Displaying images as a gallery in PHP/MySQL

[ayooshkasmth]

Hi,

I have some pictures in a MySQL database that i am able to retrieve and display. How ever, i want to display them as images in a gallery, but not able to.

The MySQL query is ok, it retrieves the images, i think the problem is with my PHP script. Any help would be greatly appreciated. Here is the code that displays a single image (which works fine)

include ("database.php");

	if (!isset($_GET['id'])) { echo "<br />ERROR: id not specified ";    } 
 
	$id = (int) $_GET['id'];
 
	if ($id <= 0) { echo "<br />ERROR: Invalid id <br />"; } 
 
	$query  = sprintf("SELECT * FROM property_images JOIN property ON property.id = property_images.property_id ORDER BY property_images.id DESC LIMIT 1");	
	$result = mysql_query($query, $db) or die('Error, query failed'); 
	
	if (mysql_num_rows($result) == 0) { echo "Error: property has no images. ";   } 

	$row = mysql_fetch_array($result);			
	extract($row);	
	
	header('Content-type: ' . $row['uploadedimage_type']);
        header('Content-length: ' . $row['uploadedimage_size']);
 
       echo $row['content'];

and this is what i've tried for displaying all images .... with no luck (blank page)

$query  = sprintf("SELECT * FROM property_images JOIN property ON property.id = property_images.property_id ORDER BY property_images.id DESC LIMIT 0 , 30");	
	$result = mysql_query($query, $db) or die('Error, query failed'); 

    if (mysql_num_rows($result) == 0) { echo "Error: property has no images. ";   }
    else {			
        while ($row = mysql_fetch_array($result)) {				
	     extract($row);	}	// end while	

    header('Content-type: ' . $row['uploadedimage_type']);
    header('Content-length: ' . $row['uploadedimage_size']);
 
    echo $row['content']; 	} // end else

[mediachicken]

easy

<?php

$query="SELECT * FROM database ";
$result=mysql_query($query);
$num=mysql_numrows($result);


$i=0;
while ($i < $num) {

// get your fields in the databse
$field1 =mysql_result($result,$i,"column");

//PUT YOUR STUFF HERE!


$i--;
}

[ayooshkasmth]

Thank you.

Could you please clarify

// get your fields in the databse
$field1 =mysql_result($result,$i,"column");

I've tried using the column name of the images in both the $field1 and column, and it doesn't work.

[vaultdweller123]

why do you have to extract() the result? when you can already traverse the recordset via mysql_fetch_array. and when you do extract you dont use arrays anymore like $row coz extract() assign values to array indexes so instead of echo $row you use $content. you may try this

<?php

    $query  = sprintf("SELECT * FROM property_images JOIN property ON property.id = property_images.property_id ORDER BY property_images.id DESC LIMIT 0 , 30");	
	$result = mysql_query($query, $db) or die('Error, query failed'); 

    if (mysql_num_rows($result) == 0) { echo "Error: property has no images. ";   }
    else {			
        while ($row = mysql_fetch_array($result)) {				
	     header('Content-type: ' . $row['uploadedimage_type']);
    header('Content-length: ' . $row['uploadedimage_size']);
 
    echo $row['content']; }	// end while	

    	} // end else

?>

header('Content-type: ' . $row['uploadedimage_type']);
    header('Content-length: ' . $row['uploadedimage_size']);
 
    echo $row['content']; 	} // end else

Why not use this within the while?

[mediachicken]

That was the point of mine, you put whatever code you want executed in the while, as for the field name it whatever you want, that's a variable name. You need to change the column to whatever your name it in the database (that you want to pull) for that entry. Therefore if you have a table and it has the values: firstname,lastname to loop through every first and last name you use my code like:

Database setup fields: firstname,lastname

$query="SELECT * FROM database ";
$result=mysql_query($query);
$num=mysql_numrows($result);
 
 
$i=0;
while ($i < $num) {
 
// get your fields in the databse
$fname =mysql_result($result,$i,"firstname");
$lname =mysql_result($result,$i,"lastname");


echo "First Name:$fname, Last Name:$lname";
 
 
$i--;
}

If there's 50 names it'll print one at a time.

[ayooshkasmth]

Thanks guys.

@ vaultdweller123 and rajarajan07 - that was the first thing i tried. I thought was only logical to loop it to get all results. But the problem is i get a link to the page as output (e.g. http://localhost/...../display_gallery.php?id=1)

Here is the code that gives me that

if (mysql_num_rows($result) == 0) { echo "Error: property has no images. ";   } // end if		
	
else {	
   while ($row = mysql_fetch_array($result)) {					
	header('Content-type: ' . $row['uploadedimage_type']);
	header('Content-length: ' . $row['uploadedimage_size']);
	echo $row['content']; 	
    }	// end while	
 } // end else

@ mediachicken - thanks for the clarification. I get the same output as above, i.e a link to the page

$num = mysql_fetch_array($result);	
  
$i = 0;	
while ($i < $num) {	
     $content = mysql_result($result, $i, "content");	
     header('Content-type: ' . $num['uploadedimage_type']);
     header('Content-length: ' . $num['uploadedimage_size']);		
	
     echo "$content";	
  
     $i--;	}	// end while

I must be doing something wrong - but i can't figure out what :(

PS. why is $i-- and not ++ ?

[daniel233]

thanks all :) Good job

[vaultdweller123]

what does your $row holds in your database? if it's an image then you dont just echo it directly, you have to put it on an <img> tag

echo "<img src='".$row['content']."' />";

i hope i get your point :)

[ayooshkasmth]

vaultdweller123 - that was it! :)

Thank you all very much guys.