檢查值是否與自定義數組元素列表中的鍵匹配並檢索其值

我在Swift中有兩個自定義數據數組。我想找到包含第二個數組中的「Destination」等於'PlaceId'的字典值的對象。這裏有一些數據結構只是爲了展示他們的樣子。檢查值是否與自定義數組元素列表中的鍵匹配並檢索其值

struct FlightInfo { 
var minPrice = "648" 
var carrier = "Canada Air ways" 
var destination = "973973" 
var origin = "983983" 
} 


struct PlaceData { 
var cityName = "Melbourne" 
var name = "Melbourne" 
var cityId = "MEL" 
var type = "blah" 
var countryName = "Australia" 
var placeId = 983983 
}

我得到這個迄今：

let destId = flightsArray[indexPath.row].destination as String 
    let results = listOfPlaces.filter { $0.placeId == Int(destId) }

，但沒有雪茄。我怎麼可以結束了類似以下內容作爲最終的結果是：

cell.destinationLabel.text = results.placeID

來源

2017-02-20 SlurBeast

'$ 0 placeId'是一個'String'，並且你試圖和'Int'（'Int（destId）'）進行比較。這可能是你的問題。而不是「{$ 0.placeId == destId}」？ – Larme

在第一行代碼中嘗試'作爲Int'，然後在第二行中將'Int（destId）'更改爲'destId' – Florensvb

首先，我會重新考慮使用字典作爲模型對象。代替使用structs,classes和enums是非常可靠的。因此，你的模式將類似於此：

enum FlightType: String { 
    case a = "A" 
    case b = "B" 
} 

enum PlaceType: String { 
    case station = "station" 
    case port = "port" 
} 

struct Flight { 
    var name: String 
    var destinationId: Int 
    var type: FlightType 
} 

struct Place { 
    var identifier: Int 
    var title: String 
    var type: PlaceType 
}

假設你有數組是這樣的：

let flights = [ 
    Flight(name: "Trenton", destinationId: 403, type: .a), 
    Flight(name: "Houston", destinationId: 288, type: .a), 
    Flight(name: "Boston", destinationId: 244, type: .a) 
] 

let places = [ 
    Place(identifier: 111, title: "Leslie", type: .station), 
    Place(identifier: 228, title: "Mary", type: .station), 
    Place(identifier: 684, title: "Joe", type: .station), 
    Place(identifier: 288, title: "Paul", type: .station), 
    Place(identifier: 465, title: "Stephanie", type: .station), 
    Place(identifier: 569, title: "Steve", type: .station), 
    Place(identifier: 663, title: "Doug", type: .station) 
]

你可以飛行的目的地標題是這樣的：

let destinationId = flights[indexPath.row].destinationId 
let destinationTitle = places.filter{ $0.identifier == destinationId }.first?.title

來源

2017-02-20 13:53:57 alexburtnik

謝謝，在我的代碼中，它們是結構體。但我確實喜歡你把它們組織成枚舉。我可能會這樣做，但都在同一個結構中。 – SlurBeast

以你的定義，以粗糙的代碼，這會給你想要的東西：

注意，在你的榜樣，你給了其他對象，所以只需將其轉換爲您的需求。

let flightsArray = 
    [[ "Name": "Trenton", "destination": 403, "type":"A" ], 
    [ "Name": "Houston", "destination": 288, "type":"A" ], 
    [ "Name": "Boston", "destination": 244, "type":"A" ]] 

var placedata = [ 
    ["PlaceId": "111", "title": "Leslie", "type": "station"], 
    ["PlaceId": "228", "title": "Mary", "type": "station"], 
    ["PlaceId": "684", "title": "Joe", "type": "station"], 
    ["PlaceId": "288", "title": "Paul", "type": "station"], 
    ["PlaceId": "465", "title": "Stephanie", "type": "station"], 
    ["PlaceId": "569", "title": "Steve", "type": "station"], 
    ["PlaceId": "663", "title": "Doug", "type": "station"], 
] 

for flightAt in flightsArray 
{ 
    let places = placedata.filter({ Int($0["PlaceId"]!)! == (flightAt["destination"] as! NSNumber).intValue }) 

    for place in places 
    { 
     print ("\(place["title"])") 
    } 
}

來源

2017-02-20 13:19:11 unkgd

你可以嘗試

let flightsArray = 
    [[ "Name": "", "destination": 403, "type":"A" ], 
[ "Name": "", "destination": 288, "type":"A" ], 
[ "Name": "", "destination": 244, "type":"A" ]] as [[String : Any]] 
var placedata = [ 
    ["PlaceId": "111", "title": "Leslie", "type": "station"], 
    ["PlaceId": "228", "title": "Mary", "type": "station"], 
    ["PlaceId": "684", "title": "Joe", "type": "station"], 
    ["PlaceId": "288", "title": "Paul", "type": "station"], 
    ["PlaceId": "465", "title": "Stephanie", "type": "station"], 
    ["PlaceId": "569", "title": "Steve", "type": "station"], 
    ["PlaceId": "663", "title": "Doug", "type": "station"], 
] 
let destId = flightsArray[indexPath.row]["destination"] as! Int 
let results = placedata.filter { $0["PlaceId"]! as String == "\(destId)" } 
cell.destinationLabel.text = results[0]["placeID"]

來源

2017-02-20 13:47:48 Ram

檢查值是否與自定義數組元素列表中的鍵匹配並檢索其值

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