我将一些信息存储到我的 Access 数据库中,一个参数是 BLOB 字段,在本例中是一个图像,加载到 Timage 中。
我正在使用此代码来保存它:
var
AStream : TMemoryStream;
begin
AStream := TMemoryStream.Create;
try
Image1.Picture.Graphic.SaveToStream(AStream);
AStream.Position := 0;
if Adotable1.Active then
begin
Adotable1.Edit;
TBlobField(Adotable1.FieldByName('Termograma')).LoadFromStream(AStream);
Adotable1.Post;
end;
finally
AStream.Free;
adotable1.Append;
adotable1['Data']:= datetimepicker1.Date;
adotable1['Temax']:= edit4.Text;
adotable1['Temin']:= edit5.Text;
adotable1['Descrição da Posição']:= memo1.Text;
adotable1['Comentários']:= memo2.Text;
adotable1.Post;
但我还通过单击“附加”部分等同一按钮来存储其他信息。
发生的情况是,当我按下保存按钮时,此信息不会存储在数据库中的同一 ID 中。
如何解决这个问题?
最佳答案
您正在编辑当前记录,将图像保存到其中,附加新记录,并将其余信息保存到该新记录。我认为您打算添加一个全新的记录,将图像和数据添加到该新记录中,然后保存这些更改。
试试这个:
var
AStream : TMemoryStream;
begin
if not AdoTable1.Active then
AdoTable1.Open;
Adotable1.Append;
AStream := TMemoryStream.Create;
try
Image1.Picture.Graphic.SaveToStream(AStream);
AStream.Position := 0;
TBlobField(Adotable1.FieldByName('Termograma')).LoadFromStream(AStream);
finally
AStream.Free;
end;
adotable1['Data']:= datetimepicker1.Date;
adotable1['Temax']:= edit4.Text;
adotable1['Temin']:= edit5.Text;
adotable1['Descrição da Posição']:= memo1.Text;
adotable1['Comentários']:= memo2.Text;
adotable1.Post;
end;
关于delphi - 保存图像进行记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14678348/