1.)我有这两个功能,我的团队负责人希望我缩短使用闭包。我是一名新开发人员,无法找到缩短这些内容的正确方法。
func sortUsers() {
var sorted = [Member]()
var sortedObjects = [PFObject]()
for j in 0...2 {
if member.count > 0 {
for i in 0...member.count - 1 {
if j == 0 {
if member[i].role == "OWNER" {
sorted.append(member[i])
sortedObjects.append(memberObject[i])
}
} else if j == 1 {
if member[i].role == "ADMIN" {
sorted.append(member[i])
sortedObjects.append(memberObject[i])
}
} else {
if member[i].role == "USER" {
sorted.append(member[i])
sortedObjects.append(memberObject[i])
}
}
}
}
}
member = sorted
memberObject = sortedObjects
}
使用此闭包几乎可以缩短。 member.sort {$0.role < $1.role }
然而它按字母顺序排列,这样就可以了,但我需要所有者永远是第一位的。
2.)第二个功能我没有想到什么。它有我们一直在做的方式,然后是一个闭包,我尝试用更换更长的方法,但它不起作用。
func filterUsers() {
sortUsers()
switch roleSelector.selectedSegmentIndex {
case 0:
filtered = member
filteredObjects = memberObject
tableView.reloadData()
case 1:
filtered = []
filteredObjects = []
for i in 0...member.count - 1 {
if member[i].role == "OWNER" {
filtered.append(member[i])
filteredObjects.append(memberObject[i])
}
}
tableView.reloadData()
case 2:
member.filter { $0.role == "ADMIN" }
tableView.reloadData()
case 3:
filtered = []
filteredObjects = []
for i in 0...member.count - 1 {
if member[i].role == "USER" {
filtered.append(member[i])
filteredObjects.append(memberObject[i])
}
}
tableView.reloadData()
default:
break
}
}
我非常感谢任何善良的灵魂可以为我提供的任何帮助:)
答案 0 :(得分:7)
首先,为Role
创建常量,例如,使用enum
:
enum Role : String {
case owner = "OWNER"
case admin = "ADMIN"
case user = "USER"
}
现在让我们声明角色的排序:
extension Role : Comparable {
var order: Int {
switch self {
case .owner:
return 0
case .admin:
return 1
case .user:
return 2
}
}
static func < (lhs: Role, rhs: Role) -> Bool {
return lhs.order < rhs.order
}
}
假设Member
看起来像这样(请注意,我使用的是Role
枚举,而不是String
。)
class Member : CustomDebugStringConvertible {
var role: Role
var name: String
init(role: Role, name: String) {
self.role = role
self.name = name
}
var debugDescription: String {
return self.name
}
}
let members: [Member] = [
Member(role: .user, name: "user1"),
Member(role: .owner, name: "owner"),
Member(role: .user, name: "user2"),
Member(role: .admin, name: "admin2"),
Member(role: .admin, name: "admin2"),
]
现在您的排序可以简化为:
let sortedMembers = members.sorted { $0.role < $1.role }
要将两个数组排序在一起,我们可以,例如,创建成员对象并对这些对进行排序:
let pairs = zip(members, objects)
let sortedPairs = pairs.sorted { $0.0.role < $1.0.role }
let sortedMembers = sortedPairs.map { $0.0 }
let sortedObjects = sortedPairs.map { $0.1 }
要获取已过滤的对象,请使用角色再次简化:
if roleSelector.selectedSegmentIndex == 0 { // all selected
filtered = member
filteredObjects = memberObject
tableView.reloadData()
return
}
let roles: [Role] = [.owner, .admin, .user]
let role = roles[roleSelector.selectedSegmentIndex - 1]
filtered = members.filter { $0.role == role }
tableView.reloadData()