Question

我有一个难以解释的不同问题这是我现在得到的结果：

╔═════════╦════════╗
║    Id   ║ Saldo  ║
╠═════════╬════════╣
║    1    ║10441,91║
╠═════════╬════════╣
║    2    ║ 441,91 ║
╚═════════╩════════╝

这就是我想要的结果（彼此相邻）：

╔═════════╦════════╗════════╗════════╗
║    Id   ║ Saldo  ║   Id   ║ Saldo  ║
╠═════════╬════════╣════════║════════║
║   1     ║10441,91║   2    ║ 441,91 ║
╚═════════╩════════╝════════╝════════╝

所以我想要的就是彼此相邻。

这是我使用的查询：

SELECT 
    o.id_shop AS id,
    SUM(CASE
        WHEN (s.amount + s.shipping_cost_amount) != '' THEN o.total_paid_tax_excl - (s.amount + s.shipping_cost_amount)
        ELSE o.total_paid_tax_excl
    END) AS Shop
FROM
    expoled.ps_orders o
        LEFT JOIN
    expoled.ps_oxoquotation_quotation q ON q.idOrder = o.id_order
        LEFT JOIN
    expoled.ps_order_slip s ON s.id_order = o.id_order
WHERE
    o.current_state IN (4 , 5, 20, 23)
        AND YEAR(q.date_add) = YEAR(UTC_TIMESTAMP())
GROUP BY o.id_shop

我尝试了什么：我已经尝试过UNION来组合结果集但是没有用。我可以把它们全部放在彼此之下但不能彼此相邻。

Answer 1

你总是只有两个可能的ID，1和2吗？如果是这样，您可以删除GROUP BY o.id_shop并以这种方式创建两个sum列：

SUM(CASE
    WHEN (s.amount + s.shipping_cost_amount) != '' AND o.id_shop = 1 THEN o.total_paid_tax_excl - (s.amount + s.shipping_cost_amount)
    WHEN (s.amount + s.shipping_cost_amount) = '' AND o.id_shop = 1 THEN o.total_paid_tax_excl
    ELSE 0
END) AS Shop_ID_1


SUM(CASE
    WHEN (s.amount + s.shipping_cost_amount) != '' AND o.id_shop = 2 THEN o.total_paid_tax_excl - (s.amount + s.shipping_cost_amount)
    WHEN (s.amount + s.shipping_cost_amount) = '' AND o.id_shop = 2 THEN o.total_paid_tax_excl
    ELSE 0
END) AS Shop_ID_2

SQL结果彼此相邻

1 个答案: