菜鳥在线求！！！！

Question

实在太菜,一直没解决图像问题,向高手求教!语言C#,现有openFileDialog1,button1,pictureBox1,在解决方案下新建Image.(就是新建在生成文件里)现在单击button1,使pictureBox1显示为openFileDialog1打开的图像,那么我做的结果是:
private void button1_Click(object sender, EventArgs e)
        {
                               
 this.openFileDialog1.ShowDialog(); 
                        if   (this.openFileDialog1.FileName.Trim()   ==   "") 
                        { 
                                return; 
                        } 
                        try 
                        { 
                                this.pictureBox1.Image   =   System.Drawing.Bitmap.FromFile(this.openFileDialog1.FileName
                        } 
                        catch(Exception   exp) 
                        { 
                                MessageBox.Show(this,"打开图片文件错误！","信息提示",MessageBoxButtons.OK,MessageBoxIcon.Information); 
        }
    }那么现在我想使pictureBox1中的图片保存在Image中,如何实现?代码如何?应该还是button1的单击事件,那么请问高手发下代码...
先谢谢了...

Answer 1

指定打开方式~~~this.pictureBox1.Image = new Bitmap(new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.ReadWrite));//保存
this.pictureBox1.Image.Save(openFileDialog1.FileName);