Ios 连接自定义对象的数组
我已经创建了一个自定义对象,其中包含id、name和shortname。我只想检索id并执行一个Ios 连接自定义对象的数组,ios,arrays,swift,Ios,Arrays,Swift,我已经创建了一个自定义对象,其中包含id、name和shortname。我只想检索id并执行一个“,”.join(),这样它将是一个字符串,例如“1,2” 那么,如何将像var recentArray=array()这样的数组转换为一个id仅用逗号分隔的字符串呢 自定义类 class Team: NSObject{ var id: Int! var name: NSString! var shortname: NSString! init(id: Int, n
“,”.join()
,这样它将是一个字符串,例如“1,2”
那么,如何将像var recentArray=array()
这样的数组转换为一个id仅用逗号分隔的字符串呢
自定义类
class Team: NSObject{
var id: Int!
var name: NSString!
var shortname: NSString!
init(id: Int, name:NSString, shortname: NSString) {
self.id = id
self.name = name
self.shortname = shortname
}
required convenience init(coder aDecoder: NSCoder) {
let id = aDecoder.decodeIntegerForKey("id")
let name = aDecoder.decodeObjectForKey("name") as! String
let shortname = aDecoder.decodeObjectForKey("shortname") as! String
self.init(id: id, name: name, shortname: shortname)
}
func encodeWithCoder(aCoder: NSCoder) {
aCoder.encodeInteger(id, forKey: "id")
aCoder.encodeObject(name, forKey: "name")
aCoder.encodeObject(shortname, forKey: "shortname")
}
}
将对象映射到字符串数组中,然后将其连接:
", ".join(recentArray.map { toString($0.id) })
您必须使用map
功能
var t1 = Team(id: 1, name: "Adria", shortname: "Ad")
var t2 = Team(id: 2, name: "Roger", shortname: "Ro")
var t3 = Team(id: 3, name: "Raquel", shortname: "Ra")
var array: [Team] = [t1, t2, t3];
var arrayMap: Array = array.map(){ toString($0.id) }
var joinedString: String = ",".join(arrayMap)
println(joinedString) // 1,2,3
您可以映射数组以仅从团队数组中获取id,并将数组缩减为字符串:
let a = Team(id: 1, name: "Greg", shortname: "G")
let b = Team(id: 2, name: "John", shortname: "J")
let c = Team(id: 3, name: "Jessie", shortname: "Je")
let d = Team(id: 4, name: "Ann", shortname: "A")
let array = [a,b,c,d]
let result = array.map({$0.id}).reduce("", combine: {result, id in return result == "" ? "\(id)" : "\(result),\(id)"})
如果要在init
中指定非可选值,为什么要使用隐式展开选项?