Question

我创建了一个包含id，name和shortname的自定义对象。我想只检索id并执行",".join()，这样它就像一个字符串，例如"1, 2"

那么如何将像var recentArray = Array<News>()这样的数组转换为只用逗号分隔id的字符串？

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")
    }    
}

Answer 1

你有使用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

Answer 2

map将对象添加到字符串数组中，然后加入：

", ".join(recentArray.map { toString($0.id) })

Answer 3

您可以映射数组以从Team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)"})

加入一组自定义对象

3 个答案: