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proctk · July 30, 2007

The below code displays an image on a page I want to somehow set it up so that when the user clicks the image it advances to the next photo in the query. any ideas how to do this thank you

$photoId = $_GET['photoID'];

//get image
$query_get_photo = ("select * from image_files WHERE image_id = '$photoId'");
$get_photo = mysql_query($query_get_photo )or die("SQL Error: $query_get_photo <br>" . mysql_error());

     $photo = mysql_fetch_assoc($get_photo);
     $foundPhoto = $photo['image_name'];
     $imageOwner = $photo['user_id'];
// Get album Author
    $query_get_author = ("select * from users WHERE user_id = '$imageOwner'");
    $get_author = mysql_query($query_get_author )or die("SQL Error: $query_get_author <br>" . mysql_error());

     $author = mysql_fetch_assoc($get_author);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
</head>
<body>
<div id="outer">
            <div id="hdr">
                  <!--Header-->             
                 <?php include('../design/topbanner.php'); ?>
</div>
            
        <div id="bar">
        <!--Subheader-->
        <h3 style="margin-left:250px; color:#848484; padding-top:10px;"><?php echo $_SESSION['fname'].' '. $_SESSION['lname']; ?></h3>
            </div>
        
        <div id="bodyblock">
          <!--body Lock-->
          
            <div id="l-col">
                        
            <?php include('../searchMembers.php'); ?>
                    </div>
            
          <div id="cont">
         <!--<p style="font-size:10px; padding:5px;"><a href="photos.php">Back to Albums</a></p>-->
          <div class="imageFull">
   <p style="font-size:10px; color:#848484; font-style:italic;">Create Date: <?php echo $photo['create_date']; ?> </p>
    <p style="text-align:center"><img src="../user_images/<?php echo $foundPhoto; ?>" /></p>
            
        <p style="text-align:right; padding-right:20px; margin-top:5px; font-size:10px;">From the Album:<br />
        <?php echo $photo['album']; ?> <span style="color:#000000">by</span> <?php echo $author['first_name'].' '.$author['last_name']; ?></p>    
        <p><?php echo $photo['details']; ?></p>

hitman6003 · July 30, 2007

How do you determine which is the next photo?

proctk · July 30, 2007

when The image is clicked it advances to the next photo in the query. like a slide show

hitman6003 · July 30, 2007

How do you determine which photo is the next one to display?

proctk · July 30, 2007

I could easily assign the values to an array.

I'm not sure how to get it to advance to the next photo when the image is clicked

//get image
$query_get_photo = ("select * from image_files WHERE album = '$album' AND user_id = '$owner_id'");
$get_photo = mysql_query($query_get_photo )or die("SQL Error: $query_get_photo <br>" . mysql_error());

  while($photo = mysql_fetch_assoc($get_photo);
 $image_id = $photo['image_id'];
/

hitman6003 · July 30, 2007

All you have to do is create a link of the current image...or add a "Next" link.

But you still haven't answered my question...how do you determine which image should be displayed as the "next" image?

Here is a one query example of your query above:

<?php

$photoId = $_GET['photoID'];

$query = "SELECT if.image_name, if.album, if.photo, if.create_date, if.details, u.first_name, u.last_name " .
     "FROM image_files if " .
     "  LEFT JOIN users u ON if.user_id = u.user_id " .
     "WHERE image_id = " . $photoId;

$result = mysql_query($query)or die("SQL Error: $query <br>" . mysql_error());
$photo = mysql_fetch_assoc($result);

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
</head>
<body>
<div id="outer">
<div id="hdr">
	  <!--Header-->             
	 <?php include('../design/topbanner.php'); ?>
</div>
            
<div id="bar">
	<!--Subheader-->
	<h3 style="margin-left:250px; color:#848484; padding-top:10px;"><?php echo $_SESSION['fname'].' '. $_SESSION['lname']; ?></h3>
</div>
        
<div id="bodyblock">
  <!--body Lock-->
  
	<div id="l-col">
		<?php include('../searchMembers.php'); ?>
	</div>

	<div id="cont">
		<div class="imageFull">
			<p style="font-size:10px; color:#848484; font-style:italic;">
				Create Date: <?php echo $photo['create_date']; ?>
			</p>

			<p style="text-align:center">
				<img src="../user_images/<?php echo $foundPhoto; ?>" />
			</p>

			<p style="text-align:right; padding-right:20px; margin-top:5px; font-size:10px;">
				From the Album:<br />
				<?php echo $photo['album']; ?>
				<span style="color:#000000">by</span>
				<?php echo $photo['first_name'] . ' ' . $photo['last_name']; ?>
			</p>    

			<p>
				<?php echo $photo['details']; ?>
			</p>

proctk · July 31, 2007

Your question is part of my problem.

I thought about assigning the image_ids to an array then some set the image url to somehow advance to the next photo

hitman6003 · July 31, 2007

Is the next photo the next id? Is it the next id in the album? Is there a date field? If so, would the next photo be the next newest one?

proctk · July 31, 2007

it would be the next image_id returned from the query. the sort order does not matter.

hitman6003 · July 31, 2007

it would be the next image_id returned from the query

By what query? Your query only returns one row.

proctk · July 31, 2007

This query could return more then one row

//get image
$query_get_photo = ("select * from image_files WHERE album = '$album' AND user_id = '$owner_id'");
$get_photo = mysql_query($query_get_photo )or die("SQL Error: $query_get_photo <br>" . mysql_error());

  while($photo = mysql_fetch_assoc($get_photo){
 $image_id = $photo['image_id'];

}

hitman6003 · July 31, 2007

An example:

<?php

$photoId = $_GET['photoID'];

$query = "SELECT if.image_name, if.album, if.photo, if.create_date, if.details, u.first_name, u.last_name " .
     "FROM image_files if " .
     "  LEFT JOIN users u ON if.user_id = u.user_id " .
     "WHERE image_id = " . $photoId;

$result = mysql_query($query)or die("SQL Error: $query <br>" . mysql_error());
$photo = mysql_fetch_assoc($result);

$query = "SELECT image_id " .
	 "FROM image_files " .
	 "WHERE album = '" . $album . "' AND user_id = " . $owner_id . " " .
	 "ORDER BY image_id ASC";

$result = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$ids[] = $row['image_id'];
}

foreach ($ids as $key => $id) {
if ($id == $photoId) {
	$next = $ids[$key + 1];
	$prev = $ids[$key - 1];
	break;
}
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
</head>
<body>
<div id="outer">
<div id="hdr">
	  <!--Header-->             
	 <?php include('../design/topbanner.php'); ?>
</div>
            
<div id="bar">
	<!--Subheader-->
	<h3 style="margin-left:250px; color:#848484; padding-top:10px;"><?php echo $_SESSION['fname'].' '. $_SESSION['lname']; ?></h3>
</div>
        
<div id="bodyblock">
  <!--body Lock-->
  
	<div id="l-col">
		<?php include('../searchMembers.php'); ?>
	</div>

	<div id="cont">
		<div class="imageFull">
			<p style="font-size:10px; color:#848484; font-style:italic;">
				Create Date: <?php echo $photo['create_date']; ?>
			</p>

			<p style="text-align:center">
				<img src="../user_images/<?php echo $foundPhoto; ?>" />
			</p>

			<p style="text-align:right; padding-right:20px; margin-top:5px; font-size:10px;">
				From the Album:<br />
				<?php echo $photo['album']; ?>
				<span style="color:#000000">by</span>
				<?php echo $photo['first_name'] . ' ' . $photo['last_name']; ?>
			</p>    

			<p>
				<?php echo $photo['details']; ?>
			</p>

			<p>
				<table width="100%">
					<tr>
						<td style="text-align: left; width: 50%;"><a href="<?php echo $_SERVER['PHP_SELF'] . "&?photoID=" . $prev; ?>">Prev</a></td>
						<td style="text-align: right; width: 50%;"><a href="<?php echo $_SERVER['PHP_SELF'] . "&?photoID=" . $next; ?>">Next</a></td>
					</tr>
				</table>
			</p>

proctk · July 31, 2007

thank you for the help. As you can tell I'm a rookie at this stuff. I'm having a hard time following your query as I have not worked with left join. I'm not able to figure out why this error is occuring

error

SQL Error: SELECT if.image_name, if.album, if.photo, if.create_date, if.details, u.first_name, u.last_name FROM image_files if LEFT JOIN users u ON if.user_id = u.user_id WHERE image_id = 184

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'if LEFT JOIN users u ON if.user_id = u.user_id WHERE image_id = 184' at line 1

$query = "SELECT if.image_name, if.album, if.photo, if.create_date, if.details, u.first_name, u.last_name " .
     "FROM image_files if " .
     "  LEFT JOIN users u ON if.user_id = u.user_id " .
     "WHERE image_id = " . $photoId;

hitman6003 · July 31, 2007

"if" is probably a reserved word....

change it to something different, "imf" or something like that, then change all of the "if.col_name" to "imf.col_name" as well.

proctk · July 31, 2007

your correct. I made the change and the error went away, however The query returns nothing my next question does your example script of to different examples. as there are duplicate variables $result, $query.

thank you

hitman6003 · July 31, 2007

Since we use the first query with the first mysql_query call, it becomes useless, and we can reuse that variable name. Same for $result...we get the values out of the first $result with mysql_fetch_assoc, which means we don't need it any more and can reuse the $result variable name.

If you aren't getting the expected results be sure to plug the queries into mysql query browser or phpMyAdmin and see if they are simply not matching anything, or if there is an error somewhere

proctk · July 31, 2007

Alright, my mistack on the query issue

hopefully last question; how do I get the image name in to the img url

thank you

hitman6003 · July 31, 2007

I'm assuming you mean in the img tag...change:

<img src="../user_images/<?php echo $foundPhoto; ?>" />

to

<img src="../user_images/<?php echo $photo['image_name']; ?>" />

proctk · July 31, 2007

thank you for the help and explanations. Would it be hard to change the foreach to loop though the image_ids and just start over when it ends and start back at image one. thank

cheers

hitman6003 · July 31, 2007

Change the foreach loop...I think this will work...

foreach ($ids as $key => $id) {
if ($id == $photoId) {
	$next = ($key + 1 > (count($ids) - 1) ? $ids[0] : $ids[$key + 1]);
	$prev = ($key - 1 < 0 ? $ids[count($ids)] : $ids[$key - 1]);
	break;
}
}

proctk · July 31, 2007

The foreach statement that you wrote for me is not working, when it comes to the end it does not start over. code it be because of the asc in the query

thank you

proctk · July 31, 2007

HI I think the issue is this from the below statement

count($ids) - 1 it the first image_id is 7. the ids run from 7-10. the link for the last image returns an image id of 6.

thank you

$next = ($key + 1 > (count($ids) - 1) ? $ids[0] : $ids[$key + 1]);

proctk · July 31, 2007

the fix was

change $ids[0] to $ids[1]

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