在最优秀的SQLite.swift中,我有
let stmt = try local.db!.prepare(..)
for row in stmt {
for tricky in row {
每个&#34;棘手&#34;是Optional<Binding>
我知道打开每个棘手问题的唯一方法就是这样
var printable:String =&#34;
if let trickyNotNil = tricky {
if let anInt:Int64 = trickyNotNil as? Int64 {
print("it's an Int")
.. use anInt
}
if let aString:String = trickyNotNil as? String {
print("it's a String")
.. use aString}
}
else {
tricky is nil, do something else
}
在这个例子中,我非常确定它只能是Int64或String(或者转到String的东西);我想有人可能需要用默认情况来涵盖其他任何可能性。
有更快捷的方式吗?
是否有办法获取Optional<Binding>
的类型?
(特别是BTW关于SQLite.swift;我可能有一种方式从doco中没有意识到&#34;得到列n&#34;的类型。那会很酷,但是,问题在于一般而言,前一段仍然存在。)
答案 0 :(得分:1)
您可以使用基于类的switch语句。这种switch语句的示例代码如下所示:
let array: [Any?] = ["One", 1, nil, "Two", 2]
for item in array {
switch item {
case let anInt as Int:
print("Element is int \(anInt)")
case let aString as String:
print("Element is String \"\(aString)\"")
case nil:
print("Element is nil")
default:
break
}
}
如果需要,您还可以在一个或多个案例中添加where子句:
let array: [Any?] = ["One", 1, nil, "Two", 2, "One Hundred", 100]
for item in array {
switch item {
case let anInt as Int
where anInt < 100:
print("Element is int < 100 == \(anInt)")
case let anInt as Int where anInt >= 100:
print("Element is int >= 100 == \(anInt)")
case let aString as String:
print("Element is String \"\(aString)\"")
case nil:
print("Element is nil")
default:
break
}
}