0

我正在努力让名称值在没有Optional(...). 我以为我正在解开它们，但我无法摆脱可选的。我在 StackOverflow 上找到了一个解决方案，但在那种情况下，涉及到了一个as?问题，这里不是这种情况。有问题的代码如下。有什么办法可以解开这些吗？我假设我遗漏了一些明显的东西。谢谢。

func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCell(withIdentifier: "Cell") as! ClientCell

    cell.setBackground()

    let clientKey = clientSectionTitles[(indexPath as NSIndexPath).section]
    if let clientValues = clientDict[clientKey] {
        if self.sortBy == "First" {
            cell.clientName.text = clientValues[(indexPath as NSIndexPath).row].firstName! + " " + clientValues[(indexPath as NSIndexPath).row].lastName!
        } else {
            cell.clientName.text = clientValues[(indexPath as NSIndexPath).row].lastName! + ", " + clientValues[(indexPath as NSIndexPath).row].firstName!
        }
    }

    return cell
}

*编辑：这是构建字典的代码：

func createClientDict() {
    var clientName: String?
    clientDict          = [String: [Client]]()
    clientSectionTitles = [String]()

    if self.sortBy == "First" {
        apiResults.sort (by: { $0.firstName < $1.firstName })
    } else {
        apiResults.sort (by: { $0.lastName < $1.lastName })
    }

    for c in apiResults {
        if self.sortBy == "First" {
            clientName = "\(c.firstName!) \(c.lastName!)"
        } else {
            clientName = "\(c.lastName!), \(c.firstName!)"
        }

        // Get the first letter of the name and build the dictionary
        let clientKey = clientName!.substring(to: clientName!.characters.index(clientName!.startIndex, offsetBy: 1))
        if var clientValues = clientDict[clientKey] {
            clientValues.append(c)
            clientDict[clientKey] = clientValues
        } else {
            clientDict[clientKey] = [c]
        }
    }

    // Get the section titles from the dictionary's keys and sort them in ascending order
    clientSectionTitles = [String](clientDict.keys)
    clientSectionTitles = clientSectionTitles.sorted { $0 < $1 }
}

*编辑：我还应该提到，这只发生在 Swift 3 中。它在 Swift 2.2 中运行良好。

4

2 回答 2

1

以下来自 Swift Programming Language Collection Types 文档：

您还可以使用下标语法从字典中检索特定键的值。因为可以请求不存在值的键，所以字典的下标返回字典值类型的可选值。如果字典包含请求键的值，则下标返回一个可选值，其中包含该键的现有值。否则，下标返回 nil：

所以clientValues[(indexPath as NSIndexPath).row]返回一个可选值

尝试

cell.clientName.text = (clientValues[(indexPath as NSIndexPath).row].firstName!)!

*已编辑 - 实际上最好使用可选绑定：

if let fname = clientValues[(indexPath as NSIndexPath).row].firstName! {
   cell.clientName.text = fname }
于 2017-03-18T06:44:26.080 回答
0

我认为这个问题与clientKey具有可选值的变量有关。在使用它之前尝试解开它，如下所示：

if let clientKey = clientSectionTitles[(indexPath as NSIndexPath).section] {
    if let clientValues = clientDict[clientKey] {
        if self.sortBy == "First" {
            cell.clientName.text = clientValues[(indexPath as NSIndexPath).row].firstName! + " " + clientValues[(indexPath as NSIndexPath).row].lastName!
        } else {
            cell.clientName.text = clientValues[(indexPath as NSIndexPath).row].lastName! + ", " + clientValues[(indexPath as NSIndexPath).row].firstName!
        }
    }
}

=== 编辑：===

 let clientKey = clientSectionTitles[(indexPath as NSIndexPath).section] 
        if let clientValues = clientDict[clientKey] {
            if self.sortBy == "First" {
                cell.clientName.text = clientValues[(indexPath as NSIndexPath).row].firstName! + " " + clientValues[(indexPath as NSIndexPath).row].lastName!
            } else {
                cell.clientName.text = clientValues[(indexPath as NSIndexPath).row].lastName! + ", " + clientValues[(indexPath as NSIndexPath).row].firstName!
            }
        }
于 2017-03-18T05:05:48.770 回答