在swift中解析XML数据
这是我的第一个iOS应用程序,我从XML获取数据时有点麻烦。我需要从如下所示的XML文件中获取歌曲名称和艺术家:在swift中解析XML数据,swift,xml,Swift,Xml,这是我的第一个iOS应用程序,我从XML获取数据时有点麻烦。我需要从如下所示的XML文件中获取歌曲名称和艺术家: <?xml version="1.0" encoding="utf-8"?> <Schedule System="Jazler"> <Event status="happening" startTime="19:14:30" eventType="song"> <Announcement Display="Now On Air:"
<?xml version="1.0" encoding="utf-8"?>
<Schedule System="Jazler">
<Event status="happening" startTime="19:14:30" eventType="song">
<Announcement Display="Now On Air:"/>
<Song title="E timpul">
<Artist name="Revers">
</Artist>
<Jazler ID="16490"/>
<PlayLister ID=""/>
<Media runTime="03:03"/>
<Expire Time="19:17:33"/>
</Song>
</Event>
</Schedule>
非常感谢您的帮助。首先将xml转换为NSData,并调用解析器对其进行解析
//converting into NSData
var data: Data? = theXML.data(using: .utf8)
//initiate NSXMLParser with this data
var parser: XMLParser? = XMLParser(data: data ?? Data())
//setting delegate
parser?.delegate = self
//call the method to parse
var result: Bool? = parser?.parse()
parser?.shouldResolveExternalEntities = true
现在,您需要在类中实现NSXMLParser委托
func parser(_ parser: XMLParser, didStartElement elementName: String, namespaceURI: String?, qualifiedName qName: String?, attributes attributeDict: [String : String] = [:]) {
currentElement = elementName
print("CurrentElementl: [\(elementName)]")
}
func parser(_ parser: XMLParser, foundCharacters string: String) {
print("foundCharacters: [\(string)]")
}
您将在xml的键下找到值。因为您的xml包含元素属性的所有值,所以您不需要实现
foundCharacters
。例如,您的解析器委托可能看起来很简单,如下所示:
var song: String?
var artist: String?
func parser(_ parser: XMLParser, didStartElement elementName: String, namespaceURI: String?, qualifiedName qName: String?, attributes attributeDict: [String : String] = [:]) {
switch elementName {
case "Song": song = attributeDict["title"]
case "Artist": artist = attributeDict["name"]
default: break
}
}
两项意见:
URLSession
,以防对请求的响应有点慢。通常,应该避免使用XMLParser(contentsOf:)
,因为这是同步执行请求的
在您的情况下,因为您是从localhost
请求数据,所以这可能不太重要。但是,始终异步执行HTTP请求是明智的
class SongParser: NSObject {
var song: String?
var artist: String?
class func requestSong(completionHandler: @escaping (String?, String?, Error?) -> Void) {
let url = URL(string: "http://localhost/jazler/NowOnAir.xml")!
let task = URLSession.shared.dataTask(with: url) { data, _, error in
guard let data = data, error == nil else {
DispatchQueue.main.async {
completionHandler(nil, nil, error)
}
return
}
let delegate = SongParser()
let parser = XMLParser(data: data)
parser.delegate = delegate
DispatchQueue.main.async {
completionHandler(delegate.song, delegate.artist, parser.parserError)
}
}
task.resume()
}
}
extension SongParser: XMLParserDelegate {
func parser(_ parser: XMLParser, didStartElement elementName: String, namespaceURI: String?, qualifiedName qName: String?, attributes attributeDict: [String : String] = [:]) {
switch elementName {
case "Song": song = attributeDict["title"]
case "Artist": artist = attributeDict["name"]
default: break
}
}
}
你会这样使用它:
SongParser.requestSong { song, artist, error in
guard let song = song, let artist = artist, error == nil else {
print(error ?? "Unknown error")
return
}
print("Song:", song)
print("Artist:", artist)
}
非常感谢你。它起作用了!刚刚在第一个函数中添加了这些行,我得到了歌曲和艺术家的名字。如果(elementName==“Song”){Song=attributeDict[“title”]!print(Song)}如果(elementName==“Artist”){Artist=attributeDict[“name”]!print(Artist)}如果你得到了答案,请接受答案,让社区知道你得到了答案Hey@Rob,我如何修改它来解析本地文件,例如从Apple Music导出的library.xml文件?@southernyanke65-重新解析本地文件,只需使用解析文件即可。但以上是一种特殊情况,其中XML的所有信息都包含在XML标记的属性中,这是非常不寻常的。通常情况下,在开始和结束标记对之间需要捕获字符(比如)。@Rob,我们可以在SwiftUI中获取XML数据吗?我需要更改什么?这里没有UI代码,所以它在SwiftUI中与在UIKit中相同。
SongParser.requestSong { song, artist, error in
guard let song = song, let artist = artist, error == nil else {
print(error ?? "Unknown error")
return
}
print("Song:", song)
print("Artist:", artist)
}