simple....help PLEASE

Anon

Hi,sorry to trouble you people with my nOOb problems, i have only just started to try and learn this stuff, so thanks for any help in advance. All i want to do (at present) is pull images out of a folder using the image_name that is stored in the db, i dont know what 2 use for this, but this doesnt work and i have no idea why....

<?php

$db = mysql_connect("localhost", "root");
mysql_select_db("antiques",$db);

$result = mysql_query("SELECT image_name FROM general ");
while ($row = mysql_fetch_assoc($result)) {

extract($row);

echo " <img src=\"images/"$image_name."\">\n";

tomhath

First thing I'd do is just put the filename into the string where you echo the img tag and see if the HTML works (don't use what came out of the database, just type in the full tag so you know that part works).

Then echo the variable you have in $image_name to see if it's the same as what works.

Reading what you have, it looks like you are missing a . and a } here:

echo " <img src=\"images/" . $image_name . "\">\n";
}

Anon

thanks, code has no errors now but is not returning the image and the image is definaetly there cause a html page displays it fine. So ?????

panreach

I'm a fresh newbie so take my advice with a grain of salt. hehe

I've read something that talks about using images in PHP. You have to have the gd library for PNGs and if you want to want to use JPEGs you have to use the jpeg-6b and recompile jpeg support or something like that. I guess gd comes bundled with PHP 4. However you can't use gifs anymore.

I think after that is working you have to use the Header() function to let the browser no that it's an image that your trying to send ie

Header ("Content-type: image/png");
ImagePng($img);

//clean up

ImageDestroy($im);

Hope that helps. Sorry if I totally mislead you in the wrong direction.

Anon

hahaha, thanks, i've had a lot of that (leading in the wrong direction thing) i think my problem, apart from not knowing anything about this stuff is that my code is a mess and it is not sending the info i.e image_name to the database when i submit the form i am using.hhhmmmm thanks for your help