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 7, Windows Vista SP1 or later, Windows XP SP3, Windows Server 2008 (Server Core not supported), Windows Server 2008 R2 (Server Core supported with SP1 or later), Windows Server 2003 SP2
See Also
