Question

我正在尝试创建一个目录并且在调用＆＃34;列出图像时遇到问题＆＃34;在结果中。问题是只有一些列表会有图像，否则如果它们没有，我希望它们使用我设置的默认图像。当我尝试添加＆＃39;图像＆＃39;表格到我的查询，它只返回有图像可用的结果（忽略其他没有图像的列表）。

这是我的代码：

   public function search($neighborhood = null, $biz_filter = null) {

    $neighborhood = $this->uri->segment(3);
    $biz_filter = $this->uri->segment(4);

    // SELECT
    $this->db->select('*');

    // MAIN TABLE TO GRAB DATA
    $this->db->from('biz');

    // TABLES TO JOIN
    $this->db->join('city', 'city.city_id = biz.biz_cityID');
    $this->db->join('zip', 'zip.zip_id = biz.biz_zipID', 'zip.zip_cityID = city.city_id');
    $this->db->join('state', 'state.state_id = city.city_stateID');
    $this->db->join('neighborhood', 'neighborhood.neighborhood_id = biz.biz_neighborhoodID');
    $this->db->join('biz_filter', 'biz_filter.bizfilter_bizID = biz.biz_id');
    $this->db->join('biz_category', 'biz_category.bizcategory_id = biz_filter.bizfilter_bizcategoryID');

    if ($neighborhood != "-" AND $biz_filter != "-") {

        $this->db->where('biz_category.bizcategory_slug', $biz_filter);
        $this->db->where('neighborhood.neighborhood_slug', $neighborhood);

    } elseif ($neighborhood != "-" AND $biz_filter == "-") {

        $this->db->where('neighborhood.neighborhood_slug', $neighborhood);  

    } elseif ($neighborhood == "-" AND $biz_filter != "-") {

        $this->db->where('biz_category.bizcategory_slug', $biz_filter);

    } else {

    }

    // ORDER OF THE RESULTS
    $this->db->group_by('biz_name asc');

    // RUN QUERY
    $query = $this->db->get();

    // IF MORE THAN 0 ROWS ELSE DISPLAY 404 ERROR PAGE
    return $query;

}

如何添加单独的表格，＆＃39; image＆＃39;它包含徽标图像（＆＃39; image.image_file＆＃39;）。＆＃39;图像＆＃39;桌子和＆＃39; biz＆＃39;表通过每个表传递的业务ID连接（image.biz_id = biz.biz_id）。

任何人都知道如何解决查询才能正常工作？

Answer 1

只需使用

$this->db->join('image', 'image.biz_id = biz.biz_id', 'left');

离开加入您的image表格。如果biz_id表中没有记录，image.image_file将具有空值。请阅读here以获取更多信息。

您可以使用COALESCE函数替换＆＃34; null＆＃34;具有预定义默认值的图像。只需将您的行$this->db->select('*');替换为此行：

// SELECT
$this->db->select("*, COALESCE(image.image_file, 'images/not_found.png') as my_image_file");

渲染输出时，请确保使用my_image_file列作为图像。

旁注：避免使用＆＃39; *＆＃39;在选择中。仅选择您实际需要的列。选择所有列会不必要地增加数据库服务器资源的负载。

MySQL并只使用一些结果

1 个答案: