Question

我正在尝试处理以下查询：

    SELECT TN.name_type_ne, WEEK(HC.date_check,1) AS date_week, YEAR(HC.date_check) AS date_year,
    SUM(IF(id_check > total_check_week, total_check_week, COUNT(id_check))) AS unique_check
    FROM HEALTH_CHECK HC
    LEFT JOIN TYPE_NE TN ON (HC.fk_id_type_ne = TN.id_type_ne AND HC.fk_id_ne IS NULL)
    LEFT JOIN CATEGORY_NE CN ON (TN.fk_id_category_ne = CN.id_category_ne)
    WHERE CN.name_category_ne = "R4"
    GROUP BY TN.name_type_ne, date_week, date_year

似乎它会为下一个选择引发错误：

SUM(IF(id_check > total_check_week, total_check_week, COUNT(id_check))) AS unique_check

我也尝试将其替换为：

SUM(CASE WHEN id_check > total_check_week THEN total_check_week ELSE COUNT(id_check) END) AS unique_check

但我一直有这个Mysql错误：

#1111 - Invalid use of group function

查询有什么问题??

Answer 1

我终于找到了解决方案。就我而言，我试图在SUM中使用聚合函数（COUNT）。似乎MySQL不允许这种做法。

以下是派生表的解决方案：

 `SELECT SUM(unique_check) FROM
    (SELECT TN.name_type_ne, WEEK( HC.date_check, 1 ) AS date_week, YEAR( HC.date_check ) AS date_year,
    CASE WHEN COUNT(id_check) > total_check_week THEN total_check_week ELSE COUNT(id_check) END unique_check
    FROM HEALTH_CHECK HC
    LEFT JOIN TYPE_NE TN ON (HC.fk_id_type_ne = TN.id_type_ne AND HC.fk_id_ne IS NULL)
    LEFT JOIN CATEGORY_NE CN ON (TN.fk_id_category_ne = CN.id_category_ne)
    WHERE CN.name_category_ne = "R4"
    GROUP BY TN.name_type_ne, date_week, date_year) As T`

使用SUM和CASE语句

1 个答案: