loading image?!?!

phphelpneeder

I'm trying to display an image using a field gathered from mySQL db.

Any ideas why it ain't working (please not im not too experienced with php/mysql so descriptions please)

<html>

<body>



<?php

include("config.inc");

// connect to MySQL

 $db =  mysql_connect($dbhost,$dbuser,$dbpasswd);

 mysql_select_db($dbname,$db);

$photo = mysql_query("SELECT * FROM testImages");
$results = "http://www.google.com/images/logo.gif";
$result = mysql_query("SELECT * FROM testImages",$db);



$foto = printf(mysql_result($result,"first"));
print "<IMG SRC=$foto>";
 ?>



</body>



</html>

svnfightsvn

                          mysql_select_db($dbname,$db);

$photo = mysql_query("SELECT * FROM testImages");

//WHAT IS THIS?
$results = "http://www.google.com/images/logo.gif";

//THIS IS THE SAME QUERY AS ABOVE - GET RID OF ONE
$result = mysql_query("SELECT * FROM testImages",$db);

//WHY printf?  - GET RID OF IT
//I WOULD ALSO USE mysql_fetch_array($results)
$foto = printf(mysql_result($result,"first"));
print "<IMG SRC=$foto>";

//TRY THIS:
//$foto = mysql_fetch_array($results);
//print "<IMG SRC=$foto[location_from_db_field]>";

phphelpneeder

all of the printf, unneeded queries, etc. are just form testing - was gonna remove them.

print "<IMG SRC=$foto[location_from_db_field]>";

I dont understand "[location_from_db_field]" bit. I actually want the script to display ALL the images in the table.

svnfightsvn

how are you storing the images? as URL's?
location_from_db is whatever field you have that stores the image location - when you use mysql_fetch_array, it puts the results of the query into an array ($foto) with each column name as an index...

then loop through the query like :

while($foto = mysql_fetch_array($results)){
    print "<IMG SRC=$foto[location_from_db]>";
}

phphelpneeder

got ya. yep - storing as urls

phphelpneeder

ignore this//

The69Pimp

i dont know if ya got it or not, but you could try this

<? 
$sql = "SELECT * FROM table ORDER BY id DESC LIMIT 1"; 
$query = mysql_query($sql); 
while($row = mysql_fetch_array($query)) 
{
?>
<img src="<? echo $row[url]; ?>" height="60" width="60">
<? } ?>

im a n00b myself, sorry if that didnt help

phphelpneeder

thats much more what i was after... thanks man