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). G.
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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The future isn't what it used to be. (It never was.)
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