Your pattern is fine but your code didn't compile. Try this instead:
Swift 4
let query = "Hello %test% how do you do %test1%"
let regex = try! NSRegularExpression(pattern:"%(.*?)%", options: [])
var results = [String]()
regex.enumerateMatches(in: query, options: [], range: NSMakeRange(0, query.utf16.count)) { result, flags, stop in
if let r = result?.range(at: 1), let range = Range(r, in: query) {
results.append(String(query[range]))
}
}
print(results) // ["test", "test1"]
NSString
uses UTF-16 encoding so NSMakeRange
is called with the number of UTF-16 code units.
Swift 2
let query = "Hello %test% how do you do %test1%"
let regex = try! NSRegularExpression(pattern:"%(.*?)%", options: [])
let tmp = query as NSString
var results = [String]()
regex.enumerateMatchesInString(query, options: [], range: NSMakeRange(0, tmp.length)) { result, flags, stop in
if let range = result?.rangeAtIndex(1) {
results.append(tmp.substringWithRange(range))
}
}
print(results) // ["test", "test1"]
Getting a substring out of Swift's native String
type is somewhat of a hassle. That's why I casted query
into an NSString
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