UISearchDisplayController错误- Swift

Question

我正在跟踪这个教程，我遇到了一些麻烦。我正在实现一个搜索栏和显示控制器来过滤结果。

下面是我所得到的错误的图片：

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主要错误是：

1. UISearchDisplayController被否决了
2. 不能下标字符串类型的值。

任何帮助，以解决这些问题将是有帮助的。

下面是我的其余代码：

    import UIKit

    class TableViewController: UITableViewController, UISearchBarDelegate, UISearchDisplayDelegate {

var candies = [Candy]()
var filteredCandies = [Candy]()
var resultSearchController = UISearchController()

override func viewDidLoad() {
    super.viewDidLoad()

    // Uncomment the following line to preserve selection between presentations
    // self.clearsSelectionOnViewWillAppear = false

    // Uncomment the following line to display an Edit button in the navigation bar for this view controller.
    // self.navigationItem.rightBarButtonItem = self.editButtonItem()

    self.candies = [Candy(category:"Chocolate", name:"Chocolate Bar"),
        Candy(category:"Chocolate", name:"chocolate chip"),
        Candy(category:"Hard", name:"lollipop"),
        Candy(category:"Hard", name:"candy cane"),
        Candy(category:"Other", name:"caramel"),
        Candy(category:"Chocolate", name:"chocolate chip")]

    self.tableView.reloadData()

}

override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.

}

func filterContentForSearchText(searchText: String, scope: String = "All"){

    self.filteredCandies = self.candies.filter({(candy: Candy) -> Bool in

        let categoryMatch = ( scope == "All") || (candy.category == scope)
        let stringMatch = candy.name.rangeOfString(searchText)
        return categoryMatch && (stringMatch != nil)
    })
}

func searchDisplayController(controller: UISearchDisplayController, shouldReloadTableForSearchString searchString: String?) -> Bool {
    let scopes = self.searchDisplayController!.searchBar.scopeButtonTitles
    let selectedScope = scopes[self.searchDisplayController!.searchBar.selectedScopeButtonIndex] as String
    self.filterContentForSearchText(searchString, scope: selectedScope)
    return true    }

func searchController(controller: UISearchController, shouldReloadTableForSearchScope searchOption: Int) -> Bool {

    let scope = self.searchDisplayController!.searchBar.scopeButtonTitles
    self.filterContentForSearchText(self.searchDisplayController!.searchBar.text!, scope: scope[searchOption])
    return true
}

// MARK: - Table view data source

override func numberOfSectionsInTableView(tableView: UITableView) -> Int {
    // #warning Incomplete implementation, return the number of sections
    return 1
}

override func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
    // #warning Incomplete implementation, return the number of rows
    if tableView == searchDisplayController!.searchResultsTableView {
        return self.filteredCandies.count
    } else {

        return self.candies.count
    }
}


override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
    let cell = self.tableView.dequeueReusableCellWithIdentifier("cell", forIndexPath: indexPath)

    // Configure the cell...
    //let candy = self.candies[indexPath.row]

    var candy : Candy

    if tableView == self.searchDisplayController!.searchResultsTableView {
        candy = filteredCandies[indexPath.row]
    }else {
        candy = candies[indexPath.row]
    }
    cell.textLabel?.text = candy.name

    return cell
}

Answer 1

1. UISearchDisplayController被否决了

UISearchDisplayController在iOS 8中被否决了，您可以使用UISearchController，它非常类似于UISearchDisplayController。例如，请查看本教程：http://www.jhof.me/simple-uisearchcontroller-implementation/

1. 不能下标字符串类型的值。

scopeButtonTitles返回可选值，因此scopes是可选的。可选数组必须在下标之前展开，强制以这种方式展开：

let scope = self.searchDisplayController!.searchBar.scopeButtonTitles
self.filterContentForSearchText(self.searchDisplayController!.searchBar.text!, scope: scope![searchOption])

或者使用任何附加的可选检查逻辑：

if let scope = scope {        
    self.filterContentForSearchText(self.searchDisplayController!.searchBar.text!, scope: scope[searchOption])
} else {
// handle error
}