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2023年月度支付次数及总额统计SQL查询需求

解决2023年各月支付统计问题

需要统计2023年每个月的支付次数和支付总额,对应输出月份名称、支付次数、格式化后的支付总额。以下是修正后的SQL方案:

SELECT
    TO_CHAR(created_at, 'FMMonth') AS Month,
    COUNT(*) AS "Payment Count",
    TO_CHAR(SUM(converted_total), 'FM999,999') AS "Payment Total"
FROM production.payments
WHERE
    created_at >= '2023-01-01'
    AND created_at < '2024-01-01'
    AND status != 'failed'
GROUP BY
    EXTRACT(YEAR FROM created_at),
    EXTRACT(MONTH FROM created_at),
    TO_CHAR(created_at, 'FMMonth')
ORDER BY
    EXTRACT(MONTH FROM created_at);

关键修改说明

  • 生效过滤条件:移除注释符号,补充完整2023年日期范围(避免误包含2024年数据),明确排除失败状态的支付记录
  • 替换聚合逻辑:用GROUP BY按月分组统计,替代原窗口函数的累计计算逻辑,直接输出每月的聚合结果
  • 格式化输出:
    • 用TO_CHAR(created_at, 'FMMonth')生成无多余空格的月份名称(如January而非January )
    • 用TO_CHAR将总额格式化为带千分位的字符串,匹配期望输出样式
  • 排序优化:按月份数字排序,保证结果按自然月份顺序展示

现有数据样例

Created_atCurrencyConverted_Subtsubtotalconverted_totalTotalYTD_Frame
2018-01-01USD20000200.00206.00206000
2018-01-01USD40000400.00410.0048000
2018-01-02USD34000340.00342.0082000
2018-01-02USD45000450.00440.0064000
2018-02-01USD50000500.00505.00164000
2019-02-01USD1420001420.001420.00112000
2019-01-18USD80000800.00807.00238200
2019-03-20USD2250002250.002260.00421400
2020-01-01USD2050002050.002050.00716600
2020-01-01USD1500001500.001520.001329000
2021-01-01USD74000740.00740.00231900
2021-01-01USD50000500.00500.0094200
2021-01-01USD40000400.00400.00198300
2022-01-01USD75000750.00752.00997700
2022-01-01USD1250001250.001250.00802000
2023-01-01USD82000820.00820.00781400
2023-06-04USD90000900.00900.00909200
2023-11-10USD1050001050.001050.00899300

期望输出格式

MonthPayment CountPayment Total
January240304,874
February262423,971
March281238,761
April207348,112
May298403,171
June240304,874
July262436,001
August282241,891
September214347,123
October301307,174

内容的提问来源于stack exchange,提问作者E A S

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最近更新时间:2026.07.14 14:38:09