具有多个连接的分页查询

Question

我有几个左外连接的查询，正如你在代码中看到的那样，我生成的对象包含 2 个填充有数组的对象，你知道我不能在这里使用“简单”分页，因为对于一个用户，我得到了几列（对于每个角色和每个组）例如，对于用户 1，我可以获得 4-5 个结果，因为用户可以拥有许多组和角色，我理解这是我不能在此查询中使用 LIMIT 和 offset 的主要原因，因为我会输一些信息。我的问题是，如何修改这个查询，让每页有 20 个不同的用户以及他们的所有角色和组。

$stmt = $mysqli->prepare("SELECT u.id, u.firstName, u.lastName, u.email, 
    u.phoneNumber, u.address, u.birthDate, ur.roleName, cg.id, cg.name FROM users as u 
    LEFT OUTER JOIN user_role as ur ON u.id = ur.userId 
    LEFT OUTER JOIN user_group as ug on ug.userId = u.id 
    LEFT OUTER JOIN control_group as cg on cg.id = ug.groupId WHERE u.id != ?");
$stmt->bind_param("i", $_SESSION["id"]);
$stmt->execute();
$stmt->bind_result($id, $firstName, $lastName, $email, $phoneNumber, 
    $address, $birthDate, $roleName, $groupId, $groupName);

$users = array();

while ($stmt->fetch()) {
    if (empty($users[$id])) {
        $users[$id] = array(
            'id' => $id,
            'firstName' => $firstName,
            'lastName' => $lastName,
            'email' => $email,
            'phoneNumber' => $phoneNumber,
            'address' => $address,
            'birthDate' => $birthDate,
            'roles' => array(),
            'groups' => array()
        );
    }
    if ($roleName) {
        $found = false;
        foreach ($users[$id]['roles'] as $role) {
            if($role['roleName'] == $roleName){
                $found = true;
                break;
            }
        }
        if($found == false)
            $users[$id]['roles'][] = array(
                'roleName' => $roleName
            );
     }

    if ($groupId) {
        $found = false;
        foreach ($users[$id]['groups'] as $group) {
            if($group['groupName'] == $groupName){
                $found = true;
                break;
            }
        }
        if($found == false)
            $users[$id]['groups'][] = array(
                'groupName' => $groupName
            );
     }
}

$res = []; 

foreach($users as $user) { 
    $res[] = $user; 
} 

$stmt->close();
$mysqli->close();
echo json_encode($res);

Answer 1

好吧，我认为您可以采用两种查询方法。 用第一个你做分页

SELECT u.id FROM users LIMIT offset, count

第二个是你拥有的 u.id = （我认为应该是 =，而不是 != 如你所见）改为 u.i​​d IN

这样的事

$stmt = $mysqli->prepare("SELECT u.id from users u limit ?, ?");
$stmt->bind_param("1", $_SESSION["offset"]);
$stmt->bind_param("2", $_SESSION["limit"]);
$stmt->execute();
$stmt->bind_result($id);

$user_ids = array();

while ($stmt->fetch()) {
    $user_ids[] = $id;
}

$stmt2 = $mysqli->prepare("SELECT u.id, u.firstName, u.lastName, u.email, 
    u.phoneNumber, u.address, u.birthDate, ur.roleName, cg.id, cg.name FROM users as u 
    LEFT OUTER JOIN user_role as ur ON u.id = ur.userId 
    LEFT OUTER JOIN user_group as ug on ug.userId = u.id 
    LEFT OUTER JOIN control_group as cg on cg.id = ug.groupId WHERE u.id in (?)");
$stmt2->bind_param("1", implode(",", user_ids);
$stmt2->execute();
$stmt2->bind_result($id, $firstName, $lastName, $email, $phoneNumber, 
    $address, $birthDate, $roleName, $groupId, $groupName);

$users = array();

while ($stmt->fetch()) {
    ...
}