C# 如何在 DataGridViewImageColumn 中添加图像？
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时间：2020-08-10 09:18:12 来源:igfitidea点击:
How do I add an image in my DataGridViewImageColumn?
提问by Flávio Costa
I have a field DataGridViewImageColumn, and for each line of the field, depending on a condition, I add a different image. Anyone know how I can do this in Windows Forms?
我有一个 field DataGridViewImageColumn，对于该字段的每一行，根据条件，我添加了不同的图像。任何人都知道如何在 Windows 窗体中执行此操作？
if (dgvAndon.Rows[e.RowIndex].Cells["urgencyOrder"].ToString() == "1")
{
//Here I want to add the image in the image property field DataGridViewImageColumn.
}
采纳答案by Manoj
- Add your image in Resources.resx under properties folder. (ex. Picture1.jpeg)
- Add a
DataGridViewImageColumnin yourDataGridView Add image this way:
for (int row = 0; row <= [YourDataGridViewName].Rows.Count - 1; row++) { ((DataGridViewImageCell)gvFiles.Rows[row].Cells[1]).Value = Properties.Resources.Picture1 }
- 在属性文件夹下的 Resources.resx 中添加您的图像。（例如图片1.jpeg）
DataGridViewImageColumn在你的添加一个DataGridView以这种方式添加图像：
for (int row = 0; row <= [YourDataGridViewName].Rows.Count - 1; row++) { ((DataGridViewImageCell)gvFiles.Rows[row].Cells[1]).Value = Properties.Resources.Picture1 }
回答by ketan italiya
use this code
使用此代码
protected void gridView1_RowDataBound(Object sender, GridViewRowEventArgs args)
{
if(args.Row.RowType == DataControlRowType.DataRow)
{
Image img = (Image) e.Row.FindControl("Image1");
img.ImageUrl = setImageURLHere;
}
}
回答by Koorosh
use this code:
使用此代码：
DataGridViewImageColumn iconColumn = new DataGridViewImageColumn();
iconColumn.Name = "AirplaneImage";
iconColumn.HeaderText = "Airplane Image";
dataGridView1.Columns.Insert(5, iconColumn);
for (int row = 0; row < dataGridView1.Rows.Count - 1; row++)
{
Bitmap bmp = new Bitmap(Application.StartupPath + "\Data\AirPlaneData\" + dt.Rows[row][4]);
((DataGridViewImageCell)dataGridView1.Rows[row].Cells[5]).Value = bmp;
}
回答by Cistelican Catalin Videographe
string FileName = null;
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.RestoreDirectory = true;
openFileDialog.Filter = "All picture files (*.BMP;*.JPG;*.GIF)|*.BMP;*.JPG;*.GIF";
if (openFileDialog.ShowDialog() == DialogResult.OK)
{
FileName = openFileDialog.FileName;
//pictureBox2.Image = Image.FromFile(FileName);
}
