Ios Swift：按字母顺序对对象数组排序

我有这个：

class Movies {
  Name:String
  Date:Int
}

还有一系列的[电影]。如何按名称的字母顺序对数组排序？我试过：

movieArr=movieArr.sorted{$0<$1}

及

movieArr=已排序（movieArr）

---

但这不起作用，因为我没有访问Movies的name属性。

在传递给

排序的闭包中，比较要排序的属性。像这样：

movieArr.sorted { $0.name < $1.name }

sortedMovies
将按顺序排列
[movieA，movieB，movieC]

Swift5更新

使用Swift 3，您可以选择以下方法之一来解决问题


1.与不符合
可比
协议的
电影
类一起使用
如果
电影
类不符合
可比
协议，则必须在闭包中指定要使用数组的
排序属性（按：​)方法

电影
类别声明：

import Foundation

class Movie: CustomStringConvertible {

    let name: String
    var date: Date
    var description: String { return name }

    init(name: String, date: Date = Date()) {
        self.name = name
        self.date = date
    }

}

import Foundation

class Movie: CustomStringConvertible, Comparable {

    let name: String
    var date: Date
    var description: String { return name }

    init(name: String, date: Date = Date()) {
        self.name = name
        self.date = date
    }

    static func ==(lhs: Movie, rhs: Movie) -> Bool {
        return lhs.name == rhs.name
    }

    static func <(lhs: Movie, rhs: Movie) -> Bool {
        return lhs.name < rhs.name
    }

}

import Foundation

class Movie: CustomStringConvertible, Comparable {

    let name: String
    var date: Date
    var description: String { return name }

    init(name: String, date: Date = Date()) {
        self.name = name
        self.date = date
    }

    static func ==(lhs: Movie, rhs: Movie) -> Bool {
        return lhs.name == rhs.name
    }

    static func <(lhs: Movie, rhs: Movie) -> Bool {
        return lhs.name < rhs.name
    }

}

用法：

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted(by: { $0.name < $1.name })
// let sortedMovies = movies.sorted { $0.name < $1.name } // also works

print(sortedMovies)

/*
prints: [Avatar, Piranha II: The Spawning, Titanic]
*/

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted(by: { $0 < $1 })
// let sortedMovies = movies.sorted { $0 < $1 } // also works
// let sortedMovies = movies.sorted(by: <) // also works

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted()

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */

用法：

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted(by: { $0.name < $1.name })
// let sortedMovies = movies.sorted { $0.name < $1.name } // also works

print(sortedMovies)

/*
prints: [Avatar, Piranha II: The Spawning, Titanic]
*/

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted(by: { $0 < $1 })
// let sortedMovies = movies.sorted { $0 < $1 } // also works
// let sortedMovies = movies.sorted(by: <) // also works

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted()

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */

用法：

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted(by: { $0.name < $1.name })
// let sortedMovies = movies.sorted { $0.name < $1.name } // also works

print(sortedMovies)

/*
prints: [Avatar, Piranha II: The Spawning, Titanic]
*/

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted(by: { $0 < $1 })
// let sortedMovies = movies.sorted { $0 < $1 } // also works
// let sortedMovies = movies.sorted(by: <) // also works

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */

let avatarMovie = Movie(name: "Avatar")
let titanicMovie = Movie(name: "Titanic")
let piranhaMovie = Movie(name: "Piranha II: The Spawning")

let movies = [avatarMovie, titanicMovie, piranhaMovie]
let sortedMovies = movies.sorted()

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */

对于使用Swift 3的用户，可接受答案的等效方法为：

movieArr.sorted { $0.Name < $1.Name }

movieArr.sorted{$0.Name<$1.Name}
让sortArray=array.sorted（按：{$0.name.lowercased（）<$1.name.lowercased（）}）
排序数组
Swift 4.2

arrayOfRaces = arrayOfItems.sorted(by: { ($0["raceName"] as! String) < ($1["raceName"] as! String) })

arrayOfRaces=arrayOfItems.sorted（按：{（$0[“raceName”]as！String）<（$1[“raceName”]as！String）}）
<代码>代码>导入基础
导入CoreData
扩展消息{
@nonobjc公共类func fetchRequest（）->NSFetchRequest{
返回NSFetchRequest（entityName:“消息”）
}
@NSManaged公共变量文本：字符串？
@NSV管理的公共var日期：日期？
@NSVAR管理的公共var朋友：朋友？
}
//这里是数组，您可以按照下面的方式对数组进行排序
var arrMessages=[Messages]（）
arrMessages.sort{（arrMessages1，arrMessages2）->Bool-in
arrMessages1.date！>arrMessages2.date！
}*
由于未能使用基于区域设置的比较进行排序，这些答案中的大多数都是错误的
谢谢，我刚开始学习Swift，我不知道闭包的用法。请使用movies.sort而不是movies.sort@Mike Sthanks进行解释。改进了我的编辑，应该可以，请看一看：）注意，对可能大写或不大写的字符串进行排序时，这不起作用。比较
。小写的
可以解决这个问题。@EricAya注意到在Swift 3中它是
排序的
：D。嘿，第二部和第三部之间有什么区别吗（电影除外）。排序（按：
排序（按：）
和
排序（）
是两种不同的方法。你可以使用
数组。排序（）
作为
array.sorted的替代方法（by:是的，只是在类定义的情况下询问我们是否需要更改什么？我的意思是，
sorted（by:）
和
sorted（）可以有相同的类定义吗）
？是的，您可以使用与类/结构相同的实现，在任何情况下，都必须符合
可比的
协议。我使用customObject的数组按升序排列时，它没有以正确的方式缩短。
“1”“14”“28”“31”“4”“42”“49”
添加.lowercased（）经过长时间的尝试，我终于找到了工作。我不明白为什么，因为所有文本都是小写的。when use lowercased（）将所有名称改为小写，以便正确排序：）