MySql當使用函數WEEK（）與模式時返回錯誤

Question

我有一個問題，我不能使功能WEEK（日期，模式）工作。

我有一個SQL我在哪裏分組通過一年的款項，然後以周：

SELECT SUM(o.total) as total_org, SUM(o.total_eur) as total_eur, 
WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) as period, YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) as smallerperiod, `s`.`domain` 
FROM `orders` `o` 
JOIN `shops` `s` ON `o`.`domain_id` = `s`.`id` 
WHERE `o`.`domain_id` = '5' 
GROUP BY YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)), WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) 
ORDER BY `smallerperiod` ASC, `period` ASC 

正是作品我多麼希望除了週數從0開始，我發現了該模式，將工作最適合我的是3號，所以我試圖改變我的SQL語句：

SELECT SUM(o.total) as total_org, SUM(o.total_eur) as total_eur, 
WEEK('DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)', 3) as period, YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) as smallerperiod, `s`.`domain` 
FROM `orders` `o` 
JOIN `shops` `s` ON `o`.`domain_id` = `s`.`id` 
WHERE `o`.`domain_id` = '5' 
GROUP BY YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)), WEEK('DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)', 3)) 
ORDER BY `smallerperiod` ASC, `period` ASC 

然而，它只是給了我的錯誤

你的SQL語法有錯誤;檢查對應於您MariaDB的服務器版本正確的語法使用 接近「）ORDER BY smallerperiod ASC手冊，period ASC

編輯： 我試圖使它沒有單引號工作過：

SELECT SUM(o.total) as total_org, SUM(o.total_eur) as total_eur, 
WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR), 3) as period, YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) as smallerperiod, `s`.`domain` 
FROM `orders` `o` 
JOIN `shops` `s` ON `o`.`domain_id` = `s`.`id` 
WHERE `o`.`domain_id` = '5' 
GROUP BY YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)), WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR), 3)) 
ORDER BY `smallerperiod` ASC, `period` ASC 

但是，它返回我同樣的錯誤

Answer 1

，因爲你要調用一個funtion，而不是把它作爲一個字符串，你必須刪除單引號角落找尋'DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)'。

你得到theeroor，監守你有一個額外的支架位置：HOUR)', 3))

所以查詢必須是：

SELECT SUM(o.total) as total_org, SUM(o.total_eur) as total_eur, 
WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR), 3) as period, YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) as smallerperiod, `s`.`domain` 
FROM `orders` `o` 
JOIN `shops` `s` ON `o`.`domain_id` = `s`.`id` 
WHERE `o`.`domain_id` = '5' 
GROUP BY YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)), WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR), 3) 
ORDER BY `smallerperiod` ASC, `period` ASC 

Answer 2

你只需要後​​。否則寄託都完美我刪除，去年支架曾嘗試過。 你也可以嘗試下面的代碼。

SELECT SUM(o.total) as total_org, SUM(o.total_eur) as total_eur, 
    WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR), 3) as period, YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)) as smallerperiod, `s`.`domain` 
    FROM `orders` `o` 
    JOIN `shops` `s` ON `o`.`domain_id` = `s`.`id` 
    WHERE `o`.`domain_id` = '5' 
    GROUP BY 
    YEAR(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR)), 
    WEEK(DATE_ADD(o.date_created_utc, INTERVAL 0 HOUR), 3) 
    ORDER BY `smallerperiod` ASC, `period` ASC