Question

var people = [Mydata]()

override func prepare(for segue: UIStoryboardSegue, sender: Any?) { 

    if segue.identifier == "thereSegue" ,
        let destinationNavigationController = segue.destination as? UINavigationController ,

    let nextScene = destinationNavigationController.topViewController as? commentView ,

        let indexPath = self.myTable.indexPathForSelectedRow {
        let selectedVehicle = people[indexPath.row]

        nextScene.postId = (selectedVehicle.value(forKey: "iD") as? String)!
        print(nextScene.postId)         
        nextScene.postType = (selectedVehicle.value(forKey: "pTYPE") as? String)!
        print(nextScene.postType)
    }
}

当我尝试

时

let jsonUrl = "https://www.zdoof.com/api/zTimeline/add_post_comment?"
let jsonUrlWithParameters = jsonUrl + "Comment=\(String(describing: self.commentText.text!))" + "&Post_id=\(postId)" + "&Post_type=\(postType)" + "&Created_by=4490" + "&Created_on=\(joinString)"
print(jsonUrlWithParameters)

正在打印

可选（8167）

可选（1）

https://www.zdoof.com/api/zTimeline/add_post_comment?Comment=hzbxhjxjx&Post_id=Optional(8167)&Post_type=Optional(1)&Created_by=4490&Created_on=18%2FAug%2F2017+17%3A42%3A43

那么如何删除＆＃34;可选＆＃34;

Answer 1

简单地说！在您想要解包的可选值之后。

例如：

var integer: Int? = 5

print(integer) // would print out Optional(5)

print(integer!) // would print out 5

你应该小心;而是使用

if let unwrappedValue = integer {
  print(unwrappedValue)
}

如果该值不是可选的，则仅打印出该值。

如何在swift3中展开可选项

1 个答案: