Question

我正在尝试根据FB的“喜欢/不喜欢”系统编写一个剧本来投票。我创建了一个php页面，其中脚本显示最喜欢的电影列表，从“投票”表中获取数据并使用“while”我将从“电影”表中选择它们，但我有一个问题：输出HTML只显示一部电影（内容最高的电影），而其他投票的电影则没有显示脚本的代码是：

if ($_GET['top_rated'] == 'show') {
   $vote= 1; //vote positive(like)
   $sql = $pdo->prepare("SELECT * FROM votes WHERE vote= :vote GROUP BY film_id");
   $swl->bindParam(":vote", $vote)
   $sql->execute();
   while ($vote_info = $sql->fetch(PDO::FETCH_ASSOC)) {
   $film_id = $vote_info['film_id'];
   $result = $pdo->prepare("SELECT * FROM films WHERE id= :film_id");
   $result->bindParam(":film_id", $film_id);
   $result->execute();
}
}

while ($film_info = $result->fetch(PDO::FETCH_ASSOC)) {
        $genre_id = $film_info['genre_id'];
        $sql = $pdo->prepare("SELECT * FROM genres WHERE id=:genre_id");
        $sql->bindParam(":genre_id", $genre_id);
        $sql->execute(); 
        $genre_info = $sql->fetch(PDO::FETCH_ASSOC);
 ?>
<div id="film-home-box">
<img class="poster" src="<?php echo $film_info['film_poster_src']; ?>"></img>
<div class="film-home-info">
Title: <?php echo $film_info['film_name'];?> <br>
Year: <?php echo $film_info['film_year'];?><br>
Genre: <?php echo $genre_info['genre_name'];?>
</div>
</div>

数据库的地图是：

my_database:
  films->id/film_name/film_year/genre_id 
  genres->id/genre_name
  votes->id/film_id/vote

提前致谢！

Answer 1

这是因为显示电影的代码只运行一次！

您应该使用某种循环来打印多个项目。我的建议是改变

while ($film_info = $result->fetch(PDO::FETCH_ASSOC)) {
        $genre_id = $film_info['genre_id'];
        $sql = $pdo->prepare("SELECT * FROM genres WHERE id=:genre_id");
        $sql->bindParam(":genre_id", $genre_id);
        $sql->execute(); 
        $genre_info = $sql->fetch(PDO::FETCH_ASSOC);
 ?>
<div id="film-home-box">
<img class="poster" src="<?php echo $film_info['film_poster_src']; ?>"></img>
<div class="film-home-info">
Title: <?php echo $film_info['film_name'];?> <br>
Year: <?php echo $film_info['film_year'];?><br>
Genre: <?php echo $genre_info['genre_name'];?>
</div>
</div>

以下

?><!-- Close the previous php block-->
<div id="film-home-box">
<img class="poster" src="<?php echo $film_info['film_poster_src']; ?>"></img>
<div class="film-home-info">
<?php
while ($film_info = $result->fetch(PDO::FETCH_ASSOC)) {
        $genre_id = $film_info['genre_id'];
        $sql = $pdo->prepare("SELECT * FROM genres WHERE id=:genre_id");
        $sql->bindParam(":genre_id", $genre_id);
        $sql->execute(); 
        $genre_info = $sql->fetch(PDO::FETCH_ASSOC);
        //print the details at each loop!
        //this should work if you have multiple items match your query...
        echo  "Title:  $film_info['film_name'] <br>
        Year: $film_info['film_year'] <br>
        Genre:  $genre_info['genre_name']";
 ?>
</div>
</div>

尝试并报告你发现的内容！

投票系统 - PHP＆amp; Mysql的

1 个答案: