Question

我有一个UITableView和UISearchBar。我在tableview中显示了一些coredata值，例如员工姓名。如果我选择特定员工姓名上的行，员工的详细信息应该显示在另一个视图控制器中，我使用这个“覆盖功能准备（对于segue：UIStoryboardSegue，发件人：任何？）”segue方法。

问题是，每当我使用搜索栏过滤员工姓名，并使用在表格视图上点击过滤后的员工姓名时，不会调用“覆盖功能准备（对于segue：UIStoryboardSegue，发件人：任何？）”。但是如果没有搜索，我可以导航到具有所选员工的详细视图控制器。但是如果使用搜索栏过滤名称，我无法实现对员工细节的segue导航。请告诉我在搜索结果之后没有导航的原因或建议解决方案。

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {

    let object : Any
    if(searchActive){
        let indexPath = self.searchDisplayController?.searchResultsTableView.indexPathForSelectedRow
        object = self.filteredEmployees[(indexPath?.row)!]
    }
    else{
        let indexPath = tableView.indexPathForSelectedRow
        object = self.employees[(indexPath?.row)!]
    }
    let controller = (segue.destination as! UINavigationController).topViewController as! DetailViewController
    //controller.detailItem = object
    controller.navigationItem.leftBarButtonItem = splitViewController?.displayModeButtonItem
    controller.navigationItem.leftItemsSupplementBackButton = true
    controller.employees = object as! EmployeeDetails

}

override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
    var cell = tableView.dequeueReusableCell(withIdentifier: "Cell")

    if cell == nil {
        cell = UITableViewCell(style: UITableViewCellStyle.value1, reuseIdentifier: "Cell")
    }

    cell?.textLabel!.numberOfLines = 0
    cell?.textLabel!.lineBreakMode = .byWordWrapping
    cell?.textLabel!.font = UIFont.systemFont(ofSize: 14.0)
    if(searchActive){
        var filter: EmployeeDetails? = nil
        filter =  self.filteredEmployees[indexPath.row]
        cell?.textLabel?.text = String(format:"%@  %@", (filter?.firstName!)!, (filter?.lastName!)!)
        cell?.detailTextLabel!.text = String(format:"%@", (filter?.title!)!)

    } else
    {
        var employee : EmployeeDetails? = nil
        employee =  self.employees[indexPath.row]
        cell?.textLabel?.text = String(format:"%@  %@", (employee?.firstName!)!, (employee?.lastName!)!)
        cell?.detailTextLabel!.text = String(format:"%@", (employee?.title!)!)

    }
    return cell!
}

Swift：prepareforsegue方法不称为

0 个答案: