如何以另一种形式显示访问querry'结果。 [英] how show access querry 'result in another form.

问题描述

你好

我有两个表单（form1和form2），每个表单都有一个Datagridview

我想运行一个查询并在datagrid中看到它在form2中的回复

**确实我想要给出当前记录的id并以新格式在网格中显示querry的结果。

i有一个名为'selectform2'的查询。

查询是这样的：select * from table where questionid =？



form1的代码是：

  public   string  a; 
  private   void  dataGridView1_CellDoubleClick（ object  sender，DataGridViewCellEventArgs e）
 {
 Form2 hh =  new  Form2（）; 
a = mokhaberat_DataSet.question [dataGridView1.CurrentRow.Index] [  id]。ToString （）; 
 hh.ss = a; 
 hh.ShowDialog（）; 
} 

表格2的代码是：

< pre lang =cs> private string _ss;
public string ss
{
set {_ss = value ; }
get { return _ss; }
}

私有 void Form2_Load（ object sender，EventArgs e）
{


mokhaberat_DataSetTableAdapters.answerTableAdapter g = new first.mokhaberat_DataSetTableAdapters.answerTableAdapter（）;
datagrid2.DataSource = g.selectform2（mokhaberat_DataSet.answer，Convert.ToInt16（ss））;


}

为什么它不能在form2上运行？

解决方案

这是关于表单协作的流行问题。最强大的解决方案是在表单类中实现适当的接口，并传递接口引用而不是引用Form的整个实例。有关更多详细信息，请参阅我以前的解决方案：如何以两种形式复制列表框之间的所有项目 [ ^ ]。



另请参阅此处的其他解决方案讨论。如果应用程序足够简单，解决方案就像在一个表单中声明一些 internal 属性并将对一个表单的实例的引用传递给另一个表单的实例一样简单形成。对于更复杂的项目，这种违反严格封装的样式和松散耦合可能会增加代码的意外复杂性并引发错误，因此封装良好的解决方案将是优惠。



另请参阅：

http://en.wikipedia.org/wiki/Accidental_complexity [ ^ ]，

http://en.wikipedia.org/wiki/Loose_coupling [ ^ ]。

-SA

hello
I have two forms (form1 and form2) and every form has a Datagridview
I want run a query and see it's reply in form2 in a datagrid
**indeed i want give id of current record and show querry's result in a grid in a new form.
i have a query to name of 'selectform2'.
that query is like this :select * from table where questionid=?

code of form1 is :

public string a;
       private void dataGridView1_CellDoubleClick(object sender, DataGridViewCellEventArgs e)
       {
           Form2 hh = new Form2();
           a = mokhaberat_DataSet.question[dataGridView1.CurrentRow.Index]["id"].ToString();
           hh.ss = a;
           hh.ShowDialog();
       }

code of form2 is :

private string _ss;
     public string ss
     {
         set { _ss = value; }
         get { return _ss; }
     }

     private void Form2_Load(object sender, EventArgs e)
     {

      
mokhaberat_DataSetTableAdapters.answerTableAdapter g = new first.mokhaberat_DataSetTableAdapters.answerTableAdapter();
       datagrid2.DataSource = g.selectform2(mokhaberat_DataSet.answer, Convert.ToInt16(ss));


     }

why it does not work on form2 ?

解决方案

This is the popular question about form collaboration. The most robust solution is implementation of an appropriate interface in form class and passing the interface reference instead of reference to a "whole instance" of a Form. Please see my past solution for more detail: How to copy all the items between listboxes in two forms[^].

Please also see other solutions in this discussion. If the application is simple enough, the solution could be as simple as declaring of some internal property in one form and passing a reference to the instance of one form to the instance of another form. For more complex projects, such violation of strictly encapsulated style and loose coupling could add up the the accidental complexity of the code and invite mistakes, so the well-encapsulated solution would be preferable.

Please see also:
http://en.wikipedia.org/wiki/Accidental_complexity[^],
http://en.wikipedia.org/wiki/Loose_coupling[^].

—SA

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