你也可以在Swift中使用stringByAppendingFormat。

var finalString : NSString = NSString(string: "Hello")
finalString = finalString.stringByAppendingFormat("%@", " World")
print(finalString) //Output:- Hello World
finalString = finalString.stringByAppendingFormat("%@", " Of People")
print(finalString) //Output:- Hello World Of People

var language = "Swift" 
var resultStr = "\(language) is a new programming language"

这也会起作用:

var string = "swift"
var resultStr = string + " is a new Programming Language"

串联是指Swift中字符串的组合。字符串可能包含文本，整数，甚至表情符号!字符串连接有很多方法。让我列举一些:

相同的字符串

使用+ =

如果我们想要添加到一个已经存在的String，这是很有用的。为此，我们的String应该是可变的或可以修改的，因此将其声明为变量。例如:

var myClassmates = "John, Jane"
myClassmates += ", Mark" // add a new Classmate
// Result: "John, Jane, Mark"

不同的字符串

如果我们想把不同的字符串组合在一起，例如:

let oldClassmates = "John, Jane" 
let newClassmate = "Mark"

我们可以使用以下任何一种:

1)使用+

let myClassmates = oldClassmates + ", " + newClassmate
// Result: "John, Jane, Mark"

注意，每个字符串可以是一个变量或常量。如果你只改变一次值，就把它声明为常量。

2)字符串插值

let myClassmates = "\(oldClassmates), \(newClassmate)"
// Result: "John, Jane, Mark"

3)添加

let myClassmates = oldClassmates.appending(newClassmate)
// Result: "John, Jane, Mark"

更多信息请参考Swift Book中的字符串和字符。

更新:在Swift 5.1上测试

快速连接字符串

关于性能的几句话

UI测试包在iPhone 7(真实设备)，iOS 14， -Onone(调试，没有优化)[关于]

var result = ""
for i in 0...count {
    <concat_operation>
}

计数= 5_000

//Append
result.append(String(i))                         //0.007s 39.322kB

//Plus Equal
result += String(i)                              //0.006s 19.661kB

//Plus
result = result + String(i)                      //0.130s 36.045kB

//Interpolation
result = "\(result)\(i)"                         //0.164s 16.384kB

//NSString
result = NSString(format: "%@%i", result, i)     //0.354s 108.142kB

//NSMutableString
result.append(String(i))                         //0.008s 19.661kB

禁用下一个测试:

加上高达100_000 ~10s 插补可达100_000 ~10s NSString高达10_000 ->内存问题

Count = 1_000_000

//Append
result.append(String(i))                         //0.566s 5894.979kB

//Plus Equal
result += String(i)                              //0.570s 5894.979kB

//NSMutableString
result.append(String(i))                         //0.751s 5891.694kB

*注意将Int转换为字符串

源代码

import XCTest

class StringTests: XCTestCase {
    let count = 1_000_000
    
    let metrics: [XCTMetric] = [
        XCTClockMetric(),
        XCTMemoryMetric()
    ]
    
    let measureOptions = XCTMeasureOptions.default
    
    override func setUp() {
        measureOptions.iterationCount = 5
    }
    
    func testAppend() {
        var result = ""
        measure(metrics: metrics, options: measureOptions) {
            for i in 0...count {
                result.append(String(i))
            }
        }
    }
    
    func testPlusEqual() {
        var result = ""
        measure(metrics: metrics, options: measureOptions) {
            for i in 0...count {
                result += String(i)
            }
        }
    }
    
    func testPlus() {
        var result = ""
        measure(metrics: metrics, options: measureOptions) {
            for i in 0...count {
                result = result + String(i)
            }
        }
    }
    
    func testInterpolation() {
        var result = ""
        measure(metrics: metrics, options: measureOptions) {
            for i in 0...count {
                result = "\(result)\(i)"
            }
        }
    }
    
    //Up to 10_000
    func testNSString() {
        var result: NSString =  ""
        measure(metrics: metrics, options: measureOptions) {
            for i in 0...count {
                result = NSString(format: "%@%i", result, i)
            }
        }
    }
    
    func testNSMutableString() {
        let result = NSMutableString()
        measure(metrics: metrics, options: measureOptions) {
            for i in 0...count {
                result.append(String(i))
            }
        }
    }
}

Xcode不接受与普通字符串一起添加的可选字符串。我写了这个扩展来解决这个问题:

extension String {
    mutating func addString(str: String) {
        self = self + str
    }
}

然后你可以这样称呼它:

var str1: String?
var str1 = "hi"
var str2 = " my name is"
str1.addString(str2)
println(str1) //hi my name is

然而，你现在也可以这样做:

var str1: String?
var str1 = "hi"
var str2 = " my name is"
str1! += str2