诠释为非可选字符串

-6

我有一个字符串00012。我需要制作一个字符串12。然后通过将字符串连接到名称来加载UIImage。所以我做了以下诠释为非可选字符串

let myString = "00012" 
let myInt = Int(myString) 
let imageName = "name_\(myInt)" 
let image = UIImage(named: imageName)

的问题是，（敏）返回一个可选的。我没有得到imageName = name_12，我得到imageName = name_optional(12)

我该如何摆脱可选？

我不能修剪从初始字符串前3个零，因为数量可00001

来源

2017-01-19 Alexey K

选配是雨燕编程语言的完全不可分割的一部分。 [第一部分，Swift编程语言指南的“基础知识”]详细介绍了它们（https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/） TheBasics.html＃// apple_ref/DOC/UID/TP40014097-CH5-ID330）。阅读。 – Alexander

使用另购的绑定：

let myString = "00012" 
if let myInt = Int(myString) { 
    let imageName = "name_\(myInt)" 
    let image = UIImage(named: imageName) 
}

选择使用正则表达式：

let myString = "00012" 
let myStringWithoutLeadingZeros = myString.replacingOccurrences(of: "^0+", with: "", options: .regularExpression) 
let imageName = "name_" + myStringWithoutLeadingZeros 
let image = UIImage(named: imageName)

来源

2017-01-19 05:52:08 vadian

你可以直接解包数值

let myString = "00012" 
let myInt = Int(myString)! // unwrapping 
let imageName = "name_\(myInt)" 
let image = UIImage(named: imageName)

或者你可以使用可选的结合

let myString = "00012" 
if let myInt = Int(myString) { // Optional binding 
    let imageName = "name_\(myInt)" 
    let image = UIImage(named: imageName) 
}

来源

2017-01-19 06:42:57

诠释为非可选字符串

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