您好,我在 SQL 查询中遇到错误,但无法找出问题所在。这是迄今为止在 Barmar 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);
我收到的错误消息如下:
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE 1 AND c.id = '1' LIMIT 0, 30' at line 6' in E:\xampp\htdocs\admin\class\pagination.php:376 Stack trace: #0 E:\xampp\htdocs\admin\class\pagination.php(376): PDOStatement->execute() #1 E:\xampp\htdocs\admin\class\pagination.php(202): pagination->excecute_query() #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} thrown in E:\xampp\htdocs\admin\class\pagination.php on line 376
最佳答案
group by
子句应位于 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";
关于php - SQLSTATE[42000] 聚合函数错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32287304/