SQLSTATE[42000] 聚合函数错误

Question

您好，我在我的 SQL 查询中遇到了一个错误，无法弄清楚是什么问题。这是迄今为止在 Barmar 的帮助下的查询。

 $query = "SELECT c.*, count(s.curso_id) as count, SUM(IF(s.status = 'aprobado', 1, 0)) AS count_approved , SUM(IF(s.status = 'cupolleno', 1, 0)) AS count_cupolleno
    , SUM(IF(s.status = 'cancelado', 1, 0)) AS count_cancelado, SUM(IF(s.status = 'noacion', 1, 0)) AS count_noacion, SUM(IF(s.status = 'ama_de_casa', 1, 0)) AS count_ama_de_casa
    , SUM(IF(s.status = 'cliente_externo', 1, 0)) AS count_cliente_externo
    FROM cursos_modulos AS c
    LEFT JOIN subscriptions AS s ON s.curso_id = c.id
    LEFT JOIN users AS u ON u.userID = s.user_id GROUP BY c.id WHERE 1";
                if (!empty($id)) { $query .= " AND c.id = '$id'"; }
                if (!empty($ciudad)) { $query .= " AND c.ciudad = '$ciudad'"; }
                if (!empty($tipo)) { $query .= " AND c.tipo = '$tipo'"; }
                if (!empty($titulo))  { $query .=" AND c.titulo = '$titulo'"; }
                if (!empty($status))  { $query .= " AND c.status = '$status'"; }
 $paginate = new pagination($page, $query, $options);

我得到的错误信息如下：

致命错误：带有消息的未捕获异常“PDOException” 'SQLSTATE[42000]: 语法错误或访问冲突：1064 你有一个 SQL 语法错误；检查与您对应的手册 MySQL 服务器版本，用于在 'WHERE 1 AND 附近使用正确的语法 c.id = '1' LIMIT 0, 30' at line 6' in E:\xampp\htdocs\admin\class\pagination.php:376 堆栈跟踪：#0 E:\xampp\htdocs\admin\class\pagination.php(376): PDOStatement->execute() #1 E:\xampp\htdocs\admin\class\pagination.php(202): 分页->执行查询（）＃2 E:\xampp\htdocs\admin\class\pagination.php(162): pagination->run(1, 'SELECT c., cou...', Array) #3 E:\xampp\htdocs\admin\search.php(146): pagination->__construct(1, 'SELECT c., cou...', Array) #4 {main} 在线抛出 E:\xampp\htdocs\admin\class\pagination.php 376

Answer 1

group by 子句应该在 where 子句之后。即：

$query = "SELECT c.*, count(s.curso_id) as count, SUM(IF(s.status = 'aprobado', 1, 0)) AS count_approved , SUM(IF(s.status = 'cupolleno', 1, 0)) AS count_cupolleno
, SUM(IF(s.status = 'cancelado', 1, 0)) AS count_cancelado, SUM(IF(s.status = 'noacion', 1, 0)) AS count_noacion, SUM(IF(s.status = 'ama_de_casa', 1, 0)) AS count_ama_de_casa
, SUM(IF(s.status = 'cliente_externo', 1, 0)) AS count_cliente_externo
FROM cursos_modulos AS c
LEFT JOIN subscriptions AS s ON s.curso_id = c.id
LEFT JOIN users AS u ON u.userID = s.user_id WHERE 1";
            if (!empty($id)) { $query .= " AND c.id = '$id'"; }
            if (!empty($ciudad)) { $query .= " AND c.ciudad = '$ciudad'"; }
            if (!empty($tipo)) { $query .= " AND c.tipo = '$tipo'"; }
            if (!empty($titulo))  { $query .=" AND c.titulo = '$titulo'"; }
            if (!empty($status))  { $query .= " AND c.status = '$status'"; } 
$query .= " GROUP BY c.id";

Answer 2

where 1 能为您做什么？尝试杀死它。

以下不会引发 1064 错误：

create table cursos_modulos
(   id int not null

);

create table subscriptions
(   curso_id int not null,
    user_id int not null,
    status varchar(100) not null
);

create table users
(   userID int not null
);

SELECT c.id,
count(s.curso_id) as count, 
SUM(IF(s.status = 'aprobado', 1, 0)) AS count_approved, 
SUM(IF(s.status = 'cupolleno', 1, 0)) AS count_cupolleno, 
SUM(IF(s.status = 'cancelado', 1, 0)) AS count_cancelado, 
SUM(IF(s.status = 'noacion', 1, 0)) AS count_noacion, 
SUM(IF(s.status = 'ama_de_casa', 1, 0)) AS count_ama_de_casa,
SUM(IF(s.status = 'cliente_externo', 1, 0)) AS count_cliente_externo
FROM cursos_modulos AS c
LEFT JOIN subscriptions AS s ON s.curso_id = c.id
LEFT JOIN users AS u ON u.userID = s.user_id 
GROUP BY c.id