我是swift和sqlite3的新手,我需要有关如何从tableview和sql db中删除的帮助。
我尝试使用reloadData(),但是它不起作用。我尝试使用tableView.deleteRows(at: [indexPath], with: .fade)
进行删除,但由于在此之前运行了sql delete语句,因此出现错误。通过下面提供的这段代码,Im成功地能够从数据库中删除该项目,但是它不会刷新tableview。我临时解决此问题的方法是,在成功删除某项后执行到上一个屏幕的搜索,当返回到tableviewcontroller时,它将被删除。
import UIKit
class TableViewController: UIViewController,UITableViewDataSource,UITableViewDelegate {
let mainDelegate = UIApplication.shared.delegate as! AppDelegate
@IBOutlet var tableView: UITableView!
func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
let tableCell = tableView.dequeueReusableCell(withIdentifier: "cell") as? SiteCell ?? SiteCell(style: .default, reuseIdentifier: "cell")
let rowNum = indexPath.row
tableCell.primaryLabel.text = mainDelegate.people[rowNum].name
tableCell.secondaryLabel.text = mainDelegate.people[rowNum].email
tableCell.myImageView.image = UIImage(named: mainDelegate.people[rowNum].avatar!)
tableCell.accessoryType = .disclosureIndicator
return tableCell
}
func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
return mainDelegate.people.count
}
func tableView(_ tableView: UITableView, heightForRowAt indexPath: IndexPath) -> CGFloat {
return 70
}
func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
let rowNum = indexPath.row
let details : String! = "Address: \(mainDelegate.people[rowNum].address!) \nPhone Num: \(mainDelegate.people[rowNum].phonenum!) \nEmail: \(mainDelegate.people[rowNum].email!) \nAge: \(mainDelegate.people[rowNum].age!) \nGender: \(mainDelegate.people[rowNum].gender!) \nDate of birth: \(mainDelegate.people[rowNum].dob!)"
let alertController = UIAlertController(title: mainDelegate.people[rowNum].name, message: details, preferredStyle: .alert
)
let cancelAction = UIAlertAction(title: "ok", style: .cancel, handler: nil)
print("TESTING ROW: \(mainDelegate.people[rowNum].id!)")
alertController.addAction(cancelAction)
present(alertController, animated: true)
}
func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCell.EditingStyle, forRowAt indexPath: IndexPath) {
var rowNum: Int = indexPath.row
if editingStyle == .delete {
print("Testing delete \(mainDelegate.people[rowNum].id!)")
print("\(indexPath.row)")
mainDelegate.removeFromDatabase(id: mainDelegate.people[rowNum].id!)
print("\(indexPath)")
// tableView.deleteRows(at: [indexPath], with: .fade)
DispatchQueue.main.async{
self.tableView.reloadData()
}
// self.performSegue(withIdentifier: "DataToInfo", sender: self)
// let mainDelegate = UIApplication.shared.delegate as! AppDelegate
// mainDelegate.removeFromDatabase(person: mainDelegate.people[indexPath.row])
}
}
override func viewDidLoad() {
super.viewDidLoad()
// Do any additional setup after loading the view.
mainDelegate.readDataFromDatabase()
}
removeFromDatabase方法
func removeFromDatabase(id : Int){
var db : OpaquePointer? = nil
if sqlite3_open(self.databasePath, &db) == SQLITE_OK{
print("Successfully opened connection to database at \(self.databasePath)")
var deleteStatement : OpaquePointer? = nil
let deleteStatementString : String = "delete from entries where id=\(id)"
if sqlite3_prepare_v2(db, deleteStatementString, -1, &deleteStatement, nil) == SQLITE_OK{
if sqlite3_step(deleteStatement) == SQLITE_DONE{
print("Deleted")
}
else{
print("Failed")
}
}else{
print("Couldn't prepare")
}
sqlite3_finalize(deleteStatement)
sqlite3_close(db)
}
}
我正在尝试从tableview以及数据库中删除它。有一次我试图
mainDelegate.people.remove(at: indexPath.row)
tableView.deleteRows(at: [indexPath], with: .fade)
然后运行removeFromDatabase,但这给了我一个错误。
答案 0 :(得分:0)
您应该更新数据源。尝试像这样重构您的commitEditing:
func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCell.EditingStyle, forRowAt indexPath: IndexPath) {
var rowNum: Int = indexPath.row
if editingStyle == .delete {
print("Testing delete \(mainDelegate.people[rowNum].id!)")
print("\(indexPath.row)")
mainDelegate.removeFromDatabase(id: mainDelegate.people[rowNum].id!)
print("\(indexPath)")
mainDelegate.readDataFromDatabase()
tableView.deleteRows(at: [indexPath], with: .fade)
}
}