我需要创建一个字符串的格式,可以转换Int, Int64,双精度等类型为字符串。使用Objective-C,我可以通过:

NSString *str = [NSString stringWithFormat:@"%d , %f, %ld, %@", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE, STRING_VALUE];

如何做同样的,但在Swift?


当前回答

有一个简单的解决方案,我学会了“我们<3 Swift”,如果你既不能导入基础,使用round()和/或不想要字符串:

var number = 31.726354765
var intNumber = Int(number * 1000.0)
var roundedNumber = Double(intNumber) / 1000.0

结果:31.726

其他回答

首先阅读Swift语言的官方文档。

答案应该是

var str = "\(INT_VALUE) , \(FLOAT_VALUE) , \(DOUBLE_VALUE), \(STRING_VALUE)"
println(str)

在这里

1)任何浮点值默认为double

EX.
 var myVal = 5.2 // its double by default;

->如果你想显示浮点值,那么你需要显式地定义像a

 EX.
     var myVal:Float = 5.2 // now its float value;

这要清楚得多。

公认的答案肯定是最好的通用解决方案(即,只需使用Foundation中的String(format:_:)方法),但…

如果你运行的是Swift≥5,你可以利用新的StringInterpolationProtocol协议为你的应用程序中常见的字符串格式化用例提供一些非常好的语法糖。

以下是官方文件对这项新协议的总结:

表示在构建字符串文字时进行插值的字符串文字的内容。

一些简单的例子:

extension String.StringInterpolation {

    /// Quick formatting for *floating point* values.
    mutating func appendInterpolation(float: Double, decimals: UInt = 2) {
        let floatDescription = String(format: "%.\(decimals)f%", float)
        appendLiteral(floatDescription)
    }

    /// Quick formatting for *hexadecimal* values.
    mutating func appendInterpolation(hex: Int) {
        let hexDescription = String(format: "0x%X", hex)
        appendLiteral(hexDescription)
    }

    /// Quick formatting for *percents*.
    mutating func appendInterpolation(percent: Double, decimals: UInt = 2) {
        let percentDescription = String(format: "%.\(decimals)f%%", percent * 100)
        appendLiteral(percentDescription)
    }

    /// Formats the *elapsed time* since the specified start time.
    mutating func appendInterpolation(timeSince startTime: TimeInterval, decimals: UInt = 2) {
        let elapsedTime = CACurrentMediaTime() - startTime
        let elapsedTimeDescription = String(format: "%.\(decimals)fs", elapsedTime)
        appendLiteral(elapsedTimeDescription)
    }
}

可用作:

let number = 1.2345
"Float: \(float: number)" // "Float: 1.23"
"Float: \(float: number, decimals: 1)" // "Float: 1.2"

let integer = 255
"Hex: \(hex: integer)" // "Hex: 0xFF"

let rate = 0.15
"Percent: \(percent: rate)" // "Percent: 15.00%"
"Percent: \(percent: rate, decimals: 0)" // "Percent: 15%"

let startTime = CACurrentMediaTime()
Thread.sleep(forTimeInterval: 2.8)
"∆t was \(timeSince: startTime)" // "∆t was 2.80s"
"∆t was \(timeSince: startTime, decimals: 0)" // "∆t was 3s"

这是由SE-0228引入的,所以请务必阅读原始提案,以更深入地了解这个新功能。最后,协议文档也很有帮助。

我认为这可以帮助你:

import Foundation

let timeNow = time(nil)
let aStr = String(format: "%@%x", "timeNow in hex: ", timeNow)
print(aStr)

结果示例:

timeNow in hex: 5cdc9c8d

有一个简单的解决方案,我学会了“我们<3 Swift”,如果你既不能导入基础,使用round()和/或不想要字符串:

var number = 31.726354765
var intNumber = Int(number * 1000.0)
var roundedNumber = Double(intNumber) / 1000.0

结果:31.726

使用下面的代码:

    let intVal=56
    let floatval:Double=56.897898
    let doubleValue=89.0
    let explicitDaouble:Double=89.56
    let stringValue:"Hello"

    let stringValue="String:\(stringValue) Integer:\(intVal) Float:\(floatval) Double:\(doubleValue) ExplicitDouble:\(explicitDaouble) "