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Little help Please


ydeonia

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Hi

Can anybody plese help me in groping the movies under same cinema in this code

 

<?php

if(count($showtimes)==0):
	echo '<div class="invalid_city">'.$textnost.'</div>';
else:
	$i = 0;	
	$ciName = '';
	$count = count($showtimes);
	foreach ($showtimes as $st):

	
?>

	
	<div class="showtime <?php echo $i+1; ?>" id="accordion">
		<h4 class="stime_heading <?php echo $i; ?>"><a href="javascript:void(0);"><span><?php echo $st['cinema_name']; ?></span> </a></h4>
		<div class="listmovies_cont">
			<?php if($address_show): ?>
				<div class="address_movies"><?php echo $st['address']; ?></div>
			<?php endif; ?>
			
			<ul class="list_movies">
			
		
				
					<li>
						<span class="tit_movie">
					
							<a class="modal" id="popup" rel="{handler: 'iframe', size: {x: 970, y: 600}}" href="http://www.imdb.com/find?s=tt&q=<?php echo $st['movie_name']; ?>"> 
								<?php echo $st['movie_name']; ?></a>
								</span></br>
								<?php echo $st['cast']; ?>

						<?php if($show_times): ?>
							<span class="time_movie"><?php echo$st['show_time']; ?>
							<?php 
					
					 		$results = $dispatcher->trigger('onBeforeDisplayContent', array (& $st, & $params));
    				 		echo $results[3];
						?>
							</span>
						<?php endif; ?>
						
					</li>			
			</ul>
		</div>
	</div>	
<?php

$i++;
$ciName = $st['cinema_name'];
	
	endforeach;
?>

 

 

Now this code displays cinema name  with single movie . But there are multiple movie running in same cinema. I want to show them in same cinema with movie name ,cast and showtime.

here in below table I want to group movies with cinema Id

 

Here is the table structure

PEWP21g.png

 

 

Thanks 

Edited by ydeonia
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If ever a table was in need of normalization it's that one ( as I have mentioned before ). If you are going to use MySql, use it as a relational database, with the advantages that brings, and not a collection of spreadsheets.

 

Just my 0.02 worth

Edited by Barand
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If ever a table was in need of normalization it's that one ( as I have mentioned before ). If you are going to use MySql, use it as a relational database, with the advantages that brings, and not a collection of spreadsheets.

 

Just my 0.02 worth

 

I didn't understand your point. :( Sorry . Yes this is mysql . I want to display the parents as cinemas and child as movies. 

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I wrote a tutorial that addresses this:

http://thewebmason.com/tutorial-parent-child-lists/

 

You'll want to store the cinema in a variable and compare the previous one to the current one.

 

Hi Jessica ,

 

After reading your blog , I created the page and it worked ! :) Loads of Love for you . Thank you

 

<?php
mysql_connect("localhost","test","test");  
mysql_select_db("test_db");  

//select the table 
$result = mysql_query("select cinema_name ,cast,show_time, cinemaId, movie_name , m.id from jos_cinema as c
INNER JOIN #__movie as m ON m.id=c.movie_id ORDER BY c.cinemaId ASC,m.id ASC") or die ('Error: '.mysql_error ()); 

while ($row = mysql_fetch_array($result)){
    if($last_cinema != $row['cinema_name']){
        echo "<h3>{$row['cinema_name']}</h3>";		
        $last_cinema = $row['cinema_name'];		
    } 
    echo "<br /><b>{$row['movie_name']}</b><br />";
    echo" {$row['cast']}<br />"; 
    echo "{$row['show_time']}";
}
?> 

My reply to previous post - view the videos in the link

 

http://forums.phpfreaks.com/topic/276568-condition-help/?do=findComment&comment=1423062

 

Hi Barand , I read your topic and I learned many new things . Thanks alot.

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You're missing the part where you define $last_cinema before you ever try to use it, if you turn on strict error reporting you'll see a warning or notice. Otherwise good. 

 

Definitely normalize the tables though, I didn't see the original screenshot before, and Barand is 100% right, you need to fix your structure.

 

Also a line like this: echo "{$row['show_time']}"; just adds extra processing. No need to put a variable in strings to echo it.

Edited by Jessica
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