SUM(subquery.total_number_before) AS total_number

如果你想要将子查询中的total_number_before列的值进行合计，可以在主查询中使用SUM函数，并将子查询结果作为表进行处理。

以下是修改后的示例查询语句：

SELECT
    subquery.week,
    SUM(subquery.total_number_before) AS total_number
FROM (
    SELECT
        CONCAT(YEAR(e.tb_time), '年第', WEEK(e.tb_time, 1), '周') AS week,
        SUM(e.number) AS total_number_before
    FROM 
        fa_tianchuang e
    GROUP BY
        YEAR(e.tb_time),
        WEEK(e.tb_time, 1)
) subquery
GROUP BY
    subquery.week

在这个查询中，我们将原始的子查询作为表subquery进行处理。然后，使用SUM函数对subquery中的total_number_before列进行合计，并将结果命名为total_number。

最后，使用GROUP BY子句对week列进行分组，以便按周数合并工程量的总和。

请根据你的实际数据表结构和需求进行适当的修改。

相关问题

请帮我修改下列sql语句的语法错误： select item_id,happen_time,subquery.recent_time from (select sd.item_id, sd.happen_time,red_flag,ROW_NUMBER() over (partition by item_id order by happen_time DESC) as recent_time from sheet_buy_detail sd left join sheet_buy_main sm on sd.sheet_id=sm.sheet_id and sd.happen_time=sm.happen_time where sm.red_flag<>2 ) as subquery where recent_time <= 3 and item_id=54209

以下是修改后的SQL语句，修复了语法错误：

SELECT item_id, happen_time, recent_time
FROM (
  SELECT sd.item_id, sd.happen_time, sd.red_flag, 
    ROW_NUMBER() OVER (PARTITION BY sd.item_id ORDER BY sd.happen_time DESC) AS recent_time
  FROM sheet_buy_detail sd
  LEFT JOIN sheet_buy_main sm ON sd.sheet_id = sm.sheet_id AND sd.happen_time = sm.happen_time
  WHERE sm.red_flag <> 2
) AS subquery
WHERE recent_time <= 3 AND item_id = 54209;

在这个修改后的示例中，我们修复了以下两个问题：

1. 在子查询中，为了准确引用字段，我们在 `SELECT` 子句中添加了 `sd.` 前缀。
2. 在主查询中，我们使用正确的别名 `recent_time` 来筛选最近的三个记录。

请确保将 `sheet_buy_detail` 和 `sheet_buy_main` 替换为实际的表名，并根据需要修改其他字段和条件。

select a.id,SUM(CASE WHEN d.shoukuan_month = '2023-05' and d.is_del = 0 then d.netincome ELSE 0 end) as one_shoukuan,SUM(CASE WHEN d.shoukuan_month = '2023-04' AND d.is_del = 0 THEN d.netincome ELSE 0 END) as two_shoukuan,(CASE WHEN b.month = '2023-05' THEN b.create_income ELSE 30000 END) as one_income,(CASE WHEN b.month = '2023-04' THEN b.create_income ELSE 30000 END) as two_income from a left join b on b.kefuid = a.id left join d on d.kefuid = a.id where a.brand = '1' and a.is_del = 0 and a.is_lizhi = 0 group by a.id这个sql如何让select 中的d.one_shoukuan >= b.one_income and d.two_shoukuan >= b.two_income

你可以将该 SQL 作为一个子查询，在外部 SELECT 语句中添加一个 WHERE 子句来实现该要求：

SELECT *
FROM (
    SELECT a.id,
        SUM(CASE WHEN d.shoukuan_month = '2023-05' AND d.is_del = 0 THEN d.netincome ELSE 0 END) AS one_shoukuan,
        SUM(CASE WHEN d.shoukuan_month = '2023-04' AND d.is_del = 0 THEN d.netincome ELSE 0 END) AS two_shoukuan,
        (CASE WHEN b.month = '2023-05' THEN b.create_income ELSE 30000 END) AS one_income,
        (CASE WHEN b.month = '2023-04' THEN b.create_income ELSE 30000 END) AS two_income
    FROM a
    LEFT JOIN b ON b.kefuid = a.id
    LEFT JOIN d ON d.kefuid = a.id
    WHERE a.brand = '1' AND a.is_del = 0 AND a.is_lizhi = 0
    GROUP BY a.id
) AS subquery
WHERE subquery.one_shoukuan >= subquery.one_income AND subquery.two_shoukuan >= subquery.two_income;

这样，子查询将返回每个客服的 `one_shoukuan`，`two_shoukuan`，`one_income` 和 `two_income` 值，而外部查询将筛选出满足 `one_shoukuan >= one_income` 和 `two_shoukuan >= two_income` 的记录。