某个字符的Swift 4字符串范围

Question

I'm trying to make a function in string that deletes some characters from a provided string.

For example i have

"‌Boll %b‌ (Teeth ‌Alligator‌ (13,8,8,5,5,3,n),20,2,ma)"

and i want to edit this to

"‌Boll %b‌"

which means to delete everything between the "(" and ")"

So i've made this function

func deleteInnerString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))
    return remainingString
}

but in this case the output string is

"‌Boll %b‌ ,20,2,ma)"

The question is, how can i scan the string to find the last closing parenthesis range??

Answer 1

Please note that your strategy won't work with non nested parenthesis.

Eg this input

a(b)c(d)e

With your approach will produce this output

ae

Which IMHO is not correct.

Solution

This solution instead will work with nested and non nested parenthesis

let text = "‌a(b)c(d)e"
var numOfNestedParentesis = 0
var indexesToRemove:Set<Int> = []

for (index, char) in text.enumerated() {

    if char == "(" {
        numOfNestedParentesis += 1
    }

    if numOfNestedParentesis > 0 {
        indexesToRemove.insert(index)
    }

    if char == ")" {
        numOfNestedParentesis -= 1
    }
}

let result = String(text.enumerated()
    .filter { !indexesToRemove.contains($0.offset) }
    .map { $0.element } )

Test

print(result) // ‌ace

Answer 2

The String.range function can be provided extra options, so you can pass it the String.CompareOptions.backwards parameters to get the last instance.

func deleteInnerString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter, options: .backwards)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))
    return remainingString
}

Answer 3

You have to use .backwards option for String.CompareOptions in range function of string. as like below

func deleteString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter, options: .backwards)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))

    return remainingString.trimmingCharacters(in: .whitespaces)
}