我正在尝试获取一个JSON响应并将结果存储在一个变量中。在Xcode8的GM版本发布之前,我已经在Swift的以前版本中使用过这种代码。我在StackOverflow上看过一些类似的帖子:Swift 2 Parsing JSON - Cannot subscript a value of type 'AnyObject'和JSON Parsing in Swift 3。
然而,那里传达的思想似乎不适用于这种情况。
如何正确解析Swift 3中的JSON响应?在Swift 3中读取JSON的方式有什么变化吗?
下面是有问题的代码(它可以在操场上运行):
import Cocoa
let url = "https://api.forecast.io/forecast/apiKey/37.5673776,122.048951"
if let url = NSURL(string: url) {
if let data = try? Data(contentsOf: url as URL) {
do {
let parsedData = try JSONSerialization.jsonObject(with: data as Data, options: .allowFragments)
//Store response in NSDictionary for easy access
let dict = parsedData as? NSDictionary
let currentConditions = "\(dict!["currently"]!)"
//This produces an error, Type 'Any' has no subscript members
let currentTemperatureF = ("\(dict!["currently"]!["temperature"]!!)" as NSString).doubleValue
//Display all current conditions from API
print(currentConditions)
//Output the current temperature in Fahrenheit
print(currentTemperatureF)
}
//else throw an error detailing what went wrong
catch let error as NSError {
print("Details of JSON parsing error:\n \(error)")
}
}
}
编辑:下面是在print(currentConditions)
之后调用的结果示例
["icon": partly-cloudy-night, "precipProbability": 0, "pressure": 1015.39, "humidity": 0.75, "precipIntensity": 0, "windSpeed": 6.04, "summary": Partly Cloudy, "ozone": 321.13, "temperature": 49.45, "dewPoint": 41.75, "apparentTemperature": 47, "windBearing": 332, "cloudCover": 0.28, "time": 1480846460]
https://stackoverflow.com/questions/39423367
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