Question

我提出了查询：

SELECT ad.spec_id, ad.height_id, ad.size_id, (SUM(ad.qty)+SUM(md.qty)) as qty, SUM(ad.qty) as arrival_qty, SUM(md.qty) as movement_qty, a.warehouse_id 
FROM arrival_data ad, movement_data md, arrivals a, movements m 
WHERE ad.size_id= md.size_id 
    AND ad.height_id = md.height_id 
    AND ad.spec_id = md.spec_id
    AND ad.arrival_id = a.arrival_id
    AND m.receiver = a.warehouse_id 
    AND md.movement_id = m.movement_id
GROUP BY ad.spec_id, ad.size_id, ad.height_id, m.receiver

但当然只有第一个和第二个表具有相同的spec_id，size_id，height_id时才会显示结果。我已经读过加入可以解决这个问题。请帮助解决正确的问题

我通过漫长而缓慢的方法解决了我的任务

 $arrival_qty = $this->db->getOne('
            SELECT SUM(ad.qty) 
            FROM arrival_data ad, arrivals a 
            WHERE ad.arrival_id = a.arrival_id AND ad.spec_id =?i AND ad.size_id =?i AND ad.height_id =?i AND a.warehouse_id = ?i', 
            $product['spec_id'], $product['size_id'], $product['height_id'], $product['warehouse_id']);

        $move_in_qty = $this->db->getOne('
            SELECT SUM(md.qty) 
            FROM movement_data md, movements m
            WHERE md.movement_id = m.movement_id AND md.spec_id =?i AND md.size_id =?i AND md.height_id =?i AND m.receiver =?i', 
            $product['spec_id'], $product['size_id'], $product['height_id'], $product['warehouse_id']);
        $move_out_qty = $this->db->getOne('
            SELECT SUM(md.qty) 
            FROM movement_data md, movements m
            WHERE md.movement_id = m.movement_id AND md.spec_id =?i AND md.size_id =?i AND md.height_id =?i AND m.sender =?i', 
            $product['spec_id'], $product['size_id'], $product['height_id'], $product['warehouse_id']);
        $invoices_qty = $this->db->getOne('
            SELECT SUM(i.qty) 
            FROM invoice_data i, invoices m
            WHERE i.invoice_id = m.invoice_id AND i.spec_id =?i AND i.size_id =?i AND i.height_id =?i AND m.warehouse_id =?i', 
            $product['spec_id'], $product['size_id'], $product['height_id'], $product['warehouse_id']);
        $arrival_qty = (!empty($arrival_qty)) ? $arrival_qty : '0';
        $products[$key]['qty'] = $arrival_qty + $move_in_qty - $move_out_qty - $invoices_qty;

也许有人可以通过一个查询给我变体

Answer 1

Joins Basics

我真的不知道，你想要做什么。但我可以找到小错误，你已经在SELECT列表中添加了一个带有聚合函数的a.warehouse_id但是没有在GROUP BY子句中添加

所以，你可能会收到这样的错误..

列'a.warehouse_id'在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中。

要解决此问题，请在GROUP BY子句中添加a.warehouse_id或将其从SELECT列表中删除

Answer 2

即使列名不同，这种语法也能正常工作：从中选择sum（x）（从arrival_data中选择sum（qty）为x 联合所有从和move_data中选择sum（qty）为x ）作为x;

来自两个表mysql的两个col的总和

2 个答案: