假设我们有一个名为imageFile的自定义类，这个类包含两个属性:
class imageFile {
var fileName = String()
var fileID = Int()
}
很多都存储在一个数组中:
var images : Array = []
var aImage = imageFile()
aImage.fileName = "image1.png"
aImage.fileID = 101
images.append(aImage)
aImage = imageFile()
aImage.fileName = "image1.png"
aImage.fileID = 202
images.append(aImage)
我如何排序的图像数组由'fileID'升序或降序?
当前回答
首先，将Array声明为类型化数组，以便在迭代时调用方法:
var images : [imageFile] = []
然后你可以简单地做:
斯威夫特2
images.sorted({ $0.fileID > $1.fileID })
斯威夫特3
images.sorted(by: { $0.fileID > $1.fileID })
斯威夫特5
images.sorted { $0.fileId > $1.fileID }
上面的例子按降序给出了结果。
其他回答
如果你不使用自定义对象，而是使用值类型来实现可比协议(Int, String等)，你可以简单地这样做:
myArray.sort(>) //sort descending order
一个例子:
struct MyStruct: Comparable {
var name = "Untitled"
}
func <(lhs: MyStruct, rhs: MyStruct) -> Bool {
return lhs.name < rhs.name
}
// Implementation of == required by Equatable
func ==(lhs: MyStruct, rhs: MyStruct) -> Bool {
return lhs.name == rhs.name
}
let value1 = MyStruct()
var value2 = MyStruct()
value2.name = "A New Name"
var anArray:[MyStruct] = []
anArray.append(value1)
anArray.append(value2)
anArray.sort(>) // This will sort the array in descending order
首先，将Array声明为类型化数组，以便在迭代时调用方法:
var images : [imageFile] = []
然后你可以简单地做:
斯威夫特2
images.sorted({ $0.fileID > $1.fileID })
斯威夫特3
images.sorted(by: { $0.fileID > $1.fileID })
斯威夫特5
images.sorted { $0.fileId > $1.fileID }
上面的例子按降序给出了结果。
我是这样做的，而且很管用:
var图像=[图像文件]() 图片社。* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
斯威夫特3
people = people.sorted(by: { $0.email > $1.email })
几乎每个人都给出了如何直接，让我来展示一下演变:
你可以使用Array的实例方法:
// general form of closure
images.sortInPlace({ (image1: imageFile, image2: imageFile) -> Bool in return image1.fileID > image2.fileID })
// types of closure's parameters and return value can be inferred by Swift, so they are omitted along with the return arrow (->)
images.sortInPlace({ image1, image2 in return image1.fileID > image2.fileID })
// Single-expression closures can implicitly return the result of their single expression by omitting the "return" keyword
images.sortInPlace({ image1, image2 in image1.fileID > image2.fileID })
// closure's argument list along with "in" keyword can be omitted, $0, $1, $2, and so on are used to refer the closure's first, second, third arguments and so on
images.sortInPlace({ $0.fileID > $1.fileID })
// the simplification of the closure is the same
images = images.sort({ (image1: imageFile, image2: imageFile) -> Bool in return image1.fileID > image2.fileID })
images = images.sort({ image1, image2 in return image1.fileID > image2.fileID })
images = images.sort({ image1, image2 in image1.fileID > image2.fileID })
images = images.sort({ $0.fileID > $1.fileID })
关于排序的工作原理的详细解释，请参见排序函数。
