Question

在iOS / Swift中，我基于Client类中的clientName属性创建了一个索引的“客户端”UITableView。我用A到Z创建了一个字典作为部分。索引的tableview效果很好。但是，当用户选择一行时，我正试图找出一种方法来确定它在原始源数组中的哪一行。我正在考虑构建某种类型的交叉引用数组，除了字典最终排序以匹配部分，所以我不知道哪个部分/行组合匹配哪个原始数组条目。处理这个问题有一个共同的方法吗？

试图澄清......

class Client {
    var clientId             : Int!
    var firstName            : String!
    var lastName             : String!
    var email                : String!
    var phone                : String!
    ...

    init() {

    }
}

var clients: [Client] = []

// clients array loaded from web service
...

// Create dictionary to be source for indexed tableview
func createClientDict() {
    clientDict          = [String: [String]]()
    clientSectionTitles = [String]()

    var clientNames:[String] = []
    for i in 0..<clients.count {
        let client = clients[i]

        let clientName = "\(client.lastName), \(client.firstName)"
        clientNames.append(clientName)
    }

    for name in clientNames {
        var client: Client  = Client()

        // Get the first letter of the name and build the dictionary
        let clientKey = name.substringToIndex(name.startIndex.advancedBy(1))
        if var clientValues = clientDict[clientKey] {
            clientValues.append(name)
            clientDict[clientKey] = clientValues
        } else {
            clientDict[clientKey] = [name]
        }
    }

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

现在，当用户在tableview中点击一行时，我可以获得section和row（indexPath）。但是，假设可能存在重复的名称，如何确定clients数组中的哪一行匹配？是否有某种方法可以动态创建映射到源数组中的行的索引节/行的交叉引用？我打算在构建字典时尝试这样做，除了字典之后排序，所以没有任何东西可以匹配。也许我应该以某种方式在字典中包含源行号？

这是tableview代码：

func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCellWithIdentifier("Cell") as! ClientCell

    let clientKey = clientSectionTitles[indexPath.section]
    if let clientValues = clientDict[clientKey] {
        cell.clientName.text = clientValues[indexPath.row]
    }

    return cell
}

func numberOfSectionsInTableView(tableView: UITableView) -> Int {
    return clientSectionTitles.count
}

func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
    let clientKey = clientSectionTitles[section]
    if let clientValues = clientDict[clientKey] {
        return clientValues.count
    }

    return 0
}

func tableView(tableView: UITableView, titleForHeaderInSection section: Int) -> String? {
    return clientSectionTitles[section]
}

func sectionIndexTitlesForTableView(tableView: UITableView) -> [String]? {
    return clientIndexTitles
}

func tableView(tableView: UITableView, sectionForSectionIndexTitle title: String, atIndex index: Int) -> Int {

    guard let index = clientSectionTitles.indexOf(title) else {
        return -1
    }

    return index
}

func tableView(tableView: UITableView, heightForHeaderInSection section: Int) -> CGFloat {
    return 20
}

func tableView(tableView: UITableView, willDisplayHeaderView view: UIView, forSection section: Int) {
    let headerView   = view as! UITableViewHeaderFooterView

    headerView.contentView.backgroundColor = UIColor ( red: 0.0, green: 0.3294, blue: 0.6392, alpha: 1.0 )
    headerView.textLabel?.textColor = UIColor.greenColor()
    headerView.textLabel?.font = UIFont(name: "Noteworthy-bold", size: 15.0)
}

func tableView(tableView: UITableView, didSelectRowAtIndexPath indexPath: NSIndexPath) {
    selectedIndex = indexPath

    // In the following prepare for segue, I need to somehow use the selected indexpath to find the correct entry
    // in the clients array and pass it along.

    performSegueWithIdentifier("clientDetailSegue", sender: self)
}

Answer 1

我明白了。我没有意识到（直到我最近尝试过）你可以在字典中嵌套任何类的数组。当我更改字典以使我的客户端数组嵌套在其中时，所有内容都已解决。我改变了我的功能，如下所示。

func createClientDict() {
    // Declared for view controller. Re-initialized here.
    clientDict          = [String: [Client]]()
    clientSectionTitles = [String]()

    clients.sortInPlace ({ $0.lastName < $1.lastName })

    for c in clients {
        let clientName = "\(c.lastName), \(c.firstName)"

        // Get the first letter of the name and build the dictionary
        let clientKey = clientName!.substringToIndex(clientName!.startIndex.advancedBy(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.sort { $0 < $1 }
}

然而，这条线是解决方案的关键：

let clientDict = [String: [Client]]()

想要索引的tableview，并引用回源数组

1 个答案: