Question

我有DBGrid我在其中加载与DataSetPeople关联的table_people的名称。我创建了一个拖放过程，用户可以拖动名称并放入ListBox。用户可以放入listbox_employees或listbox_manager。

我的问题

我想使用用户在两个列表框中放置的名称在table_people之间创建关系。我创建table_relationship employeeID与managerID匹配。

那么，如何从每个listbox获取名称，将此名称与ID相关联，并将此ID放入与employeeID或managerID相关联的DataSetRelationship到table_relationship？

Answer 1

如果您的ID是数字（整数）值，则可以在填充它时使用TListBox.Items.Objects存储它（在表单上的放置事件处理程序中）：

var
  PersonName: string;
  PersonID: Integer;
begin
  with YourDataModule do  // Gives access to the tables in the data module
  begin
    PersonName := table_people.FieldByName('employeeName').AsString;
    PersonID   := table_people.FieldByName('employeeID').AsInteger);
  end;
  listbox_employees.Items.AddObject(PersonName, TObject(PersonID);
end;

对于经理，只需更改管理员的列表框和经理的字段值。

要取回ID以用于INSERT声明：

// Make sure both listboxes have the same number of items, of course.
var
  EmployeeID: Integer;
  ManagerID: Integer;
  i: Integer;
begin
  for i := 0 to ListBox_Employees.Items.Count - 1 do
  begin
    EmployeeID := Integer(ListBox_Employees.Items.Objects[i]);
    ManagerID := Integer(ListBox_Manager.Items.Objects[i]);
    if not YourDataModule.MakeRelationship(EmployeeID, ManagerID) then
      ShowMessage('Unable to relate this employee and manager!');
  end;
end;

// If you're using a SQL query, the `INSERT` should be created like this
// somewhere, like in your datamodule's OnCreate event
//   qryRelationShips.SQL.Clear;
//   qryRelationShips.SQL.Add('INSERT INTO table_relationship (employeeID, managerID)');
//   qryRelationShips.SQL.Add('VALUES (:employeeID, :managerID)';
//
// Or you can type it into the SQL property in the Object Inspector at designtime 
//   INSERT INTO table_relationship (employeeID, managerID) VALUES (:employeeID, :managerID)


function TYourDataModule.MakeRelationship(const EmpID, MgrID: Integer):  Boolean;
begin
  Result := False;
  // INSERT into your dataset:
  qryRelationShips.ParamByName('employeeID').AsInteger := EmpID;
  qryRelationShips.ParamByName('managerID').AsInteger := MgrID;
  try
    qryRelationShips.ExecSQL;
    Result := qryRelationShips.RowsAffected;
  finally
    qryRelationShips.Close;
  end;
end;

// If you're using a table instead of a query, getting a result is harder
function TYourDataModule.MakeRelationship(const EmpID, MgrID: Integer):  Boolean;
begin
  Result := True;
  try
    tblRelationShips.Insert;
    tblRelationShips.FieldByName('employeeID').AsInteger := EmpID;
    tblRelationShips.FieldByName('managerID').AsInteger := MgrID;
    tblRelationShips.Post;
  except
    Result := False;
  end;
end;

Answer 2

我希望您使用SQLite数据库制作表格。因此，您可以通过使用查询来插入行来实现。使用db.execSQL（）; 你可以保留一个计数器来增加id值。

从ListBox到DataSet的数据

2 个答案: