将选择的图片显示在listview中,并显示filename,path和type的简单实例

时间:2022-09-18 18:42:03

代码如下:


if (openFileDialog1.ShowDialog() == DialogResult.OK) 
           { 
               listView1.Items.Clear(); 
               string[] files = openFileDialog1.FileNames; //定义一个数组,获取选择的文件 
               string[] fileinfo = new string[3];  //定义一个数组,用于存储文件信息 
               for (int i = 0; i < files.Length; i++)  //遍历文件数组 
               { 
                   string path = files[i].ToString();  //获取文件路径 
                   //截取文件名字 
                   string fileName = path.Substring(path.LastIndexOf("\\") + 1, path.Length - 1 - path.LastIndexOf("\\")); 
                   //截取文件类型 
                   string fileType = fileName.Substring(fileName.LastIndexOf(".") + 1, fileName.Length - 1 - fileName.LastIndexOf(".")); 
                   fileinfo[0] = fileName; 
                   fileinfo[1] = path; 
                   fileinfo[2] = fileType; 
                   ListViewItem lvi = new ListViewItem(fileinfo); 
                   listView1.Items.Add(lvi); 

               } 
          }

 

 

 

注释:ListView1, View 属性为Details

OpenFileDialog1,Multiselect属性为true