ios Swift：按字母顺序对对象数组进行排序

Swift: Sort array of objects alphabetically

提问by Jonah Ruffer

I have this:

我有这个：

class Movies {
  Name:String
  Date:Int
}

and an array of [Movies]. How do I sort the array alphabetically by name? I've tried:

和一系列[电影]。如何按名称的字母顺序对数组进行排序？我试过了：

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

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

and

和

movieArr = sorted(movieArr)

movieArr = sorted(movieArr)

but that doesn't work because I'm not accessing the name attribute of Movies.

但这不起作用，因为我没有访问 Movies 的 name 属性。

回答by Mike S

In the closure you pass to sort, compare the properties you want to sort by. Like this:

在传递给 的闭包中sort，比较要作为排序依据的属性。像这样：

movieArr.sorted { 
movieArr.sorted { 
class Movie {
    let name: String
    var date: Int?

    init(_ name: String) {
        self.name = name
    }
}

var movieA = Movie("A")
var movieB = Movie("B")
var movieC = Movie("C")

let movies = [movieB, movieC, movieA]
let sortedMovies = movies.sorted { 
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
    }

}
.name < .name }
sortedMovies
.name.lowercased() < .name.lowercased() }
.name < .name }

or the following in the cases that you want to bypass cases:

或在您想要绕过案例的情况下执行以下操作：

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: { 
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
    }

}
.name < .name })
// let sortedMovies = movies.sorted { 
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: { 
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
    }

}
 <  })
// let sortedMovies = movies.sorted { 
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]
 */
 <  } // also works
// let sortedMovies = movies.sorted(by: <) // also works

print(sortedMovies)

/*
 prints: [Avatar, Piranha II: The Spawning, Titanic]
 */
.name < .name } // also works

print(sortedMovies)

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

Sidenote:Typically only types start with an uppercase letter; I'd recommend using nameand date, not Nameand Date.

旁注：通常只有类型以大写字母开头；我建议使用nameand date，而不是Nameand Date。

---

---

Example, in a playground:

例如，在操场上：

let sortArray =  array.sorted(by: { 
movieArr.sorted { 
arrayOfRaces = arrayOfItems.sorted(by: { (
*import Foundation
import CoreData


extension Messages {

    @nonobjc public class func fetchRequest() -> NSFetchRequest<Messages> {
        return NSFetchRequest<Messages>(entityName: "Messages")
    }

    @NSManaged public var text: String?
    @NSManaged public var date: Date?
    @NSManaged public var friends: Friends?

}

    //here arrMessage is the array you can sort this array as under bellow 

    var arrMessages = [Messages]()

    arrMessages.sort { (arrMessages1, arrMessages2) -> Bool in
               arrMessages1.date! > arrMessages2.date!
    }*
["raceName"] as! String) < (["raceName"] as! String) })
.Name < .Name }
.name.lowercased() < .name.lowercased() })

sortedMovieswill be in the order [movieA, movieB, movieC]

sortedMovies将按顺序 [movieA, movieB, movieC]

回答by Imanou Petit

With Swift 3, you can choose one of the following ways to solve your problem.

使用 Swift 3，您可以选择以下方法之一来解决您的问题。

---

---

1. Using sorted(by:?)with a Movieclass that does not conform to Comparableprotocol

1.sorted(by:?)与Movie不符合Comparable协议的类一起使用

If your Movieclass does not conform to Comparableprotocol, you must specify in your closure the property on which you wish to use Array's sorted(by:?)method.

如果你的Movie类不符合Comparable协议，你必须在你的闭包中指定你希望使用 Arraysorted(by:?)方法的属性。

Movieclass declaration:

Movie类声明：

##代码##

Usage:

用法：

##代码##

---

---

2. Using sorted(by:?)with a Movieclass that conforms to Comparableprotocol

2.sorted(by:?)与Movie符合Comparable协议的类一起使用

However, by making your Movieclass conform to Comparableprotocol, you can have a much concise code when you want to use Array's sorted(by:?)method.

但是，通过使您的Movie类符合Comparable协议，当您想要使用 Array 的sorted(by:?)方法时，您可以获得更简洁的代码。

Movieclass declaration:

Movie类声明：

##代码##

Usage:

用法：

##代码##

---

---

3. Using sorted()with a Movieclass that conforms to Comparableprotocol

3.sorted()与Movie符合Comparable协议的类一起使用

By making your Movieclass conform to Comparableprotocol, you can use Array's sorted()method as an alternative to sorted(by:?).

通过使您的Movie类符合Comparable协议，您可以使用 Array 的sorted()方法作为sorted(by:?).

Movieclass declaration:

Movie类声明：

##代码##

Usage:

用法：

##代码##

回答by m.alqadi

##代码##

回答by Jacobo Koenig

For those using Swift 3, the equivalent method for the accepted answer is:

对于那些使用 Swift 3 的人，接受答案的等效方法是：

##代码##

回答by CSE 1994

Sorted array Swift 4.2

排序数组 Swift 4.2

##代码##

回答by Muhammad Ahmad

##代码##