OpenFileDialog用于文件选择 [英] OpenFileDialog for file selection

问题描述

您好，



我有一个带文件名的字段，我希望当我点击该字段时，文件选择窗口（资源管理器）会被打开然后选择所需的文件（必须是xls或xlsx文件）。



但是当我点击该字段时，它会打开进行编辑，没有别的。

，我有以下说明：

  this  .bomNameFld.Click + =  new  System.EventHandler（ this  .bomNameFld_Click）; 

和方法，使用以下代码

 private void bomNameFld_Click（object sender，EventArgs e）
 {
 OpenFileDialog openFile = new OpenFileDialog（）; 
 string file = openFile.FileName; 
 bomNameFld.Text = file; 
} 
 

我也试过使用Mouseclick和MouseEventhandler，结果是一样的。



任何人都可以帮助我吗？



提前谢谢。



David。

解决方案

实例化 openFile 后，必须调用ShowDialog方法对象。

 private void bomNameFld_Click（object sender，EventArgs e）
 {
 OpenFileDialog openFile = new OpenFileDialog（） ; 
 if（openFile.ShowDialog（）== DialogResult.OK）
 {
 string file = openFile.FileName; 
 bomNameFld.Text = file; 
} 
} 

另外，你应该检查从ShowDialog调用返回的布尔结果。如果用户单击确定，则返回True。如果用户单击取消，则返回False。



请参阅OpenFileDialog.ShowDialog方法 [ ^ ]

你已经创建了一个新的OpenFileDialog，但你没有做任何事情！

尝试添加

 DialogResult ds = openFile.ShowDialog（）; 
 

并使用

 if（ds == System.Windows.Forms.DialogResult.OK）

在对结果做任何事情之前

Hello,

I have a field with a file name, and I want that when I click on that field, the file selection window (explorer) gets opened and thus select the desired file (must be an xls or xlsx file).

But when I click on the field, it opens for edit, nothing else.

in the design file, I have the following instruction:

this.bomNameFld.Click += new System.EventHandler(this.bomNameFld_Click);

and the method, with the following code

private void bomNameFld_Click(object sender, EventArgs e)
{
    OpenFileDialog openFile = new OpenFileDialog();
    string file = openFile.FileName;
    bomNameFld.Text = file;
}

I have also tryed using Mouseclick and MouseEventhandler, but the result is the same.

Can anyone help me?

Thank you in advance.

David.

解决方案

You have to call the ShowDialog method after you instantiate your openFile object.

private void bomNameFld_Click(object sender, EventArgs e)
        {
            OpenFileDialog openFile = new OpenFileDialog();
            if( openFile.ShowDialog() == DialogResult.OK )
            {
               string file = openFile.FileName;
               bomNameFld.Text = file;
            }
        }

Also, you should check the Boolean result returned from the call to ShowDialog. If the user clicks OK, then True is returned. If the user clicks Cancel, then False is returned.

See OpenFileDialog.ShowDialog Method[^]

You''ve created a new OpenFileDialog but you''re not doing anything with it!
Try adding

DialogResult ds = openFile.ShowDialog();

and using

if(ds == System.Windows.Forms.DialogResult.OK)

before you do anything with the result

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