0

我目前在我的网站上运行此查询，

public function searchCandidates($type=null, $gender=null, $age=null)
{
    //die(var_dump($age));
    if($age != false) {
        $age = implode(", %", $age);
    }

    $sql = 'SELECT `candidates`.`candidate_id`, 
                    `candidates`.`first_name`, 
                    `candidates`.`surname`, 
                    `candidates`.`gender`,
                    `candidates`.`DOB`, 
                    `candidates`.`talent`, 
                    `candidates`.`availability`,
                    `candidates`.`youtube_showreel_1`,
                    `candidates`.`youtube_showreel_desc_1`,
                    `candidates`.`date_created`,
                    `candidates`.`new_talent`,
                     DATE_FORMAT(NOW(), "%Y") - DATE_FORMAT(`candidates`.`DOB`, "%Y") - (DATE_FORMAT(NOW(), "00-%m-%d") < DATE_FORMAT(`candidates`.`DOB`, "00-%m-%d")) as `age`,
                    `candidate_assets`.`url`, 
                    `candidate_assets`.`asset_size`,
                    `candidate_assets`.`weight`
            FROM `candidates`
            LEFT JOIN `candidate_assets` ON `candidate_assets`.`candidates_candidate_id` = `candidates`.`candidate_id`
            WHERE `candidates`.`show_on_site` = "urban talent" AND `candidates`.`visible` = "yes" AND `candidate_assets`.`weight` = 1';
            //is there a certain criteria
            if($type != "0") { $sql .= ' AND `candidates`.`talent` LIKE "%'.$type.'%"'; }
            if($age != false) { $sql .= ' AND `candidates`.`playing_age` LIKE "%'.$age.'%"';}
            if($gender != false){ $sql .= ' AND `candidates`.`gender` = "'.$gender.'"'; }

            $sql .= ' GROUP BY `candidates`.`candidate_id` ORDER BY `candidates`.`surname` ASC, `candidate_assets`.`weight` ASC';

    $query = $this->db->query($sql);

    //die(print_r($query->result_array()));

    //echo $this->db->last_query();
    //die();
    return $query->result_array();
}

我正在努力让这个查询按我的意愿工作，我想显示所有结果，无论用户是否有图像（candidate_assets. url），但是如果他们确实有图像，那么我只需要显示具有的图像权重为 1。目前，我的查询只带回图像权重为 1 的用户。

我想要的是返回所有用户，如果他们有一个图像（或多个图像），则返回权重为 1 的图像。

我做错了什么？

4

2 回答 2

0

如果您在 where 子句中使用第二个表中的列，则您的 LEFT JOIN 没有区别。要么允许它们在 where 子句中为 NULL，要么将它们移动到 ON 子句：

你有：

LEFT JOIN `candidate_assets` 
ON `candidate_assets`.`candidates_candidate_id` = `candidates`.`candidate_id`
WHERE `candidates`.`show_on_site` = "urban talent" 
AND `candidates`.`visible` = "yes" 
AND `candidate_assets`.`weight` = 1';

做到这一点：

LEFT JOIN `candidate_assets` 
ON `candidate_assets`.`candidates_candidate_id` = `candidates`.`candidate_id`
AND `candidate_assets`.`weight` = 1 /*moved here*/
WHERE `candidates`.`show_on_site` = "urban talent" 
AND `candidates`.`visible` = "yes"

或者可能（不那么“合乎逻辑”）

LEFT JOIN `candidate_assets` 
ON `candidate_assets`.`candidates_candidate_id` = `candidates`.`candidate_id`
WHERE `candidates`.`show_on_site` = "urban talent" 
AND `candidates`.`visible` = "yes" 
AND (`candidate_assets`.`weight` = 1 OR `candidate_assets`.`weight` IS NULL)

现在您将拥有所需的所有行。如果有选择地只选择一个图像，要么让您的应用程序代码处理它，要么使用 IF 查询稍微调整您的 SELECT 子句。

于 2012-05-29T12:06:16.660 回答
0

尝试这个 -

$sql = 'SELECT `candidates`.`candidate_id`, 
                `candidates`.`first_name`, 
                `candidates`.`surname`, 
                `candidates`.`gender`,
                `candidates`.`DOB`, 
                `candidates`.`talent`, 
                `candidates`.`availability`,
                `candidates`.`youtube_showreel_1`,
                `candidates`.`youtube_showreel_desc_1`,
                `candidates`.`date_created`,
                `candidates`.`new_talent`,
                 DATE_FORMAT(NOW(), "%Y") - DATE_FORMAT(`candidates`.`DOB`, "%Y") - (DATE_FORMAT(NOW(), "00-%m-%d") < DATE_FORMAT(`candidates`.`DOB`, "00-%m-%d")) as `age`,
                `candidate_assets`.`url`, 
                `candidate_assets`.`asset_size`,
                IF(`candidate_assets`.`url` IS NULL, 0, 1) AS `weight`
        FROM `candidates`
        LEFT JOIN `candidate_assets` ON `candidate_assets`.`candidates_candidate_id` = `candidates`.`candidate_id`
        WHERE `candidates`.`show_on_site` = "urban talent" AND `candidates`.`visible` = "yes"';
于 2012-05-29T12:06:22.287 回答