Question

我正在创建一本食谱书（只是为了继续练习数据库），再一次，我被困了......

我能够显示图像，但是方式不好（我认为）...... 到目前为止我做了什么：

在我的index.php中：

$recipes = get_recipes();
$attachments = get_attachments();
$attachments_paths = get_image_path();

echo '<h2>recipes</h2>';
echo '<table id="recipesTable" border="1">';
echo '<tr>';
echo '<th>Recipe ID</th>';
echo '<th>Recipe Name</th>';
echo '<th>Attachment ID</th>';
echo '<th>Attachment path</th>';
echo '<th>Attachment image</th>';
echo '</tr>';
foreach ($recipes as $recipe) {   
  echo '<tr>';
    echo '<td>' . $recipe['id'] . '</td>';
    echo '<td>' . $recipe['name'] . '</td>';
    echo '<td>' . $recipe['attachment_id'] . '</td>';
    foreach ($attachments_paths as $attachment_path) {
        if ($recipe['attachment_id'] === $attachment_path['id']) {
            echo '<td>' . $attachment_path['attachment_path'] . '</td>';
        }
    }
    echo '<td>' . echo display_image(); . '</td>';
  echo '</tr>';
}
echo '</table>';

的functions.php：

 function get_recipes() {
    include'db_connection.php';
    try {
        return $conn->query("SELECT * FROM recipes");
    } catch (PDOException $e) {
        echo 'Error:' . $e->getMessage() . "<br />";
        return array();
    }
    return true;
}

function get_attachments() {
    include'db_connection.php';
    try {
        return $conn->query("SELECT * FROM attachments");
    } catch (PDOException $e) {
        echo 'Error:' . $e->getMessage() . "<br />";
        return array();
    }
    return true;
}

function get_image_path() {
    include 'db_connection.php';

    $sql = 'SELECT recipes.name, attachments.id, attachments.attachment_path FROM attachments LEFT JOIN recipes ON recipes.attachment_id=attachments.id';

    try {
        $results = $conn->prepare($sql);
        $results->execute();
    } catch (PDOException $e) {
        echo 'Error: ' . $e->getMessage() . '<br />';
        return array();
    }
    return $results->fetchAll(PDO::FETCH_ASSOC);
}

function display_image() {
     foreach($attachments as $attachment) {
         $file = $attachment['attachment_path'];   
         return '<img src="' . $file . '" />';
 }

} ＆＃34;食谱＆＃34;的列表： id，name，created，duration，source，categories_id，attachment_id，chef_id

＆＃34;附件＆＃34;的列表： id，attachment_path，recipe_id

还提到，我保存了图像的路径： http://localhost/cooking_book/images/mistique.jpeg

如果有更好的方法，请告诉我！

所以，正如你所看到的，我的display_image（）根本不起作用，我知道foreach循环在那里什么也没做......我只是没有其他任何想法，我可以做，任何建议将不胜感激:)）

Answer 1

嗯，我确信这不是最好的方法，我会继续努力，但至少，我现在正在显示图像;

这个功能可以更基本......

function get_image_path() {
    include 'db_connection.php'; 

      $sql = 'SELECT recipes.name, attachments.id, attachments.attachment_path FROM attachments LEFT JOIN recipes ON recipes.attachment_id=attachments.id'; 

     try {
      $results = $conn->prepare($sql);   
      $results->execute();
     } catch(PDOException $e) {
        echo 'Error: ' . $e->getMessage() . '<br />';
        return array();
    }
      return $results->fetchAll(PDO::FETCH_ASSOC);    
}

function display_image() {
     include 'db_connection.php'; 

       return $image_path = get_image_path();       
}

为了显示它，在index.php上，我刚刚在我的表中添加了另一个字段：

echo '<td>';
    foreach ($images as $img) {
        echo '<img class="attachment" src="' . $img['attachment_path'] . '"/>';
  } 
 echo '</td>';

但是，如果有人建议选择更好的选项，请告诉我们：）

用于显示来自数据库存储路径

1 个答案: