Question

我想要一个子字符串，用另一个常量String＆＃34; _TRALLALA＆＃34;

分隔

这样做很好：

 if let lb = s2.range(of: "_TRALLALA")?.lowerBound
 { let s3 = url.substring(to: lb)
 // do something  
 }

我想在单曲中表达这一点，如果这样：

 if let s3 =  url.substring(to: s2.range(of: "_TRALLALA")?.lowerBound)
 { // do something   
 }

但我明白了：

Value of optional type String.Index? not unwrapped

如何正确表达

Answer 1

map可与可选一起使用以解包该值并使用它，或者如果可选项为nil则返回nil。使用它，您的语句可以这样写：

if let s3 = (s2.range(of: "_TRALLALA")?.lowerBound).map({ url.substring(to: $0) }) {
    // do something
}

或（略短）将lowerBound移至map：

之后

if let s3 = s2.range(of: "_TRALLALA").map({ url.substring(to: $0.lowerBound) }) {
    // do something
}

Answer 2

没有可以执行您想要的本机功能，因此您必须将其包装到函数中：

extension String {
    func substring(delimitedBy: String) -> String? {
        guard let range = self.range(of: delimitedBy) else {
            return nil
        }

        return self.substring(to: range.lowerBound)
    }
}

if let s3 = url.substring(delimitedBy: "_TRALLALA") {
    // ...
}

Answer 3

var s2 = "Hello, playground"

var url = "some url"


let delimiter  =  "_TRALLALA"
let r  = s2.range(of: delimiter) ?? s2.startIndex..<s2.startIndex // if found nil then assigning empty Range

let s3  = url.substring(to: r.lowerBound)
if s3 != "" {
// do something
}

单行

let s3  = url.substring(to: (s2.range(of: "_TRALLALA") ?? s2.startIndex..<s2.startIndex).lowerBound)
if s3 != "" {
// do something

 }

Answer 4

if let s3 = s2.range(of: "_TRALLALA").flatMap { url.substring(to: $0.lowerBound) }
{ 
    // do something   
}

Swift 3可选展开

4 个答案: