[SOLVED] Different database results using a variable and hard coding in a search value.

[php_beginner_83]

Hi

 

I'm trying to create an online photo gallery and I'm pretty new to php. 

My problem is when I try and search my database.  When I use the following code..

 

$result = mysql_query("SELECT * FROM Images 
INNER JOIN Images_In_Album 
ON Images.ID = Images_In_Album.PicID 
INNER JOIN Albums 
ON Images_In_Album.AlbumID = Albums.ID 
WHERE Albums.ID = 1");

 

I get all the photos in the album with ID = 1.  So when I click my 'Next' link to browse through the pics they are all there.  However, when I use this code...

 

$result = mysql_query("SELECT * FROM Images 
INNER JOIN Images_In_Album 
ON Images.ID = Images_In_Album.PicID 
INNER JOIN Albums 
ON Images_In_Album.AlbumID = Albums.ID 
WHERE Albums.ID = \"$albumID\"");

 

Only the first picture is there.  When I click the 'Next' link, the wee box with the red cross in it appears.

 

Do you have any idea why this happens and how I can fix it??

 

This is my complete code...  $albumID comes from the previous page and changes depending on which photo album the user has clicked.

 

<?php
$username = "root";
$password = "";
$hostname = "localhost"; 


//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) 
  or die("Unable to connect to MySQL");
//echo "Connected to MySQL<br>";

$selected = mysql_select_db("MyWebsite",$dbhandle) 
  or die("Could not select examples");

//code here to determine what link has been clicked.
$albumID = $_GET['ID'];

//sql query to get records from that album.
$result = mysql_query("SELECT * FROM Images 
INNER JOIN Images_In_Album 
ON Images.ID = Images_In_Album.PicID 
INNER JOIN Albums 
ON Images_In_Album.AlbumID = Albums.ID 
WHERE Albums.ID = \"$albumID\"");

//arrays to hold the titles, descriptions and paths separately
$titles = array();
$descriptions = array();
$paths = array();

$counter = 0;
//fetch tha data from the database
while ($row = mysql_fetch_array($result)) 
{

$titles[$counter] = $row['Title'];
$descriptions[$counter] = $row['Description'];
$paths[$counter] = substr(strrchr($row['Path'],92),1);
   
   $counter++;
   
}

$imgIndex = $_GET['img'];


if(!isset($paths[$imgIndex]))
{
    $imgIndex = 0;
}

$currentImage = "images\\" . $paths[$imgIndex];

if ($imgIndex<=0)
{
    $prev = "Previous";
}
else
{
    	$prev = "<a href=\"".$_SERVER['SCRIPT_NAME']."?page=photo&img=".($imgIndex-1)."\">Previous</a>";

}

if ($imgIndex>=(count($paths)-1))
{
    $next = "Next";
}
else
{
   	$next = "<a href=\"".$_SERVER['SCRIPT_NAME']."?page=photo&img=".($imgIndex+1)."\">Next</a>";

}

echo " albumID " . $albumID;
echo "<pre>$prev                       $next</pre><br>";
echo "<div id='photoBody'></br><h2>$titles[$imgIndex]</h2>";
echo "<img src=\"{$currentImage}\"></br></br>";
echo "<p>$descriptions[$imgIndex]</p></div>";


mysql_close($dbhandle);
?>

 

Thank you.

[mattal999]

When you pass on to the next page, you don't resend the AlbumID. You just have the page number. Add the ID part to the next page hyperlink. Something like:

 

<?php

$next = "<a href=\"".$_SERVER['SCRIPT_NAME']."?page=photo&img=".($imgIndex+1)."&ID=".$albumID."\">Next</a>";

?>

[php_beginner_83]

thanks...that worked a treat :-)