Question

当我尝试在单个查询中使用5个表从数据库获取数据时，我正在使用codeigniter项目。

在phpmyadmin中，sql查询工作正常但是codeigniter查询无效。

我的Phpmyadmin查询：

select sma_sales.date, sma_sales.reference_no, biller, sma_companies.name as DeliveryRep,customer,c.code, 
GROUP_CONCAT(CONCAT(sma_sale_items.product_name, ' (', sma_sale_items.quantity, ')') SEPARATOR '\n') as iname, 
grand_total, paid, sma_payments.amount as pyname, 
payment_status
from sma_sales
left outer join sma_sale_items on sma_sale_items.sale_id=sma_sales.id
left outer join sma_companies on sma_companies.id=sma_sales.delivered_id
left outer join sma_companies as c on c.id=sma_sales.customer_id
left outer join sma_warehouses on sma_warehouses.id=sma_sales.warehouse_id
left outer join sma_payments on sma_payments.sale_id=sma_sales.id
group by sma_sales.id

我的Codeigniter查询：

->select("date, reference_no, biller, companies.name as DeliveryRep,customer,c.code, "
                        . "GROUP_CONCAT(CONCAT(" . $this->db->dbprefix('sale_items') . ".product_name,"
                        . " ' (', " . $this->db->dbprefix('sale_items') . ".quantity, ')') SEPARATOR '\n') as iname,"
                        . " grand_total, paid,payments.amount as pyname, payment_status", FALSE)
                ->from('sales')
                ->join('sale_items', 'sale_items.sale_id=sales.id', 'left')
                ->join('companies', 'companies.id=sales.delivered_id', 'left')
                ->join('companies as c', 'c.id=sales.customer_id', 'left')
                ->join('warehouses', 'warehouses.id=sales.warehouse_id', 'left')
                ->join('payments', 'payments.sale_id=sales.id', 'left')
                ->group_by('sales.id')
$q = $this->db->get();

sma只是数据库前缀

Answer 1

您需要在表名中添加 sma ，它是表名的一部分。

<强>查询：

$this->db->select("date, reference_no, biller, sma_companies.name as DeliveryRep,customer,c.code, "
                            . "GROUP_CONCAT(CONCAT(" . $this->db->dbprefix('sale_items') . ".product_name,"
                            . " ' (', " . $this->db->dbprefix('sale_items') . ".quantity, ')') SEPARATOR '\n') as iname,"
                            . " grand_total, paid,sma_payments.amount as pyname, payment_status", FALSE)
                    ->from('sma_sales')
                    ->join('sma_sale_items', 'sma_sale_items.sale_id=sma_sales.id', 'left')
                    ->join('sma_companies', 'sma_companies.id=sma_sales.delivered_id', 'left')
                    ->join('sma_companies as c', 'c.id=sma_sales.customer_id', 'left')
                    ->join('sma_warehouses', 'sma_warehouses.id=sma_sales.warehouse_id', 'left')
                    ->join('sma_payments', 'sma_payments.sale_id=sales.id', 'left')
                    ->group_by('sales.id')
    $q = $this->db->get();

您可以通过搜索调试查询并将其设置为true，

$ db ['default'] ['db_debug'] = true;

你的config / database.php中的

结论：问题是表名的别名，修复别名后一切正常。

Answer 2

您正在使用5个表。可能在多个表中可以存在相同的列名。您可以为所有表使用别名并更改查询，如：

$this->db->select("s.date , s.reference_no, s.biller, companies.name as DeliveryRep,s.customer,c.code, "
                       . "GROUP_CONCAT(CONCAT(" . $this->db->dbprefix('sale_items') . ".product_name, "
                       . "' (', " . $this->db->dbprefix('sale_items') . ".quantity, ')') SEPARATOR '\n') as iname, "
                       . "s.grand_total, s.paid,pay.amount as pyname, s.payment_status", FALSE)
               ->from('sales as s')
               ->join('sale_items', 'sale_items.sale_id=s.id', 'left')
               ->join('companies', 'companies.id=s.delivered_id', 'left')
               ->join('companies as c', 'c.id=s.customer_id', 'left')
               ->join('warehouses', 'warehouses.id=s.warehouse_id', 'left')
               ->join('payments as pay', 'pay.sale_id=s.id', 'left')
               ->group_by('s.id')
               ->order_by('s.date desc');

MySql查询在PHPmyadmin中工作，但不能在codeigniter中工作

2 个答案: