I need to split one variable into 3 variables. For example I have a variable called Data and it contains AAA00000001. I need to have them be split between the "AAA", "0000000", and "1". I looked at the Split() function but didn't get a clear example for this situation. The reason I need to do this is because I want to store these 3 variables into fields for a MySQL database.
Are the three subvariables always of the same length?
If so, you can use Substrings:
Dim substring1 As String = Data.Substring(0, 3)
Dim substring2 As String = Data.Substring(3, 7)
Dim substring3 As String = Data.Substring(10, 1)
Assuming the string is ALWAYS the EXACT same length and need to be split at the SAME place, you can use Substring().
dim s as String = "AAA00000001"
dim s1 as String = s.Substring(0, 3)
dim s2 as String = s.Substring(3, 7)
dim s3 as String = s.Substring(10)
If they're not always the same length, you're probably going to need to use Regular Expressions.
Split will break your string apart based on a character, or group of. It's not appropriate here, that is unless you're always splitting on 0000000, which I doubt you are.
If you know that the first 3 characters will always be your first group, second 7 your next, and last character, your final group, you could do something like this.
This uses the Substring function, e.g.
Dim yourString as String = "AAA00000001"
Dim c1 As String = yourString.Substring(0, 3)
Dim c2 As String = yourString.Substring(3, 7)
Dim c3 As String = yourString.Substring(10, 1)
Related
I'm using a very particular program that doesn't have many pre loaded functions on it. I would like to split a Date and time into individual variables. 6 variables to be exact
I get the date and time like this 5/28/2014 15:34:40 so I have 3 characters "/" " " and ":" separating the 6 variables I want, but I can only get one to work at a time. I have tried many other ways but the x.Split() is the only syntax that this program seems to like (PARCView).
Dim x as String = "5/28/2014 15:34:40"
Dim y as String() = x.Split(New Char() {":"c})
So how can I add all three at once or use consecutive steps to store each number as a unique variable?
You actually have the Char Array that you need, though it only has one delineator... just fill it with the other delineators and it will split it for you.
Dim y As String() = x.Split(New Char() {":"c, "/"c, " "c})
example:
Another option is to parse it as a datetime object which will contain all the parts as properties:
Dim teststr = "5/28/2014 15:34:40"
Dim dt = DateTime.Parse(teststr)
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()
I got this url for example "http://www.yellowpages.com/manhattan-beach-ca/mip/marriott-manhattan-beach-4933923?lid=185795402"
I want to get the last digit numbers and the rest could be anything.
I need a format like this "http://www.yellowpages.com/anything.... lid=randomdigitnumbers" or as long as i get those numbers.
My knowledge is very poor in this regex thing so please guys help me.
the following did not work
Dim r As New System.Text.RegularExpressions.Regex("http://www.yellowpages.com/.*lid=d*", RegexOptions.IgnoreCase)
Dim m As Match = r.Match(txt)
If (m.Success) Then
Dim int1 = m.Groups(1)
MsgBox("(" + int1.ToString() + ")" + "")
End If
thank you in advance
Using Regular Expressions for this is a bit of overkill, IMO.
You could accomplish the same thing using string functions:
Dim url As String = "http://www.yellowpages.com/manhattan-beach-ca/mip/marriott-manhattan-beach-4933923?lid=185795402"
Dim queryString As String = url.SubString(url.IndexOf("?"), url.Length - url.IndexOF("?"))
Dim nameValuePairs As String() = queryString.Split("=")
Dim lid As String = nameValuePairs(1)
This is off the top of my head, so you may need to tweak it a bit. The basic concept is to the portion of the URL after the ? (the query string), and then split it on the = sign, taking the second element of the resulting array (the value).
Also, if the query string has more than one name value pair, they'll be separated by &, so you'll need to split on the ampersand (&) first, then the equal signs.
Just find lid= and get everything after that:
Dim url As String = "http://www.yellowpages.com/manhattan-beach-ca/mip/marriott-manhattan-beach-4933923?lid=185795402"
Dim lidIndex As Integer = url.IndexOf("lid=") + "lid=".Length
Dim lid As Integer = url.Substring(lidIndex)