I am trying to list products images , and show their own gallery with popup fancybox. A can list all the images from database products but cant filter when clicking the image. It shows all photos.
Can you help me please.
Regards
<?php $query_product = mysql_query("SELECT * FROM products WHERE status = '1' AND language_parent_id = '0' AND id = '".$row_product_id['product_id']."' ORDER BY sort_order");
$row_product = mysql_fetch_array($query_product);
$imageFile = HTTP_RESIM.'products/'.$row_product['image'];
?>
<section class="flexslider">
<ul class="slides">
<?php $imageFile = HTTP_RESIM.'products/'.$row_product['image'];
$query_resim = mysql_query("SELECT * FROM product_image WHERE product_id = '".$row_product['id']."' AND status = '1' ORDER BY sort_order");
while($row_product_resim = mysql_fetch_array($query_resim)){ ?>
<?php $imageFile2 = HTTP_RESIM.'product_image/'.$row_product_resim['image']; ?>
<li><?php echo "<a href=\"$imageFile2\" rel=\"fancybox-gallery\" ><img src=\"$imageFile\" /></a>"; ?> </li>
<?php } ?>
</ul>
</section>
Their docs say
you should use it like this.
<a class="fancybox" rel="group" href="big_image_1.jpg"><img src="small_image_1.jpg" alt="" /></a>
So you will need to add the class "fancybox" to the anchor tag.
Edit: You should not use the mysql functions, they are outdated and very unsafe.
Look into PDO (http://php.net/manual/en/book.pdo.php) or mysqli (http://php.net/manual/en/book.mysqli.php):
Related
This image shows that I assigned different house_ID's for each image. Despite that, I would like to display images that have the same house_id, instead of just displaying all of them. For example, I would like to only display images with the house_id of 4. Does anyone know how to do this?
This is my current code which just displays all of them in a slide show.
<?php
// Include the database configuration file
include_once 'dbConfig.php';
// Get images from the database
$query = $db->query("SELECT * FROM images2 ORDER BY house_id DESC");
?>
<div class="frame">
<div class="slideshow-container">
<?php if($query->num_rows > 0){
while($row = $query->fetch_assoc()){
$imageURL = 'uploads2/'.$row["file_name"];?>
<div class="mySlides fade">
<img style="height: 200px;" src="<?php echo $imageURL; ?>">
</div><?php }?>
<a class="prev" onclick="plusSlides(-1)">❮</a>
<a class="next" onclick="plusSlides(1)">❯</a>
</div></div>
i am writing a php script that retrieves images from a database and shows them on page inside an album , each album contains several images...
here is an example for the code below :
$sql = "SELECT * FROM tbl_album where status='process' ORDER BY albumid DESC LIMIT $start_from, 12";
<?php
$rs_result = mysql_query ($sql,$con);
while ($row = mysql_fetch_assoc($rs_result))
{
$aid=$row['albumid'];
$aimage=$row['image'];
$aname=$row['name'];
$astatus=$row['status'];
echo '<div class="col-lg-4 col-md-6 col-sm-6 agile_gallery_grid">';
echo '<div class="hover ehover14">';
?>
<?php
echo"<a href='#details?d=$aid' onclick='showfunction()'>";
echo "<img src='admin/acatch/$aimage' class='img-responsive' alt='$aname' width='200' height='200' data-picture_id='$aid'>";
echo'</a>';
}
?>
this code shows all the albums contained in the database ,
but what i want to do is that when i click on the album cover , i want to display the images inside it on the same page , but i am not able to pass the album id to another php script on the same page to do this,
please if anyone can help
thank u in advance
Have a look at the following code:
<?php
$sql = "SELECT * FROM tbl_album where status='process' ORDER BY albumid DESC LIMIT $start_from, 12";
$rs_result = mysql_query($sql, $con);
while($row = mysql_fetch_assoc($rs_result)):
$aid=$row['albumid'];
$aimage=$row['image'];
$aname=$row['name'];
$astatus=$row['status'];
?>
<div class="col-lg-4 col-md-6 col-sm-6 agile_gallery_grid">
<div class="hover ehover14">
<a href='#details?d=$aid' onclick='showfunction()'>
<img src="admin/acatch/<?=$aimage?>" class="img-responsive" alt="<?=$aname?>" width="200" height="200" data-picture_id="<?=$aid?>">
</a>
</div>
</div>
<div class="gallery_album" id="gallery_album_<?=$row['albumid']?>">
<?php
/**
* now collect all images associated with each album by using $aid
*/
$sql_images = "SELECT * FROM tbl_album_images WHERE fk_album_id = $aid";
$images_result = mysql_query($sql_images, $con);
while($image = mysql_fetch_assoc($images_result)):
?>
<div class="gallery_album_image"><img src="<?=$image['dir_and_filename']?>"></div>
<?php
endwhile;
?>
</div>
<?php endwhile; ?>
What we are doing here is gettign all the albums as you allready does. But we introduce a subquery inside the first while-loop to fetch all associated images for each album.
We put these images inside new div-container with a unique album id you can use to show or hide the actual album with your showfunction()
Use CSS and javascript to show or hide accordingly.
OBS! WARNING!! Your mysql queries, and the one I provided, is very very BAD!
Have a look into prepared statement using php PDO or mysqli...
I have this php array. Where in my database i have a field called body, and in that field there is some html code. Like this:
<h1>title</h1><img src="http://farm5.staticflickr.com/4075/4788694752_d03557765b_z.jpg" alt=""/>
Here is my php code:
<?php
$q = "SELECT * FROM journals ORDER BY timestamp DESC";
$r = mysqli_query($dbc, $q);
while($journal_list = mysqli_fetch_assoc($r)) { ?>
<div class="col-md-4">
<a class="list-group-item" href="journal.php?id=<?php echo $journal_list['id']; ? >">
<h4 class="list-group-item-heading"><?php echo $journal_list['body']; ´?></h4>
</a>
</div>
<?php } ?>
In the h4 im calling the body field in the database. But i only want the img in that field??
Try phpquery:
$src = phpQuery::newDocumentHTML($journal_list['body'])->find('img')->attr('src');
This is a little hack that you can use
$img = explode('img',$journal['body']);
$final_img = '<img '.$img[1];
Now echo image as
<h4 class="list-group-item-heading"><?php echo $final_img; ?></h4>
I've been trying to add a section called: "related games", in it there is a script (related.php) that will fetch 5 random related games of the same category that the online game is displayed on.
I tried this (related.php):
<?php
if(isset($_GET['genre'])){
$game_category = $_GET['genre'];
$select_games = "SELECT * FROM games ORDER BY rand() LIMIT 0,5";
$run_games = mysql_query($select_games);
while($row = mysql_fetch_array($run_games)){
$game_id = $row['game_id'];
$game_name = $row['game_name'];
$game_image = $row['game_image'];
?>
<table>
<tr>
<div class="game_grid">
<a href="game_page.php?id=<?php echo $game_id; ?>"><img src="images/games_images/<?php echo $game_image; ?>" width="120" height="120" />
<span><?php echo $game_name; ?></span></a>
</div>
<tr>
</table>
<?php } } ?>
This is the "related.php" file and I tried to implement it in the following file called: "game_page.php" which works perfectly...
For some reason no random game is showing under the current played game...
Any idea?
I fixed it xD, I simply removed the
if(isset($_GET['genre'])){
$game_category = $_GET['genre'];
strings because they are already included in the "game_page.php" file :)
That's a simple script to adult content site I'd make with pieces of another codes. I am not a expertin PHP! I just can understand the basic.
<html>
<body>
<div align="center">
<div align="center" id="box">
<div align="center" id="carrosel">
<?php
$quantidade = 64;
$pagina = (isset($_GET['pagina'])) ? (int)$_GET['pagina'] : 1;
$inicio = ($quantidade * $pagina) - $quantidade;
$sql = "SELECT * FROM teen ORDER BY RAND() ASC LIMIT $inicio, $quantidade";
$qr = mysql_query($sql) or die(mysql_error());
while($ln = mysql_fetch_assoc($qr)){
?>
So, my problem is here. In that way, when you click the thumb, the browser will open that url that is out my site. the idea is to embed the video on a new iframe page. Someone can help me?
<a href="<?php echo $ln['url_video'];?>"target="_blank">
<img src="<?php echo $ln['thumb_video'];?>" /></a>
<?php
}
?>
<div align="center">
<?php
$sqlTotal = "SELECT thumb FROM teen";
$qrTotal = mysql_query($sqlTotal) or die(mysql_error());
$numTotal = mysql_num_rows($qrTotal);
$totalPagina= ceil($numTotal/$quantidade);
for($i = 1; $i <= $totalPagina; $i++){
if($i == $pagina)
echo $i;
else
echo "$i ";
}
?>
</div><br><br>
</body>
</html>
_blank : Opens the linked document in a new window or tab
_self : Opens the linked document in the same frame as it was clicked (this is default)
_parent : Opens the linked document in the parent frame
_top : Opens the linked document in the full body of the window
Or you can use iframe .or embed some players to play the video in the same area.
Use echo to print your links
while($ln = mysql_fetch_assoc($qr)){
echo '<a href='.$ln['url_video'].'target="_blank">',
'<img src='.$ln['thumb_video'].'/></a>';
}
And I found one more thing replace your echo "$i "; with this:
echo '$i';
And one more thing, use mysqli instead of mysql
You can also use window.open function, it will open the video in a new popup,
<a href="Javascript:;" onclick="window.open('<?php echo $ln['url_video'];?>',
'View Video')" ><img src="<?php echo $ln['thumb_video'];?>" /></a>
More details for window.open at http://www.w3schools.com/jsref/met_win_open.asp