Question

像这样我的查询工作正常：

$pdo = $db->prepare('SELECT *
    FROM projects WHERE project_id = :project_id ');  
    $pdo->execute(array('project_id' => $project_id));
    while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
         echo "ok";
}

但是当我添加一个子选择时，我得到一个空白页面：

$pdo = $db->prepare('SELECT *
    (SELECT * FROM animals WHERE projects.animal=animals.id) AS animal
    FROM projects WHERE project_id = :project_id ');  
    $pdo->execute(array('project_id' => $project_id));
    while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
         echo "ok";
}

我做错了什么？

Answer 1

您可以使用INNER JOIN同时拥有两个表。

$pdo = $db->prepare('SELECT *
FROM projects
INNER JOIN animals ON projects.animal=animals.id
WHERE project_id = :project_id ');  
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
   echo "ok";
}

尝试此查询，因为此查询将返回所有具有动物关联动物的项目。

我认为这是你想要的

Answer 2

找到了这个解决方案：

$pdo = $db->prepare('SELECT *
    FROM projects 
    LEFT JOIN animals ON projects.animal=animals.id
    WHERE project_id = :project_id ');  
    $pdo->execute(array('project_id' => $project_id));
    while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
         echo "ok";
}

如何在查询中进行子选择？

2 个答案: