Question

C#中如何打开图片文件

在C#中如何用代码打开本地计算机上的图片文件，然后把这张打开的图片放在窗体上某个picturebox中？一定是用代码实现才可以，谁能实现，帮一下忙了，急用，先谢谢了！

Answer 1

这需要一个按钮，openFileDialog1控件 ，picturebox控件（base_student_picture）
private void button14_Click(object sender, EventArgs e)
{
openFileDialog1.Filter = "图片文件（*.jpg）|*.jpg|所有文件|*.*";

if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
base_student_picture.ImageLocation = openFileDialog1.FileName;
path = base_student_id.Text.Trim()+Path.GetExtension(openFileDialog1.FileName);
//MessageBox.Show(path.ToString());
if (File.Exists(Application.StartupPath + "\\image\\" + path))
{
File.Delete(Application.StartupPath + "\\image\\" + path);
File.Copy(openFileDialog1.FileName, Application.StartupPath + "\\image\\" + path);
}
else
{
File.Copy(openFileDialog1.FileName, Application.StartupPath + "\\image\\" + path);
}
}
}
这样是将你在任何盘上的图片都考到你的bin文件夹下的debug文件夹下面了，你先试试！

Answer 2

using (OpenFileDialog lvse = new OpenFileDialog())
{
lvse.Title = "选择图片";
lvse.InitialDirectory = "";
lvse.Filter = "图片文件|*.bmp;*.jpg;*.jpeg;*.gif;*.png";
lvse.FilterIndex = 1;

if (lvse.ShowDialog() == DialogResult.OK)
{
pictureBox1.Image = Image.FromFile(lvse.FileName);
}
}

Answer 3

OpenFileDialog openFile = new OpenFileDialog();
if (openFile.ShowDialog() == DialogResult.OK)
{
string path = openFile.FileName;
Image image = Image.FromFile(path);
this.pictureBox1.Image = image;
}