofnumber, i As Integer
Dim numberthatIwant As String
Do Until movingcounter = counter
startofnumber = InStr(i + 1, a, ",")
i = startofnumber
movingcounter = movingcounter + 1
Loop
endofnumber = InStr(startofnumber + 1, a, ",")
numberthatIwant = (Mid(a, startofnumber + 1, endofnumber - startofnumber - 1))
Console.WriteLine("The number that I want: " + numberthatIwant)
Console.ReadLine()
End Sub
End Module
Edit: You can make this into a procedure or function if you wish to use it in a larger program, but this code run in console mode will give the output of 12.
The solution provided by Plutonix as a comment to my question is straightforward and exactly what I was looking for, to wit:result = csv.Split(","c)(5)In my case I was incrementing a variable each time my program ran and needed to get the nth character or string after the incremented value. That is, if my program had incremented the variable 5 times, then I needed the string after the 4th comma, which of course, is the 5th string. So my solution was something like this:result = WholeString.Split(","c)(IncrementedVariable)Note that this is a zero-based variable.Thanks, Plutonix.
I have an AD username called "lastname-132" in Textbox1, this string is 12 long, so i want to add the username in to Textbox2, but shortened, in the textbox2 i only have a string length of only 10 available due to other tools this program is using, so i don't want to convert it all the time manually and want to just convert it automatically with a onleave event.
Anyone any idea how to write this?
So the End Result should look like this.
'String length can be 20 max.
Textbox1.Text = "lastname-123"
'some code to convert it to this:
'String length 10 max. Numbers and the "-" should stay the same, but remove letters if necessary.
Textbox2.Text = "lastna-123"
Here's the concept:
Split string based on '-' into 2 strings
In the example above: 'lastname' and '123'.
Check the length of the first string and cut if it is too long
the program checks 'lastname' and finds that it is too long, then
cuts it into 'lastna'
Combine 'lastna' and '123' back into a string
I hope this helps
Without more information, this will assume that there can be multiple hyphens, the number can be of variable length, and you can change the maximum length of the string by changing one variable.
Dim username As String = "lastname-123"
Dim max As Integer = 10
Dim lindex As Integer = username.LastIndexOf("-")
Dim numberLength As Integer = username.Length - lindex
Dim number As String = username.Substring(lindex)
Dim justName As String = username.Substring(0, lindex)
If justName.Length + numberLength >= max Then
username = justName.Substring(0, max - numberLength) & number
End If
If you are concentrating only on the restriction of length of characters to be accepted then you can use
Maxlength
property of the Textbox.
Ex: Maxlength="10"
restricts the Textbox to accept only 10 characters.
Try to make it fit with for example substring manipulation. See http://msdn.microsoft.com/en-us/library/dd789093.aspx for more info.
This should be fairly simple but I'm having one of those days. Can anyone advise me as to how to replace the first and third occurrence of a character within a string? I have looked at replace but that cannot work as the string could be of different lengths. All I want to do is replace the first and third occurrence.
There is an overload of the IndexOf method which takes a start position as a parameter. Using a loop you'll be able to find the position of the first and third occurences. Then you could use a combination of the Remove and Insert methods to do the replacements.
You could also use a StringBuilder to do the replacements. The StringBuilder has a Replace method for which you can specify a start index and a number of characters affected.
aspiringCoder,
Perhaps something like this might be useful to you (in line with what Meta-Knight was talking about <+1>)
Dim str As String = "this is a test this is a test this is a test"
Dim first As Integer
Dim third As Integer
Dim base As Integer = 0
Dim i As Integer
While str.length > 0
If i = 0 Then
first = str.IndexOf("test")
else if i = 2 Then
third = base + str.IndexOf("test")
end if
base = base + str.IndexOf("test")
str = str.Remove(0, str.IndexOf("test") + "test".length -1 )
i++
End While
It might have a one-off error somewhere...but this should at least get you started.
Just like the title says; I want to use something like Mid(stringName, startIndex,[integerLength]), but instead of the third argument taking a string length, I want it to take the end character index. So in an example like this
alphabet = "ABCDEFG"
partial = *method I want to use*(alphabet, 2, 4) 'partial would equal "BC"
(Forgive me if my index numbers are off, but I hope you get my point.)
Does something like this exist in VB.NET?
You'll want to use String.Substring
http://msdn.microsoft.com/en-us/library/aka44szs.aspx#Y0
dim alphabet as string = "ABCDEFG"
'partial is a reserved word!
'1,2 is the correct parameters to get 'BC'
dim partialString as string = alphabet.Substring(1, 2) 'partial would equal "BC"
Edit - Oooooh you want to do StartIndex,StopIndex not StartIndex,Length. Just apply a bit of math.
dim startIndex as integer = 1
dim stopIndex as integer = 3
'partial would equal "BC"
dim partialString as string = _
alphabet.Substring(startIndex , stopIndex-startIndex )
I'd wrap that in an extension method on string, giving it a new name of course.
Just use Mid, the math for the length is quite easy (length = endIndex - startIndex):
part = Mid(alphabet, 2, 4-2)
You could also achieve the same thing with Substring (which uses 0 based indexes rather than 1 based):
part = alphabet.Substring(1, 3-1);
targetstring=alphabet.Substring(2,4)
The above one should work..
i have a string "<PinX F='53mm'></PinX>", I want to access the 53 within the string and do some addition to it and then add the answer back into that string. I've been thinking about this and wasn't sure whether this can be done with regular expression or not? Can anybody help me out.
thanks
Yes, you can use a regular expression. This will get the digits, parse them to a number, add one to it, and put it back in the string (that is, the result is actually a new string as strings are immutable).
string s = Regex.Replace(
input,
#"(\d+)",
m => (Int32.Parse(m.Groups[1].Value) + 1).ToString()
);
Take a look at the HTML Agility Pack.
A regular expression looks like a good fit for this particular problem:
\d+
Will match one or more digits.
Int32.Parse(Regex.Match("<PinX F='53mm'></PinX>", #"\d+").Value)
Will return 53.
In this single case yes. "'(.*?)' then access the first group, but if this is part of a larger xml regular expressions should not be used. You should utilize the xml parser build into .net find the attribute with xsd and get the value.
Alternatively, here's a small routine...
' Set testing string
Dim s As String = "<PinX F='53mm'></PinX>"
' find first occurence of CHAR ( ' )
Dim a As Integer = s.IndexOf("'")
' find last occurence of CHAR ( ' )
Dim b As Integer = s.LastIndexOf("'")
' get substring "53mm" from string
Dim substring As String = s.Substring(a, b - a)
' get integer values from substring
Dim length As Integer = substring.Length
Dim c As Char = Nothing
Dim result As String = Nothing
For i = 1 To length - 1
c = substring.Chars(i)
If IsNumeric(c) Then
result = result & c
End If
Next
Console.WriteLine(Int32.Parse(result))
Console.ReadLine()