vfd Posted June 17, 2003 Share Posted June 17, 2003 Hi, I am currently working on my website and am facing a problem. I have a table of this sort... Table name: persons The fields are: name, photo, profile. This is for my family website. For each person I have stored their photo link and their profile link. For example for person1 the fields are... person1, images/person1/image1.jpg, profiles/person1/index.html When i query the database using \"phpmyadmin\" it shows all the fields. But when I use the above php script and try to print each field... it prints the name field but shows blank for photo and profile. Could you kindly let me know what i\'m doing wrong. This is the code i use... $result = mysql_query($sqlquery); if ($myrow = mysql_fetch_array($result)) { do { printf(\"%s %s %s<br>\", $myrow[\"name\"], $myrow[\"photo\"], $myrow[\"profile\"]); } while ($myrow = mysql_fetch_array($result)); } Quote Link to comment Share on other sites More sharing options...
deki Posted June 17, 2003 Share Posted June 17, 2003 Hey dude, Something i found tough when i was learning Coldfusion, which i have noticed applies the same with PHP, When displaying an image from a database it has to be in an image tag, for e.g. <img src=\"$images\"> Like that. Natrually you add the width and hight if needed. Hope it helps, but i am a newbie too. But try it out anyways. Cheers Quote Link to comment Share on other sites More sharing options...
michael yare Posted July 31, 2003 Share Posted July 31, 2003 You need to output the HTML required to display the image path too... So for each instance of your loop, you must output: echo "<img src = ""; echo $row["imagepath"]; echo "">"; Just as you would to display other HTML - data in table cells etc. Quote Link to comment Share on other sites More sharing options...
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