I want to have a string in the following format
"FAG001 FAG002 FAG003"
and want to split it into
"FAG001"
"FAG002"
"FAG003"
using a regular expression. Unfortunately my knowledge of regular expression synatax is limited to say teh least. I have tried things like
Dim result = Regex.Split(npcCodes, "([A-Z]3[0-9]3)").ToList
without luck
No need of regex here, you could use String.Split
Dim result As String() = npcCodes.Split(new Char[]{" "})
But if you really want to use regex :
Dim result = Regex.Split(npcCodes, " ").ToList()
As madgnome has pointed out you don't need regular expressions here if the string is always separated with spaces.
However for your information the error you made was that you need curly braces for numeric quantifiers:
[A-Z]{3}
And instead of Regex.Split you can uses Regex.Matches.
The regular expression to use in the Split method would be really simple:
Dim result = Regex.Split(npcCodes, " ").ToList
As the expression only matches a single character, you can just as well use the regular Split method in the String class:
Dim result = npcCodes.Split(" "C).ToList
Related
Beginner here. 2nd day into VB.
How do I capture the middle part of the string below? What method (Split, Contains, Trim, eg) to use?
Variable string message1:
"background:url(//test.image.com/bao/uploaded/i2/1951211594/TB2m1gjzYlmpuFjSZFl>XXbdQXXa_!!1951211594.jpg_40x40q90.jpg)"
To become:
//test.image.com/bao/uploaded/i2/1951211594/TB2m1gjzYlmpuFjSZFlXXbdQXXa_!!19512>11594.jpg
Dim s
s = Split(strMessage, ":url(")(1)
s = Left(s, Len(s)-2)
If you know the string will always contain "background:url( in the beginning and )" at the end, you can just use Mid$(message1, 17, Len(message1) - 18).
I have this string:
201057&channelTitle=null_JS
I want to be able to cut out the '201057' and make it a new variable. But I don't always know how long the digits will be, so can I somehow use the '&' as a reference?\
myDigits substring(0, position of &)?
Thanks
Sure, you can split the string along the &.
String s = "201057&channelTitle=null_JS";
String[] parts = s.split("&");
String newVar = parts[0];
The expected result here is
parts[0] = "201057";
parts[1] = "channelTitle=null_JS";
In production code you chould check of course the length of the parts array, in case no "&" was present.
Several programming languages also support the useful inverse operation
String s2 = parts.join("&"); // should have same value like s
Alas this one is not part of the Java standard libs, but e.g. Apache Commons Lang features it.
Always read the API first. There is an indexOf method in String that will return you the first index of the character/String you gave it.
You can use myDigits.substring(0, myDigits.indexOf('&');
However, if you want to get all of the arguments in the query separately, then you should use mvw's answer.
I was getting weird results when doing multiple splits on a string, so I decided to make a simple test to figure out what was going on
testString "1234567891011121314151617181920"
If I wanted to get whats between 10 to 20 in Javascript I would do this:
var results = testString.split("10")[1].split("20")[0]
Which would return 111213141516171819
However when I do this in VB I get 111
Split(testString,"10")(1).Split("20")(0)
It seems the 2nd split is only recognizing the first character no matter what I put.
So it's stopping when it finds the next "2" in the string, even "2abc" would have the same outcome even though that string doesn't even exist.
String.Split does not have an overload that takes only a String. The argument is a Char array or String array. Your string is probably being converted to a char array. Explicitly pass a string array like so:
testString.Split(New String() { "10" }, StringSplitOptions.None)
Try wrapping the second split so it's fashioned like the first one, i.e.:
Split( Split(testString,"10")(1), "20" )(0)"
Vb treats the delimiter argument only as a single character.
This is a tricky scenario that I have seen trip people up before, so I think it is worth a little more explanation than the other answers give. In your original format Split(testString,"10")(1).Split("20")(0), you are unknowingly using two DIFFERENT Split functions.
