Hello,
I think you embed picture path to a table. This is very easy solution
to using picture. in your code you close the your mysql connection and after
this you call the mysql_result. if you close the connection before call the
result, it happen a problem. You cannot call result set after close the
connection. You should close the connection all of process.:))
if you call the picture in HTML with img tag. in your code if you
change the echo line like this echo "<img src =\"$logos \">, you will see
the picture:)
--
Haydar TUNA
Republic Of Turkey - Ministry of National Education
Education Technology Department Ankara / TURKEY
Web: http://www.haydartuna.net
"Chris Carter" <chandan9sharma@xxxxxxxxx> wrote in message
news:9079259.post@xxxxxxxxxxxxxxxxxx
>
> I have a field in database called "logos" which has one value
> "images/logos/some_logo.jpg"
>
> In my php I am trying to call it in my php file as image. With this code.
>
> $sno = $_REQUEST['sno'];
>
> $query="SELECT logos FROM table WHERE sno = '$sno'";
>
> $result=mysql_query($query);
> $num=mysql_numrows($result);
>
> mysql_close();
> $logos=mysql_result($result,$i,"logos");
>
> echo "
> <div>
> <p> $logos </p>
> </div>";
>
> ?></div>
>
> But getting errors. What is the way to call image in php file from the
> database. I know that this is a very basic question but first time for me.
> I
> ignored this in the begining but now its right infront of me.
>
> Please advice.
>
> Chris
> --
> View this message in context:
> http://www.nabble.com/How-to-call-image-from-mySql-to-php-file-tf3266161.html#a9079259
> Sent from the Php - Database mailing list archive at Nabble.com.
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