Question

我有一个查询执行我需要的大部分内容，但不能为空结果提供表示逻辑。

考虑这个问题：

SELECT 
SUM(order_detail.price/order_detail.qty*order_detail.qty_dispatched) AS orderedPrice, 
orders.location, 
orders.id,
suppliers.supplier_name,
DAYNAME(FROM_UNIXTIME(orders.datetime)) AS Day_Name,
from_unixtime(orders.datetime) AS Day_Date
FROM orders, order_detail, products,suppliers 
WHERE (orders.datetime BETWEEN '$from_date' AND '$to_date'  
AND order_detail.order_id = orders.id  
AND orders.location LIKE '%$locations%'
AND  order_detail.product_id=products.id 
AND products.supplier_id IN ($suppliers)
AND suppliers.id IN ($suppliers)
AND products.supplier_id=suppliers.id
AND (orders.status <> 'deleted') )

如果我得到每周天的结果，我可以使用PHP发布处理结果并显示结果：

Location | Supplier Name | Mon | Tue | Wed | Thu | Fri | Sat | Sun | Total
--------------------------------------------------------------------------
Alhambra | Widget Co     | $50 | $50 | $50 | $50 | $50 | $50 | $50 | $350

但是，如果在任何一天没有订单，我显然没有结果，我无法找到一个方法来发布流程以在列中显示$ 0。

我想也许平日表可能是JOIN的答案，但我无法理解。像这样的东西？

CREATE TABLE week_days(
  week_day_num INT(11) DEFAULT NULL
);

INSERT INTO week_days(week_day_num) VALUES (1),(2),(3),(4),(5),(6),(7);

SELECT
DAYNAME(FROM_UNIXTIME(orders.datetime)) AS day_name,
SUM(order_detail.price/order_detail.qty*order_detail.qty_dispatched) AS orderedPrice,
COALESCE(SUM(order_detail.price/order_detail.qty*order_detail.qty_dispatched) AS orderedPrice , 0)
FROM week_days wd
LEFT JOIN (

SELECT 
orders.location, 
orders.id,
suppliers.supplier_name,
from_unixtime(orders.datetime) AS Day_Date
FROM orders, order_detail, products,suppliers 
WHERE (orders.datetime BETWEEN '$from_date' AND '$to_date'  
AND order_detail.order_id = orders.id  
AND orders.location LIKE '%$locations%'
AND  order_detail.product_id=products.id 
AND products.supplier_id IN ($suppliers)
AND suppliers.id IN ($suppliers)
AND products.supplier_id=suppliers.id
AND (orders.status <> 'deleted') )


) order_results
ON wd.week_day_num = DAYOFWEEK(FROM_UNIXTIME(orders.datetime))
GROUP BY
      products.supplier_id,orders.location,DAYOFWEEK(FROM_UNIXTIME(orders.datetime))

ORDER BY location, products.supplier_id,Day_Date ASC;

Answer 1

这里的语法正在变弱：

COALESCE(SUM(order_detail.price/order_detail.qty*order_detail.qty_dispatched) AS orderedPrice , 0)

AS子句需要超出COALESCE函数。

COALESCE(SUM(order_detail.price/order_detail.qty*order_detail.qty_dispatched), 0) AS orderedPrice

使用MySQL DAYNAME作为列标题，即使结果为NULL

1 个答案: