Swift 4.2 extract substring using multiple characters are delimiter - ios

Question

I'm new to Swift and after going through the Apple documentation and other sources is not clear for me how can I extract a substring using more than one character as delimiter. For example: I have a string which looks like:
A.1 value1
B.2 value2
E value3
C value4
and need to assign the values 1 - 4 to different variables.

Answer 1

• Possible solution:
1. Separate all the elements (separator: white space)
2. Iterate 2 by 2 and use a key/value system, like a Dictionary.
3. Read each values from the keys afterward
Step 1:
let string = "A.1 value1 B.2 value2 E value3 C value4"
let components = string.components(separatedBy: CharacterSet.whitespaces)
Step 2:
var dictionary: [String: String] = [:]
stride(from: 0, to: components.count - 1, by: 2).forEach({
dictionary[components[$0]] = components[$0+1]
})
or
let dictionary = stride(from: 0, to: components.count - 1, by: 2).reduce(into: [String: String]()) { (result, currentInt) in
result[components[currentInt]] = components[currentInt+1]
}
dictionary is ["A.1": "value1", "C": "value4", "E": "value3", "B.2": "value2"]
Inspiration for the stride(from:to:) that I rarely use.
Step 3:
let name = dictionary["A.1"]
let surname = dictionary["C"]
• Potential issues:
If you have:
let string = "A.1 value One B.2 value2 E value3 C value4"
You want "value One", and since there is a space, you'll get some issue because if will give a false result (since there is the separator).
You'll get: ["A.1": "value", "One": "B.2", "value2": "E", "value3": "C"] for dictionary.
So you could use instead a regex: A.1(.*)B.2(.*)E(.*)C(.*) (for instance).
let string = "A.1 value One B.2 value2 E value3 C value4"
let regex = try! NSRegularExpression(pattern: "A.1(.*)B.2(.*)E(.*)C(.*)", options: [])
regex.enumerateMatches(in: string, options: [], range: NSRange(location: 0, length: string.utf16.count)) { (result, flags, stop) in
guard let result = result,
let aValueRange = Range(result.range(at: 1), in: string),
let bValueRange = Range(result.range(at: 2), in: string),
let cValueRange = Range(result.range(at: 4), in: string),
let eValueRange = Range(result.range(at: 3), in: string) else { return }
let aValue = string[aValueRange].trimmingCharacters(in: CharacterSet.whitespaces)
print("aValue: \(aValue)")
let bValue = string[bValueRange].trimmingCharacters(in: CharacterSet.whitespaces)
print("bValue: \(bValue)")
let cValue = string[cValueRange].trimmingCharacters(in: CharacterSet.whitespaces)
print("cValue: \(cValue)")
let eValue = string[eValueRange].trimmingCharacters(in: CharacterSet.whitespaces)
print("eValue: \(eValue)")
}
Output:
$>aValue: value One
$>bValue: value2
$>cValue: value4
$>eValue: value3
Note that the trim could be inside the regex, but I don't especially like having too complex regexes.

Answer 2

I like regular expressions for this sort of thing.
I'm going to take you very literally and assume that the substrings to be found are preceded by "A.1", "B.2", "E", and "C", and are all preceded and followed by a space except for the last substring which is followed by the end of the original string. Moreover I'm going to assume very simple-mindedly that the delimiters such as "E" cannot appear in our string in any other way. Then we can capture each substring with an appropriate pattern:
let s = "A.1 harpo B.2 chico E zeppo C groucho"
let p1 = "^A\\.1 (.*) B\\.2 "
let p2 = " B\\.2 (.*) E "
let p3 = " E (.*) C "
let p4 = " C (.*)$"
let patts = [p1,p2,p3,p4]
var result = [String]()
for patt in patts {
let regex = try! NSRegularExpression(pattern: patt, options: [])
if let match = regex.firstMatch(in: s, options: [],
range: NSRange(s.startIndex..<s.endIndex, in: s)) {
let r = match.range(at: 1)
result.append((s as NSString).substring(with: r))
}
}
// result is now ["harpo", "chico", "zeppo", "groucho"]
We now have the four desired substrings extracted into an array, and dealing with them from there is trivial.
Observe that we make no assumptions about spaces. The above works perfectly well even if the target substrings contain spaces, because we are appealing only to the delimiters. For example, if the original string is
let s = "A.1 the rain B.2 in spain E stays mainly C in the plain"
then result is the array
["the rain", "in spain", "stays mainly", "in the plain"]
I should point out, however, that another way to do this sort of thing is to walk the original string with a Scanner. You might prefer this because regular expressions are not really needed here, and if you don't know regular expressions you'll find this kind of walk much clearer. So here it is rewritten to use a scanner. Note that we end up with four Optional NSString objects, because Scanner is actually an Objective-C Cocoa Foundation thing, but it isn't difficult to turn those into String objects as needed:
let s = "A.1 the rain B.2 in spain E stays mainly C in the plain"
let scan = Scanner(string: s)
scan.scanString("A.1 ", into: nil)
var r1 : NSString? = nil
scan.scanUpTo(" B.2 ", into: &r1)
scan.scanString("B.2 ", into: nil)
var r2 : NSString? = nil
scan.scanUpTo(" E ", into: &r2)
scan.scanString("E ", into: nil)
var r3 : NSString? = nil
scan.scanUpTo(" C ", into: &r3)
scan.scanString("C ", into: nil)
var r4 : NSString? =
(scan.string as NSString).substring(from: scan.scanLocation) as NSString
r1 // the rain
r2 // in spain
r3 // stays mainly
r4 // in the plain