В этом массиве сведений о заказе у меня есть 10 словарей, но мне нужно отобразить только первый словарь. Может ли кто-нибудь помочь мне, как это реализовать?
http://www.json-generator.com/api/json/get/bUKEESvnvS?indent=2
вот мой код, показанный ниже
func downloadJsonWithURL() {
let url = NSURL(string: self.url)
URLSession.shared.dataTask(with: (url as URL?)!, completionHandler: {(data, response, error) -> Void in
if let jsonObj = try? JSONSerialization.jsonObject(with: data!, options: .allowFragments) as? NSDictionary {
self.orderdetailsArray = (jsonObj!.value(forKey: "Orders detail") as? [[String: AnyObject]])!
for array in self.orderdetailsArray {
let key = "OrderId"
let value = "#1000501"
for (key,value) in array{
if let addressDict = array as? NSDictionary{
if let orderid = addressDict.value(forKey: "OrderId"){
self.orderid.append(orderid as! String)
}
if let orderdate = addressDict.value(forKey: "OrderDate"){
self.orderdate.append(orderdate as! String)
}
if let subtotal = addressDict.value(forKey: "SubTotal"){
self.subTotal.append(subtotal as! Int)
}
if let Shipping = addressDict.value(forKey: "Shipping"){
self.shippingPrice.append(Shipping as! Int)
}
if let tax = addressDict.value(forKey: "Tax"){
self.tax.append(tax as! Int)
}
if let grandtotal = addressDict.value(forKey: "GrandTotal"){
self.grandTotal.append(grandtotal as! Int)
}
if let shippingAddress = addressDict.value(forKey: "ShippingAddress"){
self.shippingAddress.append(shippingAddress as AnyObject)
}
if let shippingMethod = addressDict.value(forKey: "ShippingMethod"){
self.shippingMethod.append(shippingMethod as AnyObject)
}
if let billingAddress = addressDict.value(forKey: "BillingAddress"){
self.billingAddress.append(billingAddress as AnyObject)
}
if let paymentMethod = addressDict.value(forKey: "PayMentMethod"){
self.paymentMethod.append(paymentMethod as AnyObject)
}
self.itemsArray = addressDict.value(forKey: "Items detail") as! [[String : AnyObject]]
}
}
}
OperationQueue.main.addOperation({
self.tableDetails.reloadData()
})
}
}).resume()
}
NSDictionary
в Swift, но используйте родную альтернативу Swift,Dictionary
. 20.07.2017