Question

我想知道是否有任何捷径设定 colorPredicate 的价值

枚举颜色{ 案例黑色案例白色 }

func predicateForColor(color: Color, compoundWith compoundPredicate: NSPredicate?) -> NSPredicate {

//  NOTE: if I use the code bellow to set the value of colorPredicate, will got error: expected initial value after '='.
//    let colorPredicate =
//        switch color {
//        case .black:   return predicateForBlack()
//        case .white:   return predicateForWhite()
//        }

    func getPredicateByColor(color: Color) -> NSPredicate {
        switch color {
        case .black:    return predicateForBlack()
        case .white:    return predicateForWhite()
        }
    }

    let colorPredicate = getPredicateByColor(color: color)

    if let predicate = compoundPredicate {
        return NSCompoundPredicate(andPredicateWithSubpredicates: [predicate, colorPredicate])
    } else {
        return colorPredicate
    }
}


func predicateForBlack() -> NSPredicate {
    print("get black predicate")
    return NSPredicate(format: "color = black")
}

func predicateForWhite() -> NSPredicate {
    print("get white predicate")
    return NSPredicate(format: "color = white & someother condition")
}


print(predicateForColor(color: .black, compoundWith: nil))

Answer 1

let colorPredicate: NSPredicate = { (color: Color) -> NSPredicate in
    switch color {
        case .black:   return predicateForBlack()
        case .white:   return predicateForWhite()
    }
}(color)

<强> 更新

您的代码会产生错误，因为您需要编写：

let variable = { switch { ... } }()

而不是

let variable = switch { ... }

这样你就可以定义一个块并调用它，你不能从switch语句中分配它。

字典方法

设置：

var lookup: [Color: NSPredicate] = [:]
lookup[.black] = NSPredicate(format: "color = black")
lookup[.white] = NSPredicate(format: "color = white")

使用：

let colorPredicate = lookup[color]

带开关的swift设定值会出现错误：＆＃39; =＆＃39;之后的预期初始值

1 个答案: