I need to get substring between two strings from my text. For example, I have text "http://google.com" and I want to get substring between "://" and ".".
I don't know, how I can do that.
I try to use regular expressions, but I think, it's bad way.
A couple of options:
Regular expressions work well. See ICU User Guide: Regular Expressions
Example:
let us = "http://google.com"
let range = us.rangeOfString("(?<=://)[^.]+(?=.)", options:.RegularExpressionSearch)
if range != nil {
let found = us.substringWithRange(range!)
println("found: \(found)") // found: google
}
Notes:
(?<=://) means preceded by ://
[^.]+ means any characters except .
(?=.) means followed by .
NSScanner
is also a good method. See Apple's NSScanner Class Reference
Example:
let us = "http://google.com"
let scanner = NSScanner(string:us)
var scanned: NSString?
if scanner.scanUpToString("://", intoString:nil) {
scanner.scanString("://", intoString:nil)
if scanner.scanUpToString(".", intoString:&scanned) {
let result: String = scanned as String
println("result: \(result)") // result: google
}
}
You can use the regular Expression
://.+.
it matches to
://google.
in this code:
var yourURL: NSString = "http://google.com" // this is your input and could be any URL
var regex: NSRegularExpression = NSRegularExpression.regularExpressionWithPattern("://.+\\.", options: NSRegularExpressionOptions.fromMask(UInt(0)), error: nil) // need double backspace because of backspace in String is \\ not \
var needleRange = regex.rangeOfFirstMatchInString(yourURL, options:NSMatchingOptions.Anchored, range: NSMakeRange(0, yourURL.length))
var needle: NSString = yourURL.substringWithRange(needleRange)
Now you can remove the first 3 symbols and the last one and you got
with this code:
import Foundation
var halfURL: NSString = "://google."
var prefix: NSString = "://"
var suffix: NSString = "."
var needleRange: NSRange = NSMakeRange(prefix.length, halfURL.length - prefix.length - suffix.length)
var needle: NSString = halfURL.substringWithRange(needleRange)
// needle is now 'google'
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