Question

我试图绘制每天订单金额的总和。

SELECT DATE(FROM_UNIXTIME(timestamp)) AS timey, 
SUM(amount) AS cash 
FROM orders 
WHERE touid = :uid AND status = 1 
GROUP BY DATE(FROM_UNIXTIME(timestamp))

正常工作，除了sum = 0之外，它不会在数组中显示。我不得不手动拼接输出，将当月的结果限制为31。

我已经看过使用date_format并使用"%m"属性，但是没有任何运气可以搞清楚。

编辑：

Array ( 
    [0] => Array ( [timey] => 2015-03-22 [cash] => 0.03 ) 
    [1] => Array ( [timey] => 2015-03-23 [cash] => 0.2 ) 
    [2] => Array ( [timey] => 2015-03-29 [cash] => 0.08 ) 
    [3] => Array ( [timey] => 2015-04-03 [cash] => 0.03 ) 
    [4] => Array ( [timey] => 2015-04-04 [cash] => 0.99 ) 
    [5] => Array ( [timey] => 2015-04-06 [cash] => 1.55 ) 
    [6] => Array ( [timey] => 2015-04-07 [cash] => 0.03 ) 
    [7] => Array ( [timey] => 2015-04-10 [cash] => 4 ) 
    [8] => Array ( [timey] => 2015-04-13 [cash] => 5 ) 
    [9] => Array ( [timey] => 2015-04-14 [cash] => 8 ) 
    [10] => Array ( [timey] => 2015-04-17 [cash] => 1 ) 
    [11] => Array ( [timey] => 2015-04-18 [cash] => 4 ) 
    [12] => Array ( [timey] => 2015-05-01 [cash] => 5 ) 
    [13] => Array ( [timey] => 2015-05-02 [cash] => 2 ) 
    [14] => Array ( [timey] => 2015-05-03 [cash] => 1 ) 
    [15] => Array ( [timey] => 2015-05-05 [cash] => 3 ) 
    [16] => Array ( [timey] => 2015-05-06 [cash] => 1 ) 
    [17] => Array ( [timey] => 2015-05-07 [cash] => 12 ) 
    [18] => Array ( [timey] => 2015-05-08 [cash] => 1 ) 
    [19] => Array ( [timey] => 2015-05-11 [cash] => 0.1 ) 
    [20] => Array ( [timey] => 2015-05-13 [cash] => 1 ) 
    [21] => Array ( [timey] => 2015-05-14 [cash] => 1.9 ) 
    [22] => Array ( [timey] => 2015-05-16 [cash] => 2 ) 
    [23] => Array ( [timey] => 2015-05-17 [cash] => 0.15 ) 
    [24] => Array ( [timey] => 2015-05-18 [cash] => 1 ) 
    [25] => Array ( [timey] => 2015-05-19 [cash] => 1.09 ) 
    [26] => Array ( [timey] => 2015-05-23 [cash] => 0.5 ) 
    [27] => Array ( [timey] => 2015-05-24 [cash] => 2.5 ) 
    [28] => Array ( [timey] => 2015-05-26 [cash] => 3 ) 
    [29] => Array ( [timey] => 2015-05-27 [cash] => 2 ) 
    [30] => Array ( [timey] => 2015-05-28 [cash] => 4 ) 
)

我用PHP切片后的输出。

Answer 1

修改好吧，尝试这个概念，无耻地从here获取。相应地调整你的名字和条款，但你得到了这个概念。

CREATE TABLE orders ( id INT PRIMARY KEY AUTO_INCREMENT, order_date DATETIME, product_id INT, quantity INT, customer_id INT ); INSERT INTO orders (order_date, product_id, quantity, customer_id) VALUES ('2009-08-15 12:20:20', '1', '2', '123'), ('2009-08-15 12:20:20', '2', '2', '123'), ('2009-08-17 16:43:09', '1', '1', '456'), ('2009-08-18 09:21:43', '1', '5', '789'), ('2009-08-18 14:23:11', '3', '7', '123'), ('2009-08-21 08:34:21', '1', '1', '456');

CREATE TABLE aCalendar (datefield DATE);

DELIMITER | CREATE PROCEDURE fill_calendar(start_date DATE, end_date DATE) BEGIN DECLARE crt_date DATE; SET crt_date=start_date; WHILE crt_date < end_date DO INSERT INTO aCalendar VALUES(crt_date); SET crt_date = ADDDATE(crt_date, INTERVAL 1 DAY); END WHILE; END | DELIMITER ;

多年来一直禁止日历：

CALL fill_calendar('2007-01-01', '2009-12-31'); CALL fill_calendar('2010-01-01', '2017-12-31'); -- coffee break, 1 minute -- select count(*) from aCalendar; -- an index on aCalendar wouldn't be a bad idea

2009年8月的销售额：

SELECT aCalendar.datefield AS theDate, IFNULL(SUM(orders.quantity),0) AS total_sales FROM orders RIGHT JOIN aCalendar ON (DATE(orders.order_date) = aCalendar.datefield) WHERE (aCalendar.datefield BETWEEN DATE('2009-08-01') AND DATE('2009-08-31')) GROUP BY theDate ORDER BY theDate theDate total_sales 2009-08-01 0 2009-08-02 0 2009-08-03 0 2009-08-04 0 2009-08-05 0 2009-08-06 0 2009-08-07 0 2009-08-08 0 2009-08-09 0 2009-08-10 0 2009-08-11 0 2009-08-12 0 2009-08-13 0 2009-08-14 0 2009-08-15 4 2009-08-16 0 2009-08-17 1 2009-08-18 12 2009-08-19 0 2009-08-20 0 2009-08-21 1 2009-08-22 0 2009-08-23 0 2009-08-24 0 2009-08-25 0 2009-08-26 0 2009-08-27 0 2009-08-28 0 2009-08-29 0 2009-08-30 0 2009-08-31 0

选择一个月的每一天的总和

1 个答案: