我的原始 json 数据可能会误导您。键数组并不总是在同一索引处匹配其值。所以我重写了我的数据以反射(reflect)我的意图。
假设我们有一个 TableView 来显示带有它的 json 的歌曲:
{
"albums": [
{
"title": "A",
"id": "174172",
"artistName": "Person X"
},
{
"title": "B",
"id": "19201827",
"artistName": "Person Y"
},
{
"title": "C",
"id": "1927",
"artistName": "Person Z"
}
],
"songs": [
{
"name": "Song A",
"albumName": "A",
"albumId": "174172",
"duration": 180
},
{
"name": "Song B",
"albumName": "A",
"albumId": "174172",
"duration": 200
},
{
"name": "Song C",
"albumName": "B",
"albumId": "19201827",
"duration": 216
},
{
"name": "Song D",
"albumName": "C",
"albumId": "1927",
"duration": 216
}
]
}
我的模式是这样的:
struct Album: Decodable {
let title: String
let id: String
let artistName: String
}
struct Song: Decodable {
let name: String
let albumName: String
let albumId: String
let duration: Int
}
像这样的 View Controller 假代码:
class ViewController: UIViewController {
var songs: [Song] = []
var albums: [Album] = []
func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
return songs.count
}
func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
let cell = tableview.dequeueReusableCell(withIdentifier: "SongCell", for: indexPath) as! SongCell
let song = songs[indexPath.row]
let album = albums.first { $0.id == song.albumId }
cell.updateUI(withSong: song, album: album)
return cell
}
func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
let song = songs[indexPath.row]
let album = albums.first { $0.id == song.albumId }
pushDetailSongViewController(song, album)
}
func pushDetailSongViewController(_ song: Song, _ album: Album?) {
}
}
当我们有太多带有专辑的歌曲时,let album = albums.first { $0.id == song.albumId }
会出现严重的性能问题。
那么我们应该在这里使用什么数据结构来处理更新性能?
最佳答案
在解析键和值之后,您可以将这两个数组组合成一个字典,然后让您的 TableView 的数据源成为该字典。
首先,使您的Song
结构符合Hashable
协议(protocol):
struct Song: Hashable {
为专辑和歌曲创建一个数组:
var albums: [Album] = []
var songs: [Song] = []
然后,将 songs
数组缩减为字典,如下所示:
let data = songs.reduce([Album: Song]()) { (result, song) -> [Album: Song] in
guard let album = albums.first(where: { $0.id == song.albumID }) else { return result }
return result.merging([album: song], uniquingKeysWith: { (first, _) in first })
}
我用两个演示数组对此进行了测试:
let albums = [Album(id: "1", name: "one"), Album(id: "2", name: "two"), Album(id: "3", name: "three")]
let songs = [Song(albumID: "1", name: "ONE"), Song(albumID: "2", name: "TWO"), Song(albumID: "3", name: "THREE")]
那些把数据
变成:
[
<Album id: "1", name: "one"> : <Song albumID: "1", name: "ONE">,
<Album id: "2", name: "two"> : <Song albumID: "2", name: "TWO">,
<Album id: "3", name: "three">: <Song albumID: "3", name: "THREE">
]
额外学分
如果你想要每张专辑的所有歌曲,你必须制作data
[Album: [Song]]
:
let data = albums.reduce([Album: [Song]]()) { (result, album) -> [Album: [Song]] in
let _songs = songs.filter({ $0.albumID == album.id })
guard !_songs.isEmpty else { return result }
return result.merging([album: _songs], uniquingKeysWith: { (first, _) in first })
}
使用以下数组:
let albums = [Album(id: "1", name: "one"), Album(id: "2", name: "two"), Album(id: "3", name: "three")]
let songs = [Song(albumID: "1", name: "ONE"), Song(albumID: "2", name: "TWO"), Song(albumID: "3", name: "THREE"),
Song(albumID: "1", name: "ONE-1"), Song(albumID: "1", name: "ONE-2"), Song(albumID: "3", name: "THREE-1")]
...你会得到:
[
<Album name: three, id: 3>: [
<Song name: THREE, albumID: 3>
<Song name: THREE-1, albumID: 3>
],
<Album name: one, id: 1>: [
<Song name: ONE, albumID: 1>,
<Song name: ONE-1, albumID: 1>,
<Song name: ONE-2, albumID: 1>
],
<Album name: two, id: 2>: [
<Song name: TWO, albumID: 2>
]
]
关于ios - 这里是否存在更好的解决方案来处理 Swift 中的这个 api 问题?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53213304/