从两个父母那里打开一个孩子表格
本文关键字:一个 孩子 表格 父母 两个 那里 | 更新日期: 2023-09-27 18:32:27
好的,我正在尝试打开来自两个不同父母的表格,但我不知道如何让它工作。 我尝试对 AdminStartMenu 和 TeacherStartMenu 使用布尔值,如果从一个打开一个,则使一个为真,从另一个打开一个为假,但这似乎不起作用。
我如何让一个孩子为两个不同的父母工作?
谢谢!
public partial class EditUser : Form
{
AdminStartMenu pf;
TeacherStartMenu tp;
public bool first = true;
public int st = 0;
public bool Editting;
public bool Adding;
public bool Viewing;
public bool AdminParent;
public bool TeacherParent;
public EditUser()
{
InitializeComponent();
}
public EditUser(AdminStartMenu Parent)
{
pf = Parent;
InitializeComponent();
EditFunction();
if (pf.Adding == true)
{
BlankForm();
SaveButton.Text = "Save";
}
if (pf.Editting == true)
{
FillFormVariables();
SaveButton.Text = "Save";
}
}
public EditUser(TeacherStartMenu TParent)
{
tp = TParent;
InitializeComponent();
EditFunction();
if (tp.Adding == true)
{
BlankForm();
SaveButton.Text = "Save";
}
if (tp.Editting == true)
{
FillFormVariables();
SaveButton.Text = "Save";
}
}
尽管 Hasan 关于在两者之间耦合太多是正确的,但您也许可以简化一种形式调用另一种形式,但无论用户类型如何,都放置常见的东西......也许就像使用界面对 StartMenu 进行分类一样。 我看到它们有共同的元素,所以我你声明了一个接口,例如:
public interface IMyCommonParent
{
bool Adding { get; set; }
bool Editing { get; set; }
bool Viewing { get; set; }
}
public class AdminStartMenu : DerivedFromSomeOtherClass, IMyCommonParent
{ // being associated with IMyCommonParent, this class MUST have a declaration
// of the "Adding", "Editing", "Viewing" boolean elements
}
public class TeacherStartMenu : DerivedFromSomeOtherClass, IMyCommonParent
{ // same here }
它们都支持"IMyCommonParent"接口,您的子窗体可以接受支持该接口的任何对象。 然后,您不必具体知道哪个用于启动子窗体,并且可以将其保留到窗体的属性中
public class ChildForm : Form
{
private IMyCommonParent whichMenu
private bool IsAdminMode;
public ChildForm(IMyCommonParent UnknownParent)
{
whichMenu = UnknownParent;
// then a flag to internally detect if admin mode or not based on
// the ACTUAL class passed in specifically being the admin parent instance
IsAdminMode = ( UnknownParent is AdminStartMenu );
// the rest is now generic.
InitializeComponent();
EditFunction();
if( whichMenu.Adding )
BlankForm();
else if( whichMenu.Editing )
FillFormVariables();
SaveButton.Text = "Save";
}
}