Re: How to call image from mySql to php file

Hi, Some points already given by others just reiterating: Chris Carter wrote: > $sno = $_REQUEST['sno']; $_REQUEST is dodgy - you don't know what makes it up (cookies, post, get etc.), only use it when it when you are sure. > $query="SELECT logos FROM table WHERE sno = '$sno'"; NEVER do this. VERY dangerous. Always use mysql_real_escape or similar delimiting functions on input data. > $result=mysql_query($query); > $num=mysql_numrows($result); > > mysql_close(); You can't close it yet..... > $logos=mysql_result($result,$i,"logos"); .... as you need it here!! Also $i is not initialised, I presume you mean to use $num? Even so, if there is 1 row, you will be passing in 1 and this would not return any results! If you want the last row returned you would have to pass in ($num - 1). HOwever if $i is indicative of a for loop, you are better using the fetch associative arrays functions alreayd mentioned and a while loop. > echo " > <div> > <p> $logos </p> > </div>"; > > ?></div> You'd presumable by better with: echo "<img src=\"$logos\" />" or similar :) HTH Col