使用MySQL和PHP创建过滤功能

Question

嘿，这里是我的问题，我正在使用SQL查询来尝试返回基于帖子元数据的结果。这个元数据是用户在用户界面创建后创建的。我遇到了一个过滤器问题，它一次只能处理一个变量，但不超过这个范围（例如：如果主题被选中，它可以正常工作，但是如果选择了主题AND介质，它将不返回结果），这里是代码：

$db_build_post_filter_WHERE = array(); 
// Default to avoid errors on WHERE GROUP BY 
$db_build_post_filter_WHERE [] = 'cmsp_post.post_id > 0'; 
//$db_build_post_filter_WHERE [] = 'cmsp_post.post_id = cmsp_post_meta.post_id'; 
if (isset($gnocore_cmsp_build_topic_slug_id_array[$filter_topic])) { 
    $db_build_post_filter_WHERE [] = 'post_meta_key = "topic_id" AND post_meta_value = ' . $gnocore_cmsp_build_topic_slug_id_array[$filter_topic]; 
} 
if (isset($gnocore_cmsp_build_media_slug_id_array[$filter_media])) { 
$db_build_post_filter_WHERE [] = 'post_meta_key = "media_id" AND post_meta_value = ' . $gnocore_cmsp_build_media_slug_id_array[$filter_media]; 
} 
if (isset($gnocore_cmsp_build_author_slug_id_array[$filter_author])) { 
$db_build_post_filter_WHERE [] = 'post_meta_key = "author_id" AND post_meta_value = ' . $gnocore_cmsp_build_author_slug_id_array[$filter_author]; 
} 

// PROJECT FILTER ARRAY 
$build_post_filter_array = array(); 
gnoshare_db_select ('cmsp_post LEFT JOIN cmsp_post_meta ON cmsp_post.post_id = cmsp_post_meta.post_id','cmsp_post.post_id',implode(' AND ',$db_build_post_filter_WHERE) . ' GROUP BY cmsp_post.post_id','cmsp_post.project_id, post_name, cmsp_post.post_id','db_build_post_filter_array_num','db_build_post_filter_array_results'); 
if ($db_build_post_filter_array_num > 0) { 
    foreach ($db_build_post_filter_array_results as $db_build_post_filter_array_result) { 
     $build_post_filter_array [$db_build_post_filter_array_result->post_id] = ''; 
    } 
} 

我相信我的问题是在“PROJECT FILTER ARRAY”部分，如果有人可以协助它，将不胜感激。

干杯

编辑：改变“和” to“或”生成的简单显示所有帖子的结果，我认为这是公平地说，我的问题就出在这行代码，但是我仍然找不到一种方法来生成我正在寻找的结果。

gnoshare_db_select ('cmsp_post LEFT JOIN cmsp_post_meta ON cmsp_post.post_id = cmsp_post_meta.post_id','cmsp_post.post_id',implode(' OR ',$db_build_post_filter_WHERE) . ' GROUP BY cmsp_post.post_id','cmsp_post.project_id, post_name, cmsp_post.post_id','db_build_post_filter_array_num','db_build_post_filter_array_results'); 

编辑：这里是我的表（大约）：

+----------+---------------+-----------------+ 
| post_id | post_meta_key | post_meta_value | 
+----------+---------------+-----------------+ 
|  1 | topic_id |  1  | 
+----------+---------------+-----------------+ 
|  1 | media_id |  1  | 
+----------+---------------+-----------------+ 
|  1 | author_id |  2  | 
+----------+---------------+-----------------+ 
|  2 | media_id |  2  | 
+----------+---------------+-----------------+ 
|  2 | topic_id |  2  | 
+----------+---------------+-----------------+ 

，我想它的基础上的选择的用户已作出后（例如过滤：用户选定的主题1 ，媒体1和作者2的帖子1和媒体2的帖子2和帖子2的帖子2 （如上表））假设这个例子是真的，我希望我的网页生成后一个，如果任何以下内容被过滤器选中：按主题过滤1仅显示帖子1，按主题1过滤，媒体1仅显示帖子1，按主题过滤1媒体2将显示一条消息，指出与所选条件没有匹配，依此类推。这是否澄清了一点？

Answer 1

如果我没有错，你应该在过滤器之间使用OR并用（）括起来。例如：

$db_build_post_filter_WHERE [] = '(cmsp_post.post_id > 0)'; 
if (isset($gnocore_cmsp_build_topic_slug_id_array[$filter_topic])) { 
    $db_build_post_filter_WHERE [] = '(post_meta_key = "topic_id" 
            AND post_meta_value = ' . 
       $gnocore_cmsp_build_topic_slug_id_array[$filter_topic].')'; 
} 
if (isset($gnocore_cmsp_build_media_slug_id_array[$filter_media])) { 
    $db_build_post_filter_WHERE [] = '(post_meta_key = "media_id" 
            AND post_meta_value = ' . 
       $gnocore_cmsp_build_media_slug_id_array[$filter_media].')'; 
} 
//Other conditions 

// PROJECT FILTER ARRAY 
$build_post_filter_array = array(); 
gnoare_db_select ('cmsp_post 
        LEFT JOIN cmsp_post_meta 
          ON cmsp_post.post_id = cmsp_post_meta.post_id', 
       'cmsp_post_post.post_id', 
       implode(' OR ',$db_build_post_filter_WHERE) . 
       ' GROUP BY cmsp_post.post_id','cmsp_post.project_id, post_name, cmsp_post.post_id','db_build_post_filter_array_num','db_build_post_filter_array_results');