3
我有一個角色需要某個證書，才能將用戶分配給該角色。SQL JOIN問題未將用戶詳細信息從查詢中帶回並將它們打印到數組中
我正在嘗試從數據庫中建立一個用戶列表，這個用戶列表中有一個或多個匹配的特定角色的證書。 （我也只查看用戶獲得的最後一個證書，並且只將該證書與該角色進行比較）
只要用戶實際上擁有該角色所需的兩個證書，該查詢就可以很好地工作。
如果用戶不匹配所有需要的證書，那麼用戶詳細信息不會回來。
首先，我將說明我是運行的，以匹配用戶，則該方法我會告訴你陣列輸出結果
你將看到第一個用戶（搶劫，冬青）陣列0和1匹配角色所需的兩個證書，因此所有信息（如id，圖像和名稱）都會返回到數組的第一部分。
第三個用戶（daniel）數組（2）只匹配一個，所以我沒有在數組的第一部分中獲得他的名字或詳細信息。
第4個用戶（katie）數組（3）不匹配，所以我在最後返回一個完全空白的數組。
我需要丹尼爾和凱蒂都至少返回他們的名字和idimage所以我可以將它們打印到屏幕上。
我想這是什麼做的LEFT JOIN用戶，但即使我做了RIGHT JOIN沒有回來。如果我堅持一個FULL JOIN，則查詢完全失敗。
//$allcertificates contains ids of certificates required for role
public function getStaffByProjectRoleCertificates($allCertificates)
{
//get all active users
$allUsers = $this->allActiveUser();
$users = array();
$temp = 0;
//loop through the users to find matches
for($i = 0; $i < count($allUsers); $i++)
{
$allRows = array();
for($j = 0; $j < count($allCertificates); $j++)
{
$query = sprintf("
SELECT cert.certName, usercert.idusercert, usercert.usercertEnd, user.iduser, user.idimage, user.userForename, user.userSurname, certStatus.certStatusName FROM usercert
LEFT JOIN user ON user.iduser = usercert.iduser
INNER JOIN cert ON usercert.idcert = cert.idcert
INNER JOIN certStatus ON certStatus.idcertStatus = usercert.idcertStatus
WHERE
usercert.iduser=%s AND usercert.idcert=%s AND
NOT EXISTS
(SELECT * FROM `usercert` as u2 WHERE u2.iduser=%s AND u2.idcert=%s AND u2.idcert = usercert.idcert
AND u2.idusercert > usercert.idusercert)
",
$this->db->GetSQLValueString($allUsers[$i], "int"),
$this->db->GetSQLValueString($allCertificates[$j], "int"),
$this->db->GetSQLValueString($allUsers[$i], "int"),
$this->db->GetSQLValueString($allCertificates[$j], "int"));
$result = $this->db->query($this->db->link, $query) or die($this->db->error($this->db->link));
$numRows = $this->db->num_rows($result);
$row = $this->db->fetch_assoc($result);
array_push($allRows, $row);
if($numRows !=0)
{
$temp += 1;
}
if($j == count($allCertificates) -1)
{
array_push($users, array("iduser" => $row['iduser'], 'idimage'=>$row['idimage'], 'forename'=>$row['userForename'], 'surname'=>$row['userSurname'], "cert_no" => $temp, "data"=>$allRows));
$temp =0;
}
}
}
function sortByOrder($a, $b)
{
return $b['cert_no'] - $a['cert_no'];
}
usort($users, 'sortByOrder');
print_r($users);
return $users;
}
和outputed陣列
Array
(
[0] => Array
(
[iduser] => 90
[idimage] =>
[forename] => Holly
[surname] => Bain
[cert_no] => 2
[data] => Array
(
[0] => Array
(
[certName] => Full UK Driving License
[idusercert] => 21
[usercertEnd] => 2016-05-31
[iduser] => 90
[idimage] =>
[userForename] => Holly
[userSurname] => Bain
[certStatusName] => Expiring In 2 Months
)
[1] => Array
(
[certName] => test certificate
[idusercert] => 22
[usercertEnd] => 2016-05-31
[iduser] => 90
[idimage] =>
[userForename] => Holly
[userSurname] => Bain
[certStatusName] => Expiring In 2 Months
)
)
)
[1] => Array
(
[iduser] => 88
[idimage] => 197
[forename] => Robert
[surname] => Bain
[cert_no] => 2
[data] => Array
(
[0] => Array
(
[certName] => Full UK Driving License
[idusercert] => 24
[usercertEnd] => 2017-07-01
[iduser] => 88
[idimage] => 197
[userForename] => Robert
[userSurname] => Bain
[certStatusName] => Expiring In 2 Months
)
[1] => Array
(
[certName] => test certificate
[idusercert] => 19
[usercertEnd] => 2016-05-31
[iduser] => 88
[idimage] => 197
[userForename] => Robert
[userSurname] => Bain
[certStatusName] => Expiring In 2 Months
)
)
)
[2] => Array
(
[iduser] =>
[idimage] =>
[forename] =>
[surname] =>
[cert_no] => 1
[data] => Array
(
[0] => Array
(
[certName] => Full UK Driving License
[idusercert] => 20
[usercertEnd] => 2016-05-31
[iduser] => 86
[idimage] => 196
[userForename] => Daniel
[userSurname] => Robinson
[certStatusName] => Expiring In 2 Months
)
[1] =>
)
)
[3] => Array
(
[iduser] =>
[idimage] =>
[forename] =>
[surname] =>
[cert_no] => 0
[data] => Array
(
[0] =>
[1] =>
)
)
)
非常好的答案非常感謝你的時間。對我的代碼的其餘部分進行一些小的更改完美地工作。 –