Question

我有一个带有自定义单元格的PFQueryTableViewController，每个单元格都有1个标签和1个按钮，标签（Place）是一个文本，而按钮（Tickets），当你单击它时你会转到一个url。我已经将标签和网址的文本存储为Parse.com上的字符串，但我的问题是如何解决这个问题，让每个按钮都能动作并从我的解析数据库中打开这个存储的URL。

这是代码

class Shows: PFQueryTableViewController {


override func queryForTable() -> PFQuery {
    let query = PFQuery(className: "Shows")
    query.orderByDescending("createdAt")
    return query  
}

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath, object: PFObject?) -> PFTableViewCell? {
    let cell = tableView.dequeueReusableCellWithIdentifier("cell", forIndexPath: indexPath) as! ShowsCell
    let showplace = object?.objectForKey("Place") as? String
    let showtickets = object?.objectForKey("Tickets") as? String
    cell.place.text = showplace
    cell.tickets.tag = indexPath.row
    cell.tickets.addTarget(self, action: #selector(ticketsaction), forControlEvents: .TouchUpInside)
    return cell
    }
}

这是Button Action

   @IBAction func ticketsaction(sender: UIButton){

    let ticketsurl = String()
    let url = NSURL(string: "\(ticketsurl)")
    UIApplication.sharedApplication().openURL(url!)
  }

Answer 1

您可以使用其单元格的indexPath.row标记每个按钮，并从您用于填充列表的阵列中获取特定的URL。只需将button.tag等同于indexPath.row。

Answer 2

最后我成功了！我在自定义单元格中添加var ticketsurl: String!，在ticketaction之前，现在看起来像这样：

    var ticketsurl: String!

@IBAction func ticketsaction(sender: UIButton) {
            let url = NSURL(string: "\(ticketsurl)")
            UIApplication.sharedApplication().openURL(url!)
           }

和PFQueryTableView Controller现在看起来像这样：

class Shows: PFQueryTableViewController {


override func queryForTable() -> PFQuery {
let query = PFQuery(className: "Shows")
query.orderByDescending("createdAt")
return query  
}

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath, object: PFObject?) -> PFTableViewCell? {
let cell = tableView.dequeueReusableCellWithIdentifier("cell", forIndexPath: indexPath) as! ShowsCell
let showplace = object?.objectForKey("Place") as? String
let showtickets = object?.objectForKey("Tickets") as? String
cell.place.text = showplace
cell.tickets.tag = indexPath.row
cell.ticketsurl = showtickets
return cell
   }
}

从Parse到按钮动作的url

2 个答案: