Swift如何从字符串中获取整数并将其转换为整数

Question

我需要从字符串中提取数字并将它们放入Swift中的新数组中.

var str = "I have to buy 3 apples, 7 bananas, 10eggs"

我试图循环每个字符,我不知道比较字符和Int.

Answer 1

斯威夫特3/4

let string = "0kaksd020dk2kfj2123"
if let number = Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()) {
    // Do something with this number
}

您还可以进行如下扩展:

extension Int {
    static func parse(from string: String) -> Int? {
        return Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
    }
}

然后用它像:

if let number = Int.parse(from: "0kaksd020dk2kfj2123") { 
    // Do something with this number
} 

Answer 2

首先,我们拆分字符串,以便我们可以处理单个项目.然后我们NSCharacterSet用来选择数字.

import Foundation

let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let strArr = str.split(separator: " ")

for item in strArr {
    let part = item.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()

    if let intVal = Int(part) {
        print("this is a number -> \(intVal)")
    }
}

斯威夫特4:

let string = "I have to buy 3 apples, 7 bananas, 10eggs"
let stringArray = string.components(separatedBy: CharacterSet.decimalDigits.inverted)
for item in stringArray {
    if let number = Int(item) {
        print("number: \(number)")
    }
}

Answer 3

let str = "Hello 1, World 62"
let intString = str.componentsSeparatedByCharactersInSet(
    NSCharacterSet
        .decimalDigitCharacterSet()
        .invertedSet)
    .joinWithSeparator("")

这将为您提供一个包含所有数字的字符串,然后您可以这样做:

let int = Int(intString)

只需确保打开它,因为它let int = Int(intString)是可选的.

Answer 4

使用Swift提取正则表达式中的"正则表达式辅助函数" 匹配:

func matchesForRegexInText(regex: String!, text: String!) -> [String] {

    let regex = NSRegularExpression(pattern: regex,
        options: nil, error: nil)!
    let nsString = text as NSString
    let results = regex.matchesInString(text,
        options: nil, range: NSMakeRange(0, nsString.length))
        as! [NSTextCheckingResult]
    return map(results) { nsString.substringWithRange($0.range)}
}

你可以轻松实现这一目标

let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let numbersAsStrings = matchesForRegexInText("\\d+", str) // [String]
let numbersAsInts = numbersAsStrings.map { $0.toInt()! }  // [Int]

println(numbersAsInts) // [3, 7, 10]

该模式"\d+"匹配一个或多个十进制数字.

当然,如果您因为某种原因而不喜欢使用辅助函数,则可以完成相同的操作:

let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let regex = NSRegularExpression(pattern: "\\d+", options: nil, error: nil)!
let nsString = str as NSString
let results = regex.matchesInString(str, options: nil, range: NSMakeRange(0, nsString.length))
    as! [NSTextCheckingResult]
let numbers = map(results) { nsString.substringWithRange($0.range).toInt()! }
println(numbers) // [3, 7, 10]

没有正则表达式的替代方案

let str = "I have to buy 3 apples, 7 bananas, 10eggs"

let digits = "0123456789"
let numbers = split(str, allowEmptySlices: false) { !contains(digits, $0) }
    .map { $0.toInt()! }
println(numbers) // [3, 7, 10]

Answer 5

对我来说，将它作为字符串扩展更有意义，可能这是一个品味问题：

extension String {
 func parseToInt() -> Int? {
    return Int(self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
 }
}

所以可以这样使用：

if let number = "0kaksd020dk2kfj2123".parseToInt() { 
// Do something with this number
} 

Answer 6

根据@flashadvanced 的回答，我发现以下内容对我来说更短更简单。

let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let component = str.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let list = component.filter({ $0 != "" }) // filter out all the empty strings in the component
print(list)

在操场上试过，效果很好

希望能帮助到你 ：）