LEFT JOIN 只返回右表的一行

Question

我有以下疑问；

"SELECT goals_challenges.*, 
    products_services.id as psid,
    products_services.url,
    products_services.feature_benefit
    FROM goals_challenges 
        LEFT JOIN products_services ON goals_challenges.id = products_services.goal_challenge_id
    WHERE persona_id = :persona_id"

两个表都有一个“id”列，因此是“psid”别名。

但是，尽管 products_services 表中有两条记录与 goal_challenge_id 匹配，但只有第一行作为结果集的一部分返回。

编辑：正确的数据

目标挑战

id    persona_id    title            item_category         solution
173   14            Lead Gen         business challenge    advertising

产品服务

id    goal_challenge_id     url                feature_benefit
1     173                   www.testurl.com    good for testing, mobile
2     173                   www.google.com     good for searching, well known

PHP 代码，包括查询；

public function findByPersonaId($persona_id)
{
    try {
        $this->dblayer->beginTransaction();
        $stmt = $this->dblayer->prepare("SELECT goals_challenges.*, products_services.id as psid, products_services.url, products_services.feature_benefit from goals_challenges LEFT JOIN products_services ON goals_challenges.id = products_services.goal_challenge_id WHERE goals_challenges.persona_id = :persona_id");
        $stmt->bindParam(':persona_id', $persona_id);
        $stmt->execute();

        $this->dblayer->commit();

        $result_set = array();

        while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
            $result_set[] = $this->mapObject($row); 
        }

        return $result_set;

    } catch (PDOException $e) {
        $this->dblayer->rollBack();
        echo $e->getMessage();
        exit;
    }
}

public function mapObject(array $row)
{
    $entry = new GoalChallenge();
    $entry->setId($row['id']);
    $entry->setPersonaId($row['persona_id']);
    $entry->setTitle($row['title']);
    $entry->setItemCategory($row['item_category']);
    $entry->setDescription($row['description']);
    $entry->setSolution($row['solution']);
    $entry->setProductService(new ProductService($row['psid'], $row['id'], $row['url'], explode(',', $row['feature_benefit'])));
    $entry->SetResearchChecklist($row['research_checklist']);
    $entry->setSubtopics($row['subtopics']);
    $entry->setKeywords($row['keywords']);
    $entry->setStatus($row['status']);

    return $entry;
}

我得到的回报

Array
(
    [id] => 173
    [persona_id] => 14
    [title] => Lead Gen
    [item_category] => Business Challenge    
    [solution] => Advertising
    [product_service] => 
    [research_checklist] => 0,0,0,0,0,0
    [psid] => 1
    [url] => www.google.com
    [feature_benefit] => good for testing, mobile
)

编辑：好的，所以我已经计算出我期望的结果，它与另一个不在同一个 goalChallenge 对象中 - 显然是 PHP 中的东西 - 有什么想法吗？

我从goals_challenges 表中获取所有数据，但仅从products_services 表中获取第一行（id 1）。

我的查询有问题吗？我已经尝试添加“GROUP BY targets_challenges.id”，但它不会改变结果。

Answer 1

您的查询看起来不错。 在 ON 条件下，列：-“goal_condition_id”是否具有主键。 检查两列是否都有主键。

谢谢 阿曼

Answer 2

"SELECT gc.*, 
    ps.id as psid,
    ps.url,
    ps.feature_benefit
    FROM product_services ps 
        LEFT JOIN goals_challenges gc ON gc.id = ps.goal_challenge_id
    WHERE ps.persona_id = :persona_id"

由于您的 goal_challenges 表有 1 行，您正试图找到与 product_services 表的任何适当匹配项。但只有一排，

选择意味着在这些条件下为我区分行并且 限制对我来说不会重复行。