On Friday 08 February 2002 16:26, phantom wrote:
> I have this really cool script that grabs image data stored in a mysql
> bin field and echo's the data into an image file.
>
> 01: /* QUERY DB AND LOAD IMAGE DATA */
> 02: /* must get values for ImgType and ThmData */
> 03: $Results = mysql_query($Query, $Link)
> 04: or die ("SL3-".mysql_errno().": ".mysql_error());
> 05: $Num_rows = mysql_num_rows($Results);
> 06: if ($Num_rows==1) { // show image;
> 07: $ImgType = mysql_result($Results,0,ImgType);
> 08: $ThmData = mysql_result($Results,0,ThmData);
> 09: header("Content-Type: " . $ImgType); //$ImgType shoud be
> image/jpeg;
> 10: echo $ThmData;
> 11: }
>
> However, I have a simple mysql database counter that counts how many
> times this image is loaded. $UpdateQuery = "UPDATE Counter SET Ct=Ct=1
> WHERE Img = '${Img}'";
Where is this query being used? You're not inadvertently using it twice?
Could you post the complete code?
> However everytime I run this script the counter increments by 2 (not
> 1). GRRRR.
>
> After dicing this script up, the line that is suspect is Line 09:
>
> If I comment out this line (Line 09), the counter increments properly
> (by 1).
>
> If I comment out Line 10 and leave Line 09 in, the counter increases by
> 2 and the image fails to display. Also Line 09 echos the URL of the
> image. Could this URL be resubmitting the script again, thus
> incrementing the counter an extra count???
I don't see how it can "resubmit" your script again, it's not using Location:.
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
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