詮釋爲非可選字符串

-6

我有一個字符串00012。我需要製作一個字符串12。然後通過將字符串連接到名稱來加載UIImage。所以我做了以下詮釋爲非可選字符串

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

的問題是，（敏）返回一個可選的。我沒有得到imageName = name_12，我得到imageName = name_optional(12)

我該如何擺脫可選？

我不能修剪從初始字符串前3個零，因爲數量可00001

來源

2017-01-19 Alexey K

選配是雨燕編程語言的完全不可分割的一部分。 [第一部分，Swift編程語言指南的「基礎知識」]詳細介紹了它們（https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/） TheBasics.html＃// apple_ref/DOC/UID/TP40014097-CH5-ID330）。閱讀。 – Alexander

使用另購的綁定：

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

選擇使用正則表達式：

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

來源

2017-01-19 05:52:08 vadian

你可以直接解包數值

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

或者你可以使用可選的結合

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

來源

2017-01-19 06:42:57

詮釋爲非可選字符串

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