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请尽可能帮助我。当用户使用removeValue方法调用delete editingStyle函数时,我想删除tableView中的数据。到目前为止,我有一类项目:从Firebase数据库中删除子数据Swift 3
class Item : NSObject {
var itemName: String!
var itemDate: String!
var itemID: String!
init(itemName: String, itemDate: String, itemID: String) {
self.itemName = itemName
self.itemDate = itemDate
self.itemID = itemID
}
init(snapshot: FIRDataSnapshot) {
let snapshotValue = snapshot.value as? NSDictionary
self.itemName = snapshotValue!["itemName"] as! String
self.itemDate = snapshotValue!["itemDate"] as! String
self.itemID = snapshotValue!["itemID"] as! String
}
func toAnyObject() -> [String: AnyObject] {
return ["itemName": itemName as AnyObject, "itemDate": itemDate as AnyObject, "itemID": itemID as AnyObject]
}
}
现在我想删除用户ITEMID内的值,但我不知道在哪里何去何从:
override func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCellEditingStyle, forRowAt indexPath: IndexPath) {
guard let uid = FIRAuth.auth()?.currentUser?.uid else {
return
}
let item = itemArray[indexPath.row]
let itemID = databaseRef.child("ItemID")
databaseRef.child("users").child(uid).child("usersList").child(itemID).removeValue()
这不起作用它说错误说“表达式类型'虚拟'(又名'()')是不明确的没有更多的上下文。 – Chace
哪一行是显示的错误? @Chace – Trev14
它显示在databaseRef.child的最后一行... @ Trev14 – Chace