I am having difficulties trying to figure out the syntax. I have a line of binary in a variable and i want to know how to to split that string every 7th character so i can put it in an ascii table to find its equivalent.
Based on your scenario, you can also do a substring to satisfy your requirement.
Dim theString As String = "w3rs0fd90as90f9sda0faf"
Dim FirstSevenCharacters as String= theString.Substring(0, 7)
Dim CharactersAfterTheFirstSeven as String= theString.Substring(7, theString.Length - 7)
Related
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"))
I am trying to split a string with multiple characters. The string might sometimes contain a - or a /. What I have achieved is the hyphen but I am not able to search for the slash. Any thoughts on how to split the string based on both characters at once ? Once I split after - I add the value after the - to the result list as a separate index and I would like to accomplish the same for '/'.
So For example the Split string has Jet-blue, the below code will add Jet in the result list with index(0) and blue with index(1). In addition to splitting with '-' I would also like to split with '/'. Any suggestions ?
Code:
Dim result As New List(Of String)()
For Each str_get As String In Split
Dim splitStr = str_get.Split("-")
For Each str_split As String In splitStr
result.Add(str_split) ' Enter into result list
' result.TrimExcess()
Next
result.Remove("")
Next
You can either use this or this overload of the Split method.
The first one takes an array of Char:
"Hello World".Split({"e"c, "o"c}) ' Notice the c!
The second one takes an array of String and StringSplitOptions:
"Hello World".Split({"el", "o"}, StringSplitOptions.None)
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)
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)