mysql - 在 debian 上执行 mysql 查询时出错

Question

我有一个在本地运行良好的 mysql 查询。我的计算机在 Windows 10 上，我正在使用此版本的 mysql:mysql Ver 8.0.11 for Win64 on x86_64 (MySQL Community Server - GPL)

这是查询

SELECT conso_ht.Date_conso AS myDate, sum(conso_ht.Conso_Tranche_Jour) AS jour, sum(conso_ht.Conso_Tranche_Pointe) AS pointe, sum(conso_ht.Conso_Tranche_Nuit) AS nuit
FROM conso_ht
WHERE conso_ht.Compteur = "37178777837" 
GROUP BY year(conso_ht.Date_conso) ASC, month(conso_ht.Date_conso) ASC;

我服务器的 mysql 版本是 mysql Ver 8.0.15 for Linux on x86_64 (MySQL Community Server - GPL)

这是我遇到的错误

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ASC, month(conso_ht.Date_conso) ASC' at line 1`

我该如何解决？？

Answer 1

如果您需要为此添加订单，则分组依据不需要 ASC 或 DESC

您也不能使用不同级别的聚合来尊重选择和分组依据例如选择一个日期并按年月分组

SELECT year(conso_ht.Date_conso)   , month(conso_ht.Date_conso)  
, sum(conso_ht.Conso_Tranche_Jour) AS jour
, sum(conso_ht.Conso_Tranche_Pointe) AS pointe
, sum(conso_ht.Conso_Tranche_Nuit) AS nuit
FROM conso_ht
WHERE conso_ht.Compteur = "37178777837" 
GROUP BY year(conso_ht.Date_conso)   , month(conso_ht.Date_conso) 
ORDER BY  year(conso_ht.Date_conso)  ASC , month(conso_ht.Date_conso)  ASC

或

SELECT conso_ht.Date_conso  AS myDate
, sum(conso_ht.Conso_Tranche_Jour) AS jour
, sum(conso_ht.Conso_Tranche_Pointe) AS pointe
, sum(conso_ht.Conso_Tranche_Nuit) AS nuit
FROM conso_ht
WHERE conso_ht.Compteur = "37178777837" 
GROUP BY conso_ht.Date_conso
ORDER BY  year(conso_ht.Date_conso)  ASC , month(conso_ht.Date_conso)  ASC