我正在尝试更新codeigniter中的多个连接表,但它给出以下错误:
错误号:1064您的SQL语法中有错误;查看与您的MySQL服务器版本对应的手册,以便在'.carro_id附近使用正确的语法INNER JOIN a ON ac.atributo_id = a.id INNER JOIN m ON c .modelo_'在第1行
UPDATE `carro c, atributo_carro ac, atributos a, modelos m, marcas ma
INNER JOIN ac ON c`.`id = ac`.`carro_id INNER JOIN a ON
ac`.`atributo_id = a`.`id INNER JOIN m ON c`.`modelo_id = m`.`id INNER
JOIN ma ON m`.`marca_id =` `ma`.`id` SET `marcas`.`marca` = 'Opel',
`modelos`.`modelo` = 'AMmmmmm', `carro`.`kilometragem` = '12000',
`carro`.`cor` = 'Azul', `carro`.`ano` = '1992', `carro`.`descricao` =
'dfghjk', `carro`.`valor` = '1288.00', `atributos`.`atributo` =
'rapido' WHERE `carro`.`id` = '10'
这是我的更新功能 .
public function update_carro_query($id, $data)
{
// print_r($data); die();
$query =
$this->db->set('marcas.marca', $data['marca']);
$this->db->set('modelos.modelo', $data['modelo']);
$this->db->set('carro.kilometragem', $data['kilometragem']);
$this->db->set('carro.cor', $data['cor']);
$this->db->set('carro.ano', $data['ano']);
$this->db->set('carro.descricao', $data['descricao']);
$this->db->set('carro.valor', $data['valor']);
$this->db->set('atributos.atributo', $data['atributo']);
$this->db->where('carro.id', $id);
$this->db->update('carro c, atributo_carro ac, atributos a, modelos m, marcas ma INNER JOIN ac ON c.id = ac.carro_id INNER JOIN a ON ac.atributo_id = a.id INNER JOIN m ON c.modelo_id = m.id INNER JOIN ma ON m.marca_id = ma.id');
return $query;
}
这是我的数据库表和关系
查询由函数生成 . 什么可能导致语法错误?我错过了什么吗?
1 回答
update() method只需要一个表名作为第一个参数,而不是包含连接的完整语句等 . 所以它会将第一个参数包装成反引号 . 这会导致您遇到sql语法错误 .
最好使用query()来生成(不是简单的)更新语句 .