Question

我已加入三个表进行查询。问题是显示每个类名称然后显示狗，但不会将狗名称链接到特定节目。它在每个节目中显示相同的dognames列表。我试图在我的狗表中使用show_id，并将它链接到show table中的show_id。

三个表是：表演，结果，狗。

任何人都可以帮助我吗？

<?php
if(isset($_GET['s_id'])) {
    $the_show_id = $_GET['s_id'];

    $view_query = "UPDATE shows SET show_view_count = show_view_count + 1 WHERE  show_id = $the_show_id ";
    $send_query = mysqli_query($connection, $view_query);

    if(!$send_query) {

        die("query failed" );
    }


    if(isset($_SESSION['user_role']) && $_SESSION['user_role'] == 'admin' ) {

        $query = "SELECT * FROM shows WHERE show_id = $the_show_id ";

    } else {

        $query = "SELECT * FROM shows WHERE show_id = $the_show_id AND  show_status = 'published' ";
    }

    $select_all_shows_query = mysqli_query($connection,$query);

    if(mysqli_num_rows($select_all_shows_query) < 1) {

        echo "<h1 class='text-center'>No shows available</h1>";

    } else { 

        while($row = mysqli_fetch_assoc($select_all_shows_query)) {
            $show_name = $row['show_title'];
            $show_author = $row['show_author'];
            $show_date = $row['show_date'];
            $show_content = $row['show_content'];

            ?>

            <h1 class="page-header">
                Shows
            </h1>
            <!-- First Blog Post -->
            <h2>
                <a href="#"><?php echo $show_name ?></a>
            </h2>
            <p class="lead">
                by <a href="index.php"><?php echo $show_author ?></a>
            </p>
            <h4>Show Date: <span class="glyphicon glyphicon-time"> </span>
 <? php echo $show_date ?></h4>
            <hr>
            <div class="row">
                <div class="col-xs-6 col-sm-3"><h4>Class Name</h4></div>    
                <div class="col-xs-6 col-sm-2"><h4>Placement</h4></div>    
                <div class="col-xs-6 col-sm-7"><h4>Dog Name</h4></div>    
            </div><hr>
            <?php  

            $query = "SELECT result.class_name, result.placement, dogs.dog_name 
            FROM result 
            INNER JOIN dogs on result.resultID = dogs.resultIDD
            INNER JOIN shows on dogs.show_id = shows.show_id
            WHERE dog_name NOT LIKE 'absent' GROUP BY shows.show_id";

            $result = mysqli_query($connection, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($connection), E_USER_ERROR);

            if($result) {
                while($row = mysqli_fetch_assoc($result)) {
                    $dog_name = $row['dog_name'];
                    $placement = $row['placement'];
                    $class_name = $row['class_name'];
                    ?> 

                    <div class="row">
                        <div class="col-xs-6 col-sm-3"><p><?php echo $class_name ?></p></div>
                        <div class="col-xs-6 col-sm-2"><p><?php echo $placement ?></p></div>
                        <div class="col-xs-6 col-sm-7"><p><?php echo $dog_name ?></p></div>
                    </div>
                    <?php
                }
            }
        }
        ?>

Answer 1

好吧，假设所有其他事情都正常工作，您可能只需要将dogs.show_id（或者show.show_id两种方式）添加到select中。您是否可以尝试下面的查询，然后如果没有工作报告您的表格结构（您可以使用＆＃34; show create table TBL_NAME＆＃34;为三个表格中的每一个执行此操作。

<?php
    $query = "
        SELECT
            dogs.show_id,
            result.class_name, result.placement,
            dogs.dog_name
        FROM
            result INNER JOIN
            dogs on result.resultID = dogs.resultIDD INNER JOIN
            shows on dogs.show_id = shows.show_id
        WHERE
            dog_name NOT LIKE 'absent'
    "; 
?>

更新：好的，现在我们知道你已经在一个告诉我们show_id的循环中，你只需要更新查询以仅为该节目拉狗（假设show_id实际上在狗表中......除非你将dog + show_id作为唯一的复合键，否则它并不属于它所属的位置。

看看这里我将$ show_id添加到初始show查询的声明列表中，然后在查询中使用它来限制来自该查询的狗。此外，内部联接在这里并不是最好的... imho。

...
while ($row = mysqli_fetch_assoc($select_all_shows_query)) {

  // WE ALREADY KNOW THE SHOW WE'RE IN
  $show_id = $row['show_id']; 

  $show_name = $row['show_title'];
  $show_author = $row['show_author'];
  $show_date = $row['show_date'];
  $show_content = $row['show_content'];
    ?>
      <h1 class="page-header">Shows</h1>

        <!-- First Blog Post -->
        <h2><a href="#"><?php echo $show_name ?></a></h2>
        <p class="lead"> by <a href="index.php"><?php echo $show_author ?></a></p>
        <h4>Show Date: <span class="glyphicon glyphicon-time"> </span> <?php echo $show_date ?></h4>
        <hr>
        <div class="row">
          <div class="col-xs-6 col-sm-3"><h4>Class Name</h4></div>
          <div class="col-xs-6 col-sm-2"><h4>Placement</h4></div>
          <div class="col-xs-6 col-sm-7"><h4>Dog Name</h4></div>
        </div>
        <hr>
      <?php

        // SO UPDATE THE QUERY TO ONLY PULL THAT SHOW'S DOGS
        $query = "SELECT result.class_name, result.placement, dogs.dog_name
                  FROM result
                  LEFT JOIN dogs on result.resultID = dogs.resultIDD
                  WHERE dogs.show_idd = $show_id AND dog_name NOT LIKE 'absent'";

         $result = mysqli_query($connection, $query) or trigger_error
           ("Query Failed! SQL: $query - Error: ". mysqli_error
           ($connection), E_USER_ERROR);

         if ($result) {
           while ($row = mysqli_fetch_assoc($result)) {
             $dog_name = $row['dog_name'];
             $placement = $row['placement'];
             $class_name = $row['class_name'];
             ?>

               <div class="row">
                 <div class="col-xs-6 col-sm-3"><p><?php echo $class_name ?></p></div>
                 <div class="col-xs-6 col-sm-2"><p><?php echo $placement ?></p></div>
                 <div class="col-xs-6 col-sm-7"><p><?php echo $dog_name ?></p></div>
               </div>

已连接三个表但需要链接ID

1 个答案: