Creates a new image cell with the specified render style.
Syntax
| Visual Basic (Declaration) | |
|---|
Public Function New( _
ByVal style As RenderStyle _
) |
Parameters
- style
- RenderStyle
Example
This example creates an image cell.
| C# |  Copy Code |
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FarPoint.Win.Spread.CellType.ImageCellType icelltype =new FarPoint.Win.Spread.CellType.ImageCellType(FarPoint.Win.RenderStyle.Stretch);
icelltype.TransparencyColor = Color.Black;
icelltype.TransparencyTolerance = 100;
fpSpread1.Sheets[0].Rows[0].CellType =icelltype;
System.Drawing.Image image = System.Drawing.Image.FromFile("D:\\alphaomega3.jpg");
System.IO.MemoryStream stream = new System.IO.MemoryStream();
byte[] bytes;
string str;
image.Save(stream,System.Drawing.Imaging.ImageFormat.Jpeg);
stream.Seek(0, System.IO.SeekOrigin.Begin);
bytes = stream.GetBuffer();
str = System.Convert.ToBase64String(bytes);
fpSpread1.Sheets[0].Cells[0,0].Value = image;
fpSpread1.Sheets[0].Cells[0,1].Value = bytes;
fpSpread1.Sheets[0].Cells[0,2].Value = str;
|
| Visual Basic | Copy Code |
|---|
Dim icelltype As New FarPoint.Win.Spread.CellType.ImageCellType(FarPoint.Win.RenderStyle.Stretch)
icelltype.TransparencyColor = Color.Black
icelltype.TransparencyTolerance = 100
FpSpread1.Sheets(0).Rows(0).CellType = icelltype
Dim image As System.Drawing.Image = System.Drawing.Image.FromFile("D:\alphaomega3.jpg")
Dim stream As New System.IO.MemoryStream
Dim bytes As Byte()
Dim str As String
image.Save(stream, System.Drawing.Imaging.ImageFormat.Jpeg)
stream.Seek(0, System.IO.SeekOrigin.Begin)
bytes = stream.GetBuffer()
str = System.Convert.ToBase64String(bytes)
FpSpread1.Sheets(0).Cells(0, 0).Value = image
FpSpread1.Sheets(0).Cells(0, 1).Value = bytes
FpSpread1.Sheets(0).Cells(0, 2).Value = str |
Requirements
Target Platforms: Windows 2000 Professional (SP4), Windows 2000 Server, Windows 2003 Server (SP1), Windows 2008, Windows XP (SP2), Windows Vista, Windows 7, Windows 8
See Also
