调用下标时没有完全匹配 [Swift]

Question

在这下面你可以看到我的代码。我评论了导致错误的行。错误消息：“调用下标时没有完全匹配”。你知道我怎样才能避免这个错误吗？感谢您的帮助！

        let dic = ["a": 1, "b": 2, "c": 3, "d": 4, "e": 5, "f": 6, "g": 7, "h": 8, "i": 9, "j": 10, "k": 11, "l": 12, "m": 13, "n": 14, "o": 15, "p": 16, "q": 17, "r": 18, "s": 19, "t": 20, "u": 21, "v": 22, "w": 23, "x": 24, "y": 25, "z": 26]
        var newwrd = ""
        for var i in str ?? "" {
            let ci = dic[i]   // This line causes the error
        }

Answer 1

实际上你应该得到错误

无法使用类型为“String.Element”（也称为“字符”）的参数为类型为“[String : Int]”的值添加下标

str显然String?，枚举字符串时的元素类型是Character，但订阅类型必须是String。

let dic = ["a": 1, "b": 2, "c": 3, "d": 4, "e": 5, "f": 6, "g": 7, "h": 8, "i": 9, "j": 10, "k": 11, "l": 12, "m": 13, "n": 14, "o": 15, "p": 16, "q": 17, "r": 18, "s": 19, "t": 20, "u": 21, "v": 22, "w": 23, "x": 24, "y": 25, "z": 26]
for i in str ?? "" { // no need for var i
   let ci = dic[String(i)] ?? 0
   print(ci)
}

如果字符串包含字典中没有的字符，则结果为 0。

有一种更短的方法，无需辅助字典

for i in str ?? "" where ("a"..."z") ~= i {
    let ci = Int(i.asciiValue!) - 96
    print(ci)
}