php - 从数组中抓取然后忽略 MYSQL

Question

下面我有这个，它根据以前的选择创建一个选择菜单。这些选择以数组形式保存在我的数据库中，其中 portfolio_tags = hfw_portfolio_tags id。表格格式如下:

+---------------------+---------+----------------+
| hfw_portfolio_items |         |                |
+---------------------+---------+----------------+
| id                  | name    | portfolio_tags |
| 1                   | name #1 | ["1","3","4"]  |
| 2                   | name #2 | ["2","3","4"]  |
+---------------------+---------+----------------+

+--------------------+----------+--+
| hfw_portfolio_tags |          |  |
+--------------------+----------+--+
| id                 | tag_name |  |
| 1                  | tag #1   |  |
| 2                  | tag #1   |  |
| 3                  | tag #1   |  |
| 4                  | tag #1   |  |
+--------------------+----------+--+

<?php 
$table_name_tags = $wpdb->prefix . "HFW_portfolio_tags";
$rows = $wpdb->get_results("SELECT * FROM"." $table_name_tags"." ORDER BY id ASC");
foreach($rows as $rowsTags) { 
    $portfolio_tags = stripslashes ($rowsTags->id);
    $tagName = stripslashes ($rowsTags->tag_name);
    $queryHere = 'SELECT * FROM '.$table_name.' WHERE id='.$id.'';
    $tagRows = $wpdb->get_results( $wpdb->prepare( $queryHere, APP_POST_TYPE));
    //exit(var_dump($queryHere));
    foreach ($tagRows as $tagFromDB ){
        $items_select = stripslashes ($tagFromDB->portfolio_tags);
        $arr = json_decode($items_select,true);                 
        $tagId = stripslashes ($tagFromDB->id);

        if (is_array($arr)){  
            foreach ($arr as $eachSelectedDB ){
                if($eachSelectedDB == $portfolio_tags){
                    echo '<option value="'.$portfolio_tags.'"  selected>'.$tagName.'</option>';
                }
                elseif($portfolio_tags !== $eachSelectedDB){

                }
            }
        }
        echo '<option value="'.$portfolio_tags.'">'.$tagName.'</option>';
    }
}

?>

问题是当我输出时 - 我通过 foreach 运行的标签被输出两次，作为整个 select MYSQL 的输出，并且输出也将它们计为存在。有没有办法在 hfw_portfolio_tags 表的最终输出中跳过这些？

Answer 1

试试这个:

<?php 
        $table_name_tags = $wpdb->prefix . "HFW_portfolio_tags";
        $rows = $wpdb->get_results("SELECT * FROM"." $table_name_tags"." ORDER BY id ASC");
        foreach($rows as $rowsTags) { 
                    $portfolio_tags = stripslashes ($rowsTags->id);
                    $tagName = stripslashes ($rowsTags->tag_name);
                    $queryHere = 'SELECT * FROM '.$table_name.' WHERE id='.$id.'';
                    $tagRows = $wpdb->get_results( $wpdb->prepare( $queryHere, APP_POST_TYPE));
                    //exit(var_dump($queryHere));
                        foreach ($tagRows as $tagFromDB ){
                            $items_select = stripslashes ($tagFromDB->portfolio_tags);
                            $arr = json_decode($items_select,true);                 
                            $tagId = stripslashes ($tagFromDB->id);

                            echo '<option value="'.$portfolio_tags.'" '.(is_array($arr) && in_array($portfolio_tags, $arr) ? 'selected' : '' ) .'>'.$tagName.'</option>';
                        }
        }

?>

问题是使用:(is_array($arr) && in_array($portfolio_tags, $arr) ? '选定' : '' )而不是 foreach