This is what I used to do what you're trying:
header("Content-Type: image/" . $imagetype); // $imagetype is jpeg or gif
header("Content-Length: " . strlen($image));
echo $image;
Note that "Type" is capitalized in "Content-Type", and include the
"Content-Length" as well.
-Steve
On Thursday, December 6, 2001, at 02:45 PM, phantom wrote:
> I have successfully placed images (jpg,gif,png) into a MySQL database
> BLOB field, now I want to be able to pull the data out and diplay it.
>
> based on tutorial at
> http://www.zdnet.com/devhead/stories/articles/0,4413,2644827,00.html
>
> after Querying the DB I have the following lines:
>
> $ImgFile = mysql_result($Results,0,ImgFile); \\ image data;
> $ImgType = mysql_result($Results,0,ImgType); \\ image type (image/jpeg,
> image/gif);
> header("Content-type: ${ImgType}");
> echo "$ImgFile";
>
> Problem: This method, when it pulls up an image, it shows no image but a
> bunch of greek code. Example
> http://www.phantomcougar.com/strawberrypie_com/mem/show_img.php?PNum=8
> ((be sure to view page source)
>
>
>
>
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