获取Picturebox文件名 [英] Get Picturebox file name

问题描述

  public   class  photo 
 
 picture pbx =  new  picturebox（）; 
 
  private  voide btnok（ object  sender，EventArgs e）
 {
  string  [] files = System.IO.Directory.GetFiles（ @  d：\Building_Case）; 
 
  for  each（ string  file  in  files）
 {
 pbx.image = image.fromfile（file）; 
 pbx.imagelocation = file; 
 
} 
 pbx.DoubleClick + = pbx_DoubleClick; 
} 
 
  void  pbx_Double_Click（ object  sender，EventArgs e） 
 {
  string  path = pbx.imagelocation; 
 messagebox.show（path）; 
} 

现在，路径返回最后一个索引。我想在双击每个图片框时显示每个路径名称。所以，PLZ帮助我



添加了代码块，索引改进了，最后删除了额外的}[/ edit]

解决方案

只有在调用Load方法时，PictureBox.ImageLocation才会相关： http://msdn.microsoft.com/en-us/library/system.windows .forms.picturebox.imagelocation（v = vs.110）.aspx [ ^ ] - 在设置图像属性时从不使用它，因为图像可能根本没有路径，即使它确实如此，Image控件可能也不会意识到它（有很多种方法可以将图像转换为Image类实例，而且很少有直接转换文件的方法）。



相反，改变你的装载公司de：

  每个（ string 文件在文件中）
 {
 pbx.image = image.fromfile（file）; 
 pbx.Tag = file; 
} 

并在点击方法中使用Tag属性：

  string  path = pbx.Tag  as   string ; 
  if （！String.IsNullOrWhiteSpace（path））
 {
 MessageBox.Show（path）; 
} 
 

  foreach  （ string  file  in  files）
 {
 pbx.image = image.fromfile （文件）; 
 FileInfo info =  new  FileInfo（file）; 
  //  从信息对象获取文件信息 
  string  FileName = info.Name; 
} 

您好b $ b尝试此代码..

添加评论，请检查。

  private   void  button1_Click（ object  sender，EventArgs e）
 {
  string  [] files = System.IO.Directory.GetFiles（ @  D：\Building_Case ）; 
  foreach （ var 文件 in 文件）
 {
 pbx.Image = Image.FromFile（file）; 
 pbx.ImageLocation = file;  //  将路径存储在此属性中 
 
} 
 
 pbx.DoubleClick + =  new  EventHandler（pbx_DoubleClick）; 
} 
 
  void  pbx_DoubleClick（ object  sender，EventArgs e） 
 {
 
  string  path =（sender  as  PictureBox）。 ImageLocation;  //  从此属性获取路径 
 MessageBox.Show（path）; 
 System.Diagnostics.Process.Start（path）;  //  在Windows Image Viewer中打开图像 
} 

public class photo

picture pbx =new picturebox();

private voide btnok(object sender, EventArgs e)
{
   string[] files=System.IO.Directory.GetFiles(@"D:\Building_Case");

   for each(string file in files)
   {
      pbx.image=image.fromfile(file);
      pbx.imagelocation=file;

   }
   pbx.DoubleClick+=pbx_DoubleClick;
}

void pbx_Double_Click(object sender, EventArgs e)
{           
   string path=pbx.imagelocation;
   messagebox.show(path);
}  

Now, path returns last index. I want to show each path name when I double click on each picturebox. So, plz help me

[edit]Code block added, indexation improved and extra "}" at the end deleted[/edit]

解决方案

The PictureBox.ImageLocation is only ever relevant when you have called the Load method: http://msdn.microsoft.com/en-us/library/system.windows.forms.picturebox.imagelocation(v=vs.110).aspx[^] - it is never used when you set the Image poperty, because the Image may not have a path at all, and even if it does, the Image control may not be aware of it (there are a huge number of ways to get an image into an Image class instace, and very few of them directly inmvolve files).

Instead, change your loading code:

for each(string file in files)
   {
   pbx.image=image.fromfile(file);
   pbx.Tag = file;
   }

And use the Tag property in your click method:

string path=pbx.Tag as string;
if (!String.IsNullOrWhiteSpace(path))
   {
   MessageBox.Show(path);
   }

foreach(string file in files)
   {
      pbx.image=image.fromfile(file);
      FileInfo info = new FileInfo(file);
  //Get file infromation from info object
      string FileName = info.Name;
   }

Hi Try this code..
added comments , pls check it.

private void button1_Click(object sender, EventArgs e)
       {
           string[] files = System.IO.Directory.GetFiles(@"D:\Building_Case");
           foreach (var file in files)
           {
               pbx.Image = Image.FromFile(file);
                pbx.ImageLocation = file; // store the path also in this Property

           }

           pbx.DoubleClick += new EventHandler(pbx_DoubleClick);
       }

       void pbx_DoubleClick(object sender, EventArgs e)
       {

           string path = (sender as PictureBox).ImageLocation;  // get the path from this property
           MessageBox.Show(path);
           System.Diagnostics.Process.Start(path);  // To open the image in Windows Image Viewer
       } 

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