Question

您好我的SQL查询中出现错误，无法弄清楚出错了什么。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);

我得到的错误消息如下：

致命错误：未捕获的异常＆＃39; PDOException＆＃39;与消息＆＃39; SQLSTATE [42000]：语法错误或访问冲突：1064您有 SQL语法错误;查看与您的手册相对应的手册 MySQL服务器版本，用于在＆＃39; WHERE 1 AND附近使用正确的语法 c.id =＆＃39; 1＆＃39;限制0,30和＃39;在第6行＆＃39;在 E：\ xampp \ htdocs \ admin \ class \ pagination.php：376堆栈跟踪：＃0 E：\ XAMPP \ htdocs中\ ADMIN \类\ pagination.php（376）： PDOStatement-＆gt; execute（）＃1 E：\ XAMPP \ htdocs中\ ADMIN \类\ pagination.php（202）：分页 - ＆gt; excecute_query（）＃2 E：\ XAMPP \ htdocs中\ ADMIN \类\ pagination.php（162）： pagination-＆gt; run（1，＆＃39; SELECT c。，cou ...＆＃39;，Array）＃3 E：\ XAMPP \ htdocs中\ ADMIN \ search.php中（146）： pagination-＆gt; __ construct（1，＆＃39; SELECT c。，cou ...＆＃39;，Array）＃4 {main} 在E：\ xampp \ htdocs \ admin \ class \ pagination.php中抛出 376

Answer 1

where 1为你做了什么？试着杀了那个。

以下内容不会引发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

Answer 2

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";

SQLSTATE [42000]聚合函数出错

2 个答案: