使用窗体名称公开赛形式的winform appliaction [英] Open form with Form Name in winform appliaction

问题描述

我想请问我应该怎么做，以开放的形式在WinForm的C＃的帮助或类的名字吗？

我有三个不同的形式

• UserManagement


• GroupsManagement


• LocationManagement

我得到许可从数据库中这三种形式

在菜单中单击我填写标签属性与表单的名称像这样

  tsmMain.Tag = item.PermissionName 
 tsmMain.Click + =新的EventHandler（tsmMain_Click）; 
  

我想要做的是在点击按钮打开动态的形式和条件，如果除去这些？
我能做到这一点与反思要不然??

  ToolStripMenuItem AA =发件人为ToolStripMenuItem; 
 var标记= aa.Tag; 
如果（标签==用户管理）
 {
 UserManagement oUserForm =新UserManagement（）; 
 oUserForm.Show（）; 
} 
如果（标签==组管理）
 {
 GroupManagement oGroupForm =新GroupManagement（）; 
 oGroupForm.Show（）; 
} 
  

解决方案

您可能能够做到这样的事情，用你的表单的名字，作为一个字符串参数：

  VAR形式=（表格）Activator.CreateInstance（ Type.GetType（YourNameSpace.UserManagement））; 
 form.Show（）; 
  

I want to ask what should i do to open form with the help or class name in winform c#?

I have three different forms

• UserManagement
• GroupsManagement
• LocationManagement

I get permission from database for these three forms

in menu click i fill tag Property with the name of form like this

tsmMain.Tag = item.PermissionName
tsmMain.Click += new EventHandler(tsmMain_Click);

what i want to do is to open form dynamically in button click and to remove these if condition? Can i do this with reflection or else??

ToolStripMenuItem aa = sender as ToolStripMenuItem;
        var tag = aa.Tag;
        if (tag == "User Management")
        {
            UserManagement oUserForm = new UserManagement();
            oUserForm.Show();
        }
        if (tag == "Groups Management")
        {
            GroupManagement oGroupForm = new GroupManagement();
            oGroupForm.Show();
        }

解决方案

You may be able to do something like this, using the name of your form, as a string argument:

var form = (Form)Activator.CreateInstance(Type.GetType("YourNameSpace.UserManagement"));
form.Show();

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