如何在Swift连接字符串?

在Objective-C中

NSString *string = @"Swift";
NSString *resultStr = [string stringByAppendingString:@" is a new Programming Language"];

or

NSString *resultStr=[NSString stringWithFormat:@"%@ is a new Programming Language",string];

但我想用swift语言来写。


当前回答

来打印组合字符串

Println("\(string1)\(string2)")

或String3存储2个字符串组合的输出

let strin3 = "\(string1)\(string2)"

其他回答

你现在可以在Swift中使用stringByAppendingString。

var string = "Swift"
var resultString = string.stringByAppendingString(" is new Programming Language")

我刚从Objective-C切换到Swift(4),我发现我经常使用:

let allWords = String(format:"%@ %@ %@",message.body!, message.subject!, message.senderName!)

在Swift 5中,apple引入了使用#符号的原始字符串。

例子:

print(#"My name is "XXX" and I'm "28"."#)
let name = "XXX"
print(#"My name is \#(name)."#)

符号#必须在\后面。常规的\(name)将被解释为字符串中的字符。

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))
            }
        }
    }
}