在两个字典数组中查找重复键以使用新数组字典更新旧数组字典

Question

我想用新词典更新已保存的词典。如果不存在值，则需要将新值附加到字典中。还使用字典的新值更新旧的现有值。

我保存的旧字典数组：

 var savedDict = [
             ["id":"1","pic":"Alice.png","name":"Alice Smith","position":"Nurse"],
             ["id":"2","pic":"brad.png","name":"Brad Smith MD","position":"Primary Doctor"],
             ["id":"12","pic":"bob.png","name":"Bob Smith PhD","position":"Hospital Coordinator"],
            ]

新下载的字典数组：

let newDict =   [
             ["id":"1","pic":"alice.png","name":"Alice Smith","position":"Nurse"],
             ["id":"2","pic":"brad.png","name":"Brad Smith MD","position":"Primary Doctor"],
             ["id":"3","pic":"user.png","name":"Dr. Quam","position":"Immunologist"],
             ["id":"4","pic":"jennifer.jpg","name":"Jennifer Johnson","position":"Case Manager"],
             ["id":"5","pic":"user.png","name":"John Banks MD","position":"Cardiologist"],
             ["id":"6","pic":"tammie.png","name":"Tammie Summers","position":"Case Manager"]
            ]

我想：
1.将“Alice.png”更新为“alice.png”，“id”=1，
2.ignore "id" = 2 因为两个字典数组相同，
3. 将 newDict 中的所有新项目追加到 savedDict

最终的字典应该是：

 upToDateDict = [
             ["id":"1","pic":"alice.png","name":"Alice Smith","position":"Nurse"],
             ["id":"2","pic":"brad.png","name":"Brad Smith MD","position":"Primary Doctor"],
             ["id":"12","pic":"bob.png","name":"Bob Smith PhD","position":"Hospital Coordinator"],
             ["id":"3","pic":"user.png","name":"Dr. Quam","position":"Immunologist"],
             ["id":"4","pic":"jennifer.jpg","name":"Jennifer Johnson","position":"Case Manager"],
             ["id":"5","pic":"user.png","name":"John Banks MD","position":"Cardiologist"],
             ["id":"6","pic":"tammie.png","name":"Tammie Summers","position":"Case Manager"]
            ]

到目前为止我已经试过了：

 import UIKit
 import Foundation

 func arrayContains(array:[[String:String]], value:[String:String]) -> Bool {
     for item in array {
         if item == value {
             return true
         }
     }
     return false
 }

 var upToDateDict = savedDict//Array<Dictionary<String,String>>()

 //Save all of the dictionaries from the 1st array (savedDict) that aren't in the 2nd array (newDict)
 for item in newDict {
     if !arrayContains(array: savedDict, value: item) {
         upToDateDict.append(item)
     }
 }

 print(upToDateDict)

 //find duplicate keys and check if need to update saved dict with new dict

 let key = "id"

 for dict1 in savedDict {
     if let value = dict1[key] {
         for dict2 in newDict {
             if dict2[key] == value { //if true duplicate "id" key's  found

                 if dict1 != dict2 { //not duplicate for all keys, so update saved with new dictionary

                     print("found \(key):\(value) in both arrays")
                     print("dict1:\(dict1)")
                     print("dict2:\(dict2)")

                     savedDict[dict1] = dict2 //what I want to do but will not build 

                 } else {}//all keys duplicate in dict1 and dict2

             }
         }
     }
 }

我只想用 new 覆盖这个字典，所以我将 ["id":"1","pic":"Alice.png"... 更新为 ["id":"1","pic": “alice.png”...但这不起作用：

 savedDict[dict1] = dict2 

Answer 1

你应该创造

class Employee {
    var id: String
    var pic: String
    var name: String
    var position: String

    init(id: String, pic: String, name: String, position: String) {
        self.id = id
        self.pic = pic
        self.name = name
        self.position = position
    }
}

那么你就有了

var oldEmployes: [Employee] = [Employee(id: "1", pic: "pic.png", name: "Name", position: "Position"), 
                               Employee(id: "2", pic: "pic2.png", name: "Name2", position: "Position2")]
var newEmployes: [Employee] = [Employee(id: "2", pic: "pic21.png", name: "Name21", position: "Position2"), 
                               Employee(id: "3", pic: "pic3.png", name: "Name3", position: "Position3")]

更新数组做一些类似的事情（使用高级函数可以更美观，但我认为这样看起来很清楚）

for employee in newEmployes {
     var isNew = true
     for oldEmployee in oldEmployes {
           if oldEmployee.id == employee.id {
               oldEmployee.name = employee.name
               oldEmployee.pic = employee.pic
               oldEmployee.position = employee.position
               isNew = false
           }
     }
     if isNew {
         oldEmployes.append(employee)
     }
}