Question

我想使用select mysql显示data_order是时间戳

首先只包括最后7行（包含列price），仅包括今天的每一天，包括今天的每一天。假设我每天（过去7天）单个订单很简单，但是当我在同一天有更多订单时我不知道如何制作mysql语句

+-------------------------------------------------------------+
| table name  `comanda`                                       |
+-------------------------------------------------------------+
| id_comanda | keyUnique_comanda | pret_comanda| data_comanda |
+------------+-------------------+-------------+--------------+
| 1          | a1234             | 150         | today        |
+------------+-------------------+-------------+--------------+
| 2          | b4321             | 300         | yesturday    |
+------------+-------------------+-------------+--------------+
| 3          | b4321             | 200         | yesturday    |
+------------+-------------------+-------------+--------------+
| 4          | c4321             | 100         | yesturday    |
+------------+-------------------+-------------+--------------+
| 5          | d234              | 50          | 3 day ago    |
+------------+-------------------+-------------+--------------+
| 6          | e453              | 200         | 9 day ago    |
+------------+-------------------+-------------+--------------+

我尝试了一些东西......但是没有给我看好

SELECT DAY(data_comanda) AS period, SUM(pret_comanda) AS total 
FROM comanda 
WHERE data_comanda >= CURDATE() + INTERVAL 1  WEEK 
GROUP BY period



-- --------------------------------------------------------

--
-- Table structure for table `comanda`
--

CREATE TABLE `comanda` (
  `id_comanda` int(11) NOT NULL,
  `pret_comanda` varchar(255) NOT NULL,
  `pret_comanda_total` int(11) NOT NULL,
  `comanda` text NOT NULL,
  `nume_comanda` varchar(255) NOT NULL,
  `prenume_comanda` varchar(255) NOT NULL,
  `email_comanda` varchar(255) NOT NULL,
  `telefon_comanda` varchar(19) NOT NULL,
  `judet_comanda` varchar(255) NOT NULL,
  `adresa_comanda` varchar(2555) NOT NULL,
  `clientIP` varchar(255) NOT NULL,
  `localitate_comanda` varchar(255) NOT NULL,
  `detalii_comanda` varchar(3000) CHARACTER SET utf8 NOT NULL,
  `data_comanda` timestamp NOT NULL,
  `data_comanda_finalizare` datetime DEFAULT NULL,
  `produse_diferite_comanda` varchar(10) NOT NULL,
  `keyUnique_comanda` varchar(25) NOT NULL,
  `status_comanda` varchar(255) NOT NULL DEFAULT 'asteptare',
  `admin_validare_comanda` varchar(255) NOT NULL DEFAULT 'neModificat'
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

Answer 1

而不是

WHERE data_comanda >= CURDATE() + INTERVAL 1  WEEK

你应该使用

WHERE data_comanda >= CURDATE() - INTERVAL 1  WEEK

但这包括今天和过去7天（总共8天）。如果您只需要7天，则应使用

WHERE data_comanda >= CURDATE() - INTERVAL 6 DAY

查询现在将返回

+--------+-------+
| period | total |
+--------+-------+
|      1 |   600 |
|      2 |   150 |
|     27 |    50 |
+--------+-------+

http://rextester.com/QSDUY96504

这可能会造成混淆，因为DAY(data_comanda)会返回该月的某一天。因此，您可能希望使用DATE代替DAY，并获得如下结果：

+------------+-------+
|   period   | total |
+------------+-------+
| 27.02.2017 |    50 |
| 01.03.2017 |   600 |
| 02.03.2017 |   150 |
+------------+-------+

http://rextester.com/XOECI72414

要包含缺失的日期，您可以使用0到6之间的数字创建内联表

SELECT *
FROM (
    SELECT 0 as diff UNION ALL
    SELECT 1 UNION ALL
    SELECT 2 UNION ALL
    SELECT 3 UNION ALL
    SELECT 4 UNION ALL
    SELECT 5 UNION ALL
    SELECT 6
) diffs

现在LEFT JOIN你的桌子

SELECT CURDATE() - INTERVAL diffs.diff DAY AS period,
       COALESCE(SUM(pret_comanda), 0) AS total
FROM (
    SELECT 0 as diff UNION ALL
    SELECT 1 UNION ALL
    SELECT 2 UNION ALL
    SELECT 3 UNION ALL
    SELECT 4 UNION ALL
    SELECT 5 UNION ALL
    SELECT 6
) diffs
LEFT JOIN comanda c ON DATE(c.data_comanda) = CURDATE() - INTERVAL diffs.diff DAY
GROUP BY period;

请注意，SUM(pret_comanda)将在缺失的日期返回NULL。要将其转换为0，您可以使用COALESCE(SUM(pret_comanda), 0)

现在结果看起来像

+------------+-------+
|   period   | total |
+------------+-------+
| 24.02.2017 |     0 |
| 25.02.2017 |     0 |
| 26.02.2017 |     0 |
| 27.02.2017 |    50 |
| 28.02.2017 |     0 |
| 01.03.2017 |   600 |
| 02.03.2017 |   150 |
+------------+-------+

http://rextester.com/WJD33053

mysql选择最近7天的查询，如果是更多查询在一天的总和，并只显示总和

1 个答案: