Question

我正在学习斯威夫特并在操场上瞎逛。我有以下词典：

var person = [
    "first": "John",
    "last": "Smith",
    "age": 21
]

我使用以下行打印输出：

"Your first name is \(person["first"]) and you are \(person["age"]) years old."

使用此代码，我得到以下输出：

// -> "Your first name is Optional(John) and you are Optional(21) years old."

我希望收到以下内容作为输出：

// -> "Your first name is John and you are 21 years old."

可选来自哪里？为什么这不是简单地在指定键上打印值？我需要做些什么来解决这个问题？

Answer 1

从字典中检索给定键的值始终是可选的，因为键可能不存在，然后值为=INDEX(V2, ,)-this is the formula I wrote to Index column V, thus creating column AV =IFERROR(VLOOKUP("Biscuits",V2:AW2,28,FALSE),"")-This is a formula I wrote to look for "Biscuits" and replace it with the contents of column AW。使用字符串插值nil可选包含在文字字符串中。

要避免字符串插值中的文字"\(...)"，您必须以安全的方式打开首选的选项

Optional(...)

Answer 2

您的字符串尚未解开并且是可选的，这就是您看到单词optional和括号的原因。如果你想让它们消失，你可以放一个！打开它。但是，我建议以不同的方式处理它，这样你就不会尝试打开零值。

例如，

var person = [
"first": "John",
"last": "Smith",
"age": 21
]

print("Your first name is \(person["first"]) and you are \(person["age"]) years old.")
// prints: "Your first name is Optional(John) and you are Optional(21) years old."

print("Your first name is \(person["first"]!) and you are \(person["age"]!) years old.")
// prints: "Your first name is John and you are 21 years old."

let name = person["first"]!
let age = person["age"]!
print("Your first name is \(name) and you are \(age) years old.")
// prints: "Your first name is John and you are 21 years old."

Vadian有一个很好的例子，说明如何正确打印出来，因为如果你得到的是零，他的例子不会崩溃。

Answer 3

尝试像这样做

"Your first name is \(person["first"]!) and you are \(person["age"]!) years old."

然后尝试在这里查看完美的解释Printing optional variable

为什么单词＆＃34;可选＆＃34;当我使用字典条目作为变量打印字符串时添加？

3 个答案: