我有一个问题,我真的不知道自己在做什么。我想要创建的是这样的。
我有4张桌子。
movies | movie_categories
series | show_categories
现在我想创建一个将连接所有这四个表并在代码中返回结果的查询,这是我到目前为止所尝试的,并且它无法正常工作
<?php
if(isset($_GET['category_id']) && !empty($_GET['category_id']))
{
$kategorija = $_GET['category_id'];
/*$query = $connection->prepare('SELECT * FROM movies LEFT JOIN movie_categories ON movie_categories.movie_hash = movies.hash WHERE movie_categories.category LIKE ? ORDER BY movies.id DESC');*/
$query = $connection->prepare("
SELECT
movie_categories.movie_hash,
movie_categories.movie_category,
show_categories.serie_hash,
show_categories.serie_category
FROM movie_categories
INNER JOIN show_categories ON movie_categories.movie_hash != show_categories.serie_hash
LEFT JOIN movies on movie_categories.movie_hash = movies.hash
LEFT JOIN series on show_categories.serie_hash = series.hash
WHERE movie_categories.movie_category LIKE ?
OR WHERE show_categories.serie_category LIKE ?
");
$query->bindValue(1, "%$kategorija%", PDO::PARAM_STR);
$query->bindValue(2, "%$kategorija%", PDO::PARAM_STR);
$query->execute();
if($query->rowCount() > 0)
{
$check = $query->fetchAll(PDO::FETCH_ASSOC);
foreach($check as $row)
{
echo '
<div class="row" id="content">
<div class="col-md-2 col-sm-4 col-xs-6 animacija animated fadeInDown">
<div class="film">
<a href="watch.php?movie=',$row['hash'],'" title="',$row['name'],'">
<img alt="',$row['name'],'" width="165" height="250" class="img-responsive" src="',$row['thumb_location'],'">
<span class="play animated flip">
<i class="fa fa-play-circle fa-4x"></i>
</span>
</a>
<div class="opis">
<a href="watch.php?movie=',$row['hash'],'" title="',$row['name'],' "><h2>',$row['name'],' </h2></a>
<div class="tekst">
</div>
<p>
<span>
<a href="godina.php?movie=',$row['year'],'" title="',$row['year'],'">',$row['year'],'</a>
</span>
</p>
</div>
</div>
</div>
';
}
}
else
{
echo '<h3>U ovoj kategoriji nema filmova.</h3>';
}
}
else
{
header("location: index.php");
exit();
}
?>