自动完成搜索列表问题在Swift中

时间:2016-03-02 20:59:37

标签: ios swift uisearchcontroller uisearchresultscontroller

你好,我的搜索功能有一个非常奇怪的问题。我已经成功实现了搜索功能。我从后端服务获取数据。问题是在某个阶段,根据键入的关键字,数据无法在建议区域(tableView)中准确加载。我也在控制台上打印结果以检查我是否正在获得关键字的准确结果,并且控制台显示准确的结果,只是建议区域有时不会加载确切的结果。例如在我的应用程序中如果我想搜索城市“拉合尔”。我输入了完整的字母“拉合尔”

显示此enter image description here

但是当我按x图标或退格键删除“e”时,它会显示准确的结果

enter image description here

我只是作为一个例子展示它。这种情况几乎一直都在发生。你能看看我的代码,看看我做错了什么。

class CountryTableViewController: UITableViewController, UISearchResultsUpdating {

    var dict = NSDictionary()
    var filteredKeys = [String]()

    var resultSearchController = UISearchController()

    var newTableData = [String]()

    override func viewDidLoad() {
        super.viewDidLoad()

        self.resultSearchController = ({

            let controller  = UISearchController(searchResultsController: nil)
            controller.searchResultsUpdater = self
            controller.dimsBackgroundDuringPresentation = false
            controller.searchBar.sizeToFit()
            self.tableView.tableHeaderView = controller.searchBar
            return controller


        })()

        self.tableView.reloadData()
    }

    override func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {

        if (self.resultSearchController.active) {

            return self.filteredKeys.count
        } else {

            return dict.count
        }

    }

    override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {

        let cell = tableView.dequeueReusableCellWithIdentifier("cell", forIndexPath: indexPath) as! CountryTableViewCell

        if(self.resultSearchController.active){





                let cityName = (((self.dict["\(indexPath.row)"] as?NSDictionary)!["Country"] as?NSDictionary)!["city_name"] as?NSString)

               let stateName  = (((self.dict["\(indexPath.row)"] as?NSDictionary)!["Country"] as? NSDictionary)!["state_name"] as? NSString)

                 let shortName  = (((self.dict["\(indexPath.row)"] as?NSDictionary)!["Country"] as? NSDictionary)!["short_country_name"] as? NSString)


            if (cityName != "-" || shortName != "-"){
                cell.stateNameLabel.text = stateName as? String
                cell.cityNameLabel.text = cityName as? String
                 cell.shortNameLabel.text = shortName as? String

            }

                      return cell

        }else{
            if let cityName = (((self.dict["\(indexPath.row)"] as?NSDictionary)!["Country"] as?NSDictionary)!["city_name"] as?NSString){
            cell.cityNameLabel.text = cityName as String
            }
            return cell
        }



    }



    func updateSearchResultsForSearchController(searchController: UISearchController) {

        let searchWord = searchController.searchBar.text!

        getCountriesNamesFromServer(searchWord)

        self.filteredKeys.removeAll()

        for (key, value) in self.dict {

            let valueContainsCity: Bool = (((value as? NSDictionary)?["Country"] as? NSDictionary)?["city_name"] as? String)?.uppercaseString.containsString(searchWord.uppercaseString) ?? false

            let valueContainsCountry: Bool = (((value as? NSDictionary)?["Country"] as? NSDictionary)?["country_name"] as? String)?.uppercaseString.containsString(searchWord.uppercaseString) ?? false

            if valueContainsCity || valueContainsCountry{ self.filteredKeys.append(key as! String) }




        }

        self.tableView.reloadData()
    }




    func getCountriesNamesFromServer(searchWord:String){


        let url:String = "http://localhost"
        let params = ["keyword":searchWord]



        ServerRequest.postToServer(url, params: params) { result, error in

            if let result = result {
                print(result)



                self.dict = result

            }
        }

    }

}

1 个答案:

答案 0 :(得分:2)

您在请求启动后重新加载表,而不是在完成时,因此您的dict仍然具有上次运行查询的结果。

执行updateSearchResults..

后,将getCountriesNamesFromServer方法中的所有内容移至您的网络请求的完成处理程序中

您的新方法将是:

self.dict = result