Выбор суммы для каждого дня за месяц

Question

Я пытаюсь рассчитать сумму для количества заказов за каждый день.

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)) 

Работает отлично, за исключением случаев, когда сумма = 0, она не отображается в массиве. Мне пришлось вручную объединить вывод, чтобы ограничить его 31 результатом за месяц.

Я видел использование date_format и используя атрибут "%m", но не повезло с этим выяснить.

Edit:

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) 
) 

выход после того, как я уже нарезанного его в PHP.

Answer 1

Редактировать: ok попробуйте эту концепцию, бесстыдно взято из here. Подстройте свои имена и предложения соответственно, но вы получите концепцию.

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 ; 

календарных дней запихать в течение многих лет:

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 

продаж для Aug2009:

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