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Posted on 2004-07-30
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Hi,

I am using MySQL database and in one table I have created 3 BLOB field type where I am storing JPG Images.

I able to store Images.

Using Following code :

ImageTable.Append;
ImageTable.FieldByName('ScopyNo').asInteger := 1;
TBlobField(ImageTable.FieldByName('Image')).LoadFromFile('C:\'Crop.jpg');
TBlobField(ImageTable.FieldByName('FinalImage')).LoadFromFile('C:\Crop.jpg');
 ImageTable.Post;

But Now I want to EDIT existing BLOB field value, So I using Following code:

ImageTable.Edit;
ImageTable.FieldByName('ScopyNo').asInteger := 2;
TBlobField(ImageTable.FieldByName('Image')).LoadFromFile('C:\'Crop.jpg');
TBlobField(ImageTable.FieldByName('FinalImage')).LoadFromFile('C:\Crop.jpg');
 ImageTable.Post;

But both BLOB fiels unable to store new image.

Please guide me what is the problem or is there other way to make changes in the field value.

Thanks & Regards,

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Question by:smartvanl5
3 Comments
 
LVL 27

Accepted Solution

by:
kretzschmar earned 200 total points
ID: 11701088
?? usual it should work this way

two possibilities:
- no commit
- you store the same picture again
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