Question

我正在尝试从字典和数组中获取数据到我的自定义结构中进行保存，但是当我打印它时，它的值是在optional（value）中给出的，并且没有将其保存到TableView单元格中出现此问题：（

func findAllShops(){
    Alamofire.request(NEARBY_STORES, method: .get, parameters: nil, encoding: JSONEncoding.default, headers: headers).responseJSON{ (response) in

        if response.result.error == nil {

            do
            {
                guard let data = response.data else { return }
                if let json = try JSON(data: data).array {
                    for item in json {
                        let storeName = item["name"].stringValue
                        let storeCover = item["cover"].stringValue
                        let storeLogo = item["logo"].stringValue
                        let storeSpeed = item["delievery_time"].stringValue
                        let storeRating = item["average_rating"].stringValue
                        let minimumOrder = item["min_order_amt"].stringValue
                        let shopList = shopModel(shopName: storeName , shopBGImage: storeCover, shoplogo: storeLogo , speed: storeSpeed, accuracy: storeRating, minimumOrder: minimumOrder)
                        self.shopLists.append(shopList)
                    }
                    print(self.shopLists[1])
                }
            }
            catch let error as NSError {
                // error
                print("Error in the parsing")
                debugPrint(response.result.error as Any)
            }
        }
    }
}

预期的输出->

shopModel（shopName：（“ Madina Cash and Carry（F11）”），shopBGImage：（“ uploads / resturant_cover / resturant_4.jpg”），shopLogo：（“上载/餐厅/resturant_4.jpg”），速度：（“ 24h”），准确性：（“ 2.5”），minimumOrder：（“ 1000”））

实际输出->

shopModel（shopName：Optional（“ Madina Cash and Carry（F11）”）， shopBGImage：可选（“ uploads / resturant_cover / resturant_4.jpg”）， shopLogo：可选（“ uploads / resturants / resturant_4.jpg”），速度：可选（“ 24h”），准确性：可选（“ 2.5”），最小顺序：可选（“ 1000”））

Answer 1

您应该重做shopModel的struct / class，以便其vars为非可选。

或像下面这样重写代码：

let storeName = item["name"] ?? "" // converts optionals to String
...

Answer 2

方法：1

当您从API收到default值时，只需分配nil值即可。

代码：

  let storeName = item["name"].stringValue ?? ""
  let storeCover = item["cover"].stringValue ?? ""
  let storeLogo = item["logo"].stringValue ?? ""
  let storeSpeed = item["delievery_time"].stringValue ?? ""
  let storeRating = item["average_rating"].stringValue ?? ""
  let minimumOrder = item["min_order_amt"].stringValue ?? ""

方法：2

如果API中所有值都是必需的，则遵循此方法。如果您已从API收到nil值，则 应用程序崩溃。

代码：

  let storeName = item["name"].stringValue!
  let storeCover = item["cover"].stringValue!
  let storeLogo = item["logo"].stringValue!
  let storeSpeed = item["delievery_time"].stringValue!
  let storeRating = item["average_rating"].stringValue!
  let minimumOrder = item["min_order_amt"].stringValue!

Answer 3

//尝试

    let storeName = String.init(format: "%@", item["name"] as! String)
    let storeCover = String.init(format: "%@", item["cover"] as! String)
    let storeLogo = String.init(format: "%@", item["logo"] as! String)
    let storeSpeed = String.init(format: "%@", item["delievery_time"] as! String)
    let storeRating = String.init(format: "%@", item["average_rating"] as! String)
    let minimumOrder =String.init(format: "%@", item["min_order_amt"] as! String)

OR

只需替换您的代码

let shopList = shopModel(shopName: storeName! , shopBGImage: storeCover!, shoplogo: storeLogo! , speed: storeSpeed!, accuracy: storeRating!, minimumOrder: minimumOrder!)

如何展开可选的不展开显示---精度：可选（“ 2.5”）---

3 个答案: