[英]Missing argument for parameter in call
New to swift (3) and Xcode (8) and I'm using firebase to load some data in a tableView
. swift (3) 和 Xcode (8) 的新手,我正在使用 firebase 在
tableView
加载一些数据。 When I try to build the app, I get the error: "Missing argument for parameter name in call"
in my fetchWhiskey function on the line when I call an instance of WhiskeyItem.当我尝试构建应用程序时,当我调用 WhiskeyItem 的实例时,在我的 fetchWhiskey 函数中出现错误:
"Missing argument for parameter name in call"
。 I can't figure out why this error is happening.我无法弄清楚为什么会发生此错误。 can anyone help me out?
谁能帮我吗?
Here's my class:这是我的课:
import UIKit
class WhiskeyItem {
let wName: String
let wType: String
init(wName: String, wType: String) {
self.wName = wName
self.wType = wType
}
}
and here's the tableView
that I'm trying to load the data in:这是我试图在其中加载数据的
tableView
:
import UIKit
import Firebase
import FirebaseDatabase
class FirstViewTableViewController: UITableViewController, UISearchBarDelegate {
let whiskeySearchBar = UISearchBar()
var ref: FIRDatabaseReference?
var refHandle: UInt!
var whiskeyList = [WhiskeyItem]()
let cell = "cell"
override func viewDidLoad() {
super.viewDidLoad()
createWhiskeySearchBar()
//Display Firebase whiskey data:
ref = FIRDatabase.database().reference()
fetchWhiskey()
}
func createWhiskeySearchBar() {
whiskeySearchBar.showsCancelButton = false
whiskeySearchBar.placeholder = "Search whiskeys"
whiskeySearchBar.delegate = self
self.navigationItem.titleView = whiskeySearchBar
}
override func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
return whiskeyList.count
}
override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
let cell = tableView.dequeueReusableCell(withIdentifier: "Cell", for: indexPath)
// Configure the cell...
cell.textLabel?.text = whiskeyList[indexPath.row].wName
return cell
}
func fetchWhiskey() {
refHandle = ref?.child("whiskey").observe(.childAdded, with: { (snapshot) in
if let dictionary = snapshot.value as? [String : AnyObject] {
print(dictionary)
let whiskeyItemInstance = WhiskeyItem()
whiskeyItemInstance.setValuesForKeys(dictionary)
self.whiskeyList.append(whiskeyItemInstance)
DispatchQueue.main.async {
self.tableView.reloadData()
}
}
})
}
Your initializer has two parameters which are required when calling it.您的初始化程序有两个调用它时需要的参数。
Calling it properly would look something like this:正确调用它看起来像这样:
let whiskeyItemInstance = WhiskeyItem(wName: "name", wType: "type")
If you don't want to pass parameters to the initializer, you could provide default params:如果您不想将参数传递给初始值设定项,您可以提供默认参数:
init(wName: String = "default name", wType: String = "default type") {
or use an initializer with no parameters at all:或者使用一个没有参数的初始化程序:
init() {
self.wName = "wName"
self.wType = "wType"
}
or call the initializer you already created like so:或者像这样调用您已经创建的初始化程序:
convenience init() {
self.init(wName: "default name", wType: "default type")
}
Or you could forgo initializers altogether:或者你可以完全放弃初始化器:
class WhiskeyItem {
let wName: String = "asdf"
let wType: String = "asdf"
}
If you want to inheritance class initialize then you can do it like this way如果你想继承类初始化那么你可以这样做
class Room {
var type: String
var client: Client
init(type: String, client: Client) {
self.type = type
self.client = client
}
let unknown = Client() // Need to define it more precisely
convenience init(type: String) {
self.type = type
self.client = unknown
}
}
Apart from this you can also do these ways除此之外你还可以通过这些方式
The initializer for Room has two parameters (type and client) and neither parameter has a specified default argument, so you must pass an argument of the correct type to each parameter. Room 的初始值设定项有两个参数(类型和客户端),并且两个参数都没有指定的默认参数,因此您必须将正确类型的参数传递给每个参数。 Same story for Client.
客户的相同故事。
let theClient = Client(name: "Thelonius", surname: "Monk")
let oneRoom = Room(type: "single", client: theClient)
For more updates: https://developer.apple.com/forums/thread/67555更多更新: https : //developer.apple.com/forums/thread/67555
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