Question

目前我正在检查字典中的某个键是否包含某个字符串值：

if dict["eventCode"] == "OFF" || dict["eventCode"] == "DHD" || dict["eventCode"] == "SBY"
  {
     // do something
  }
else if dict["eventCode"] == "DHT" || dict["eventCode"] == "SLV" || dict["eventCode"] == "CLV"
 {
   // do something else
}

可能的值列表大约为10，if语句工作但很长，容易出现类型错误。

折射这种“if语句”的最佳方法是什么？

Answer 1

String上的

switch

guard let val = dict["eventCode"] as? String else { return }

switch val {
case "OFF", "DHD", "SBY":
    print("something")

case "DHT", "SLV", "CLV":
    print("something else")

default:
    print("default")
}

Answer 2

let check1 = ["OFF", "DHD", "SBY"]
let check2 = ["DHT", "SLV", "CLV"]

if check1.contains(dict["eventCode"] ?? "") {
    print("found in 1")
} else if check2.contains(dict["eventCode"] ?? "") {
    print("found in 2")
}

但在我看来，这看起来很丑..

Answer 3

我通常使用switch声明：

switch dict["eventCode"] {
    case let ec? where ec == "OFF" || ec == "DHD" || ec == "SBY":
        // do something
    case let ec? where ec == "DHT" || ec == "SLV" || ec == "CLV":
        // do something else
    default:
        break 
}

虽然它不一定比一系列if-else分支更少的代码，但我发现逻辑更清晰，更容易阅读。

编辑：

@ Eendje的答案也非常好，可以与switch结合使用：

let check1 = ["OFF", "DHD", "SBY"]
let check2 = ["DHT", "SLV", "CLV"]

switch dict["eventCode"] {
    case let ec? where check1.contains(ec):
        // do something
    case let ec? where check2.contains(ec):
        // do something else
    default:
        break 
}

Swift如何检查多个字符串条件的变量

3 个答案: