如何加入 2 个表，以便我可以使用过滤器来查询这两个表

Question

我有 2 个表，一个叫做“ads”，一个叫做“users”。 我在每个中存储了不同的数据。 ads.ad_user 列使用用户表中的 users.user_id 填充。 如何构建查询以便可以从两个表中过滤数据。

这是我构建的查询，但它不允许我独立地从两个表中过滤数据。

<div class="col-xs-6 col-sm-6">
    <div class="form-group">
    <label for="ethnicity">ETHNICITY</label>
    <select class="form-control ethnicitySelect" id="ethnicity" onchange="searchFilter();">
        <option selected disabled>Ethnicity...</option> 
        <option value="ethnicityall">All</option>
        <option value="other">Other</option>                                                
    </select>
</div>

<div class="form-group" style="padding:5px;color:#333;">
    <label for="citySelect">LOCATION</label>
    <select class="form-control citySelect" id="citySelect" onchange="searchFilter();">                                     
        <optgroup label="State">
        <option disabled selected>Select A City</option>
    </select>
</div>


$querySelectCount = "SELECT COUNT(ads.ad_id) as rowNum,
            ads.ad_id,
            ads.ad_title,
            ads.ad_content,
            ads.ad_date,
            ads.ad_user,
            ads.ad_photo,
            ads.ad_photo_thumb,
            ads.ad_age,
            ads.ad_city,
            ads.ad_rate,
            ads.ad_plan,
            ads.ad_approved,
            ads.adminPost,
            users.user_id,
            users.username,
            users.user_picture,
            users.userAge,
            users.height,
            users.hair,
            users.ethnicity,
            users.eyeColor,
            users.type              

            FROM
                ads
            LEFT JOIN
                users
            ON
                ads.ad_user = users.user_id                 

            $usersSQL

            WHERE 
                ads.ad_approved = '0'               
            $adsSQL $orderSQL                   
            ";

这是我正在使用的几个过滤器。

if(isset($_POST['city']) && !empty($_POST['city']) && $_POST['city'] !== 'null'){   
    $city = $filter->filter($_POST['city']);
    $whereSQL .= " AND ads.ad_city = :city "; 
}else{
    $city = 0;
    $whereSQL .= " AND ads.ad_city >= :city "; 
}


if(isset($_POST['ethnicity']) && !empty($_POST['ethnicity']) && $_POST['ethnicity'] !== 'null'){
    $ethnicity = $filter->filter($_POST['ethnicity']);
    if($ethnicity == "ethnicityall"){
        $whereSQL .= "AND users.ethnicity != :ethnicity ";
    }else{
        $whereSQL .= "AND users.ethnicity = :ethnicity ";
    }
}else{
    $ethnicity = 0;
    $whereSQL .= "AND users.ethnicity >= :ethnicity ";
}

我正在使用多个下拉列表来过滤数据，例如：当我选择一个城市时，除非我从用户表中选择了种族或身高或眼睛颜色等选项，否则不会显示任何内容......实际上，我需要选择所有在显示任何数据之前。 查询中的另一个 SELECT 会帮助我实现我想要做什么吗？ 谢谢。

另外这是我正在应用的过滤器

这是从 ads 表中选择一个项目和从 users 表中选择一个项目后查询的图片

这是更新后的查询的打印输出

object(PDOStatement)#5 (1) { ["queryString"]=> string(973) "SELECT COUNT(ads.ad_id) as rowNum, ads.ad_id, ads.ad_title, ads.ad_content, ads.ad_date, ads. ad_user, ads.ad_photo, ads.ad_photo_thumb, ads.ad_age, ads.ad_city, ads.ad_rate, ads.ad_plan, ads.ad_approved, ads.adminPost, users.user_id, users.username, users.user_picture, users.userAge, users.height, users.hair, users.ethnicity, users.eyeColor, users.type FROM ads LEFT JOIN users ON ads.ad_user = users.user_id AND users.ethnicity != :ethnicity AND users.eyeColor != :eyeColor AND users .hair != :hairColor AND users.height >= 0 WHERE ads.ad_approved = '0' AND ads.ad_city = :city AND ads.ad_rate >= 0 AND ads.ad_content LIKE :keywords ORDER BY ad_date DESC " }

Answer 1

可能您需要使用 LEFT JOIN 而不是 INNER，然后您可以将 users 表的条件添加到 ON 语句

LEFT JOIN users ON ads.ad_user = users.user_id AND users.ethnicity = :ethnicity
WHERE ads.ad_city = city

然后他们将独立工作