The first Split(testString,"10") is using the Microsoft.VisualBasic.Strings.Split function, which takes String type parameters. http://msdn.microsoft.com/en-us/library/microsoft.visualbasic.strings.split(v=vs.110).aspx
The second .Split("20")(0) is using System.String.Split method, which does not have an overload that takes a String parameter. http://msdn.microsoft.com/en-us/library/System.String.Split(v=vs.110).aspx
So what was happening is:
Split(testString,"10") uses Microsoft.VisualBasic.Strings.Split, which
returns new String() {"123456789", "11121314151617181920"}
(1) means get 1st position of the returned array, which is "11121314151617181920"
"11121314151617181920".Split("20")(0) uses System.String.Split, and attempts to split on string separator "20"
NOTE: The string "20" param gets implicitly converted to a char "2" because the only single parameter overload of String.Split has a signature of Public Function Split (ParamArray separator As Char()) As String(). The ParamArray parameter option allows you to pass a comma delimited list of values into the function, similar to how String.Format works with a dynamic # of replacement values. http://msdn.microsoft.com/en-us/library/538f81ec.aspx
Step 3 code becomes "11121314151617181920".Split(new Char() {CChar("20")})(0), which using literal values is "11121314151617181920".Split(new Char() {"2"c})(0). The result is {"111", "13141516171819", "0"}. Get the 0th position, returns "111".
So to avoid confusion, you should convert your code to use the same version of Split on both sides.
Either of the 2 examples below should work:
Example 1: Using Microsoft.VisualBasic.Strings.Split:
Split( Split(testString,"10")(1), "20" )(0)
Example 2: Using System.String.Split:
testString _
.Split(New String() {"10"}, StringSplitOptions.None)(1) _
.Split(New String() {"20"}, StringSplitOptions.None)(0)
I am trying to perform a Substring function on a image filename.
The name format is in "images.png".
I tried using Substring it only allow me to indicate the first character till the "n" character to perform the function.
Such that SubString(1,6).
But what I want is to get any character before the ..
For example "images.png":
After the Substring function I should get "images".
You can use LastIndexOf in conjunction with Substring:
myString.Substring(0, myString.LastIndexOf('.'))
Though the Path class has a method that will do this in a strongly typed manner, whether the passed in path has directories or not:
Path.GetFileNameWithoutExtension("images.png")
How about using the Path class.
Path.GetFileNameWithoutExtension("filename.png");
In general for such string manipulations you can use:
mystring.Split("."c)(0)
But specifically for getting a filename without extension, it's best to use this method:
System.IO.Path.GetFileNameWithoutExtension
Dim fileName As String = "images.png"
fileName = IO.Path.GetFileNameWithoutExtension(fileName)
Debug.WriteLine(fileName)
http://msdn.microsoft.com/en-us/library/system.io.path.getfilenamewithoutextension.aspx
string s = "images.png";
Console.WriteLine(s.Substring(0, s.IndexOf(".")));
I'm new to regular expressions, and have no clue where to start, it's like a diff language to me. But I need one quick to accomplish a task.
I need to take
http://www.domain.com/folder1/folder2/file_path.txt
and get just
/folder1/folder2/file_path.txt
from it.
Thanks!
construct a URI object from it and one of the properties of it will have what you want.
I think that regex should work:
^http://.*?/(.*)$
(tested with Python)
Since VB.NET is in the tag for this question, I assume you have access at the server side to the Request object:
Dim instance As HttpRequest
Dim value As String
value = instance.Path
This should give you exactly what you asked for.
Edit: On second thought - you could be parsing URLs from some input string... in which case, regex will only help if you have a simple (regular) set of inputs:
Do you know all the possible domains? i.e. are "http://www.ABC.com" and "http://www.DEF.com" the only possible domains?
Then here:
Dim text As String = "http://www.ABC.com/folder1/folder2/file.txt"
Dim pattern As String = "(?:http://www.ABC.com|http://www.DEF.com)(.*)"
Dim r As Regex = new Regex(pattern, RegexOptions.IgnoreCase)
' Match the regular expression pattern against a text string.
Dim m As Match = r.Match(text)
Dim g as Group = m.Groups(2) 'Gives the string matched by the capturing parentheses
Supporting more protocols and making the protocol optional too.
((https?|ftp)://)?(.*?)/(.*)