SQL中LEFT JOIN如何按条件筛选出最新时间对应的单条记录
解决方案
你可以通过窗口函数ROW_NUMBER()对时间戳子查询的记录按创建时间倒序排序,筛选出每个匹配ID对应的最新记录即可,修改后的完整SQL如下:
SELECT distinct a.transaction_id, b.currency_id, h.SIG_CURRENCY_CODE, b.charge_id, c.charge_type_id, d.charge_group, e.account_name, f.sig_entity_label, e.trading_userid, e.exchange_id, e.tradeable_instr_name, g.underlying_sym_bloomberg, g.fut_expiration_date, e.trade_date, -- 若需要返回格式化后的时间可取消下方注释 -- timestamp_table.timeformated, a.rate * a.basis_value AS dma_fee, SUM(A.RATE * A.BASIS_VALUE) OVER () as DMA_FEE_TOTAL FROM repldoadm.ire_estimate_trans_map a LEFT JOIN repldoadm.ire_charges_estimate b ON a.estimate_id = b.estimate_id LEFT JOIN repldoadm.ire_charges_lu c ON b.charge_id = c.charge_id LEFT JOIN repldoadm.ire_charge_types_lu d ON c.charge_type_id = d.charge_type_id LEFT JOIN repldoadm.vw_ire_trade_transactions e ON a.transaction_id = e.pb_trans_id LEFT JOIN ( SELECT to_char(create_ts, 'HH24:MI:SS') AS timeformated, trans_id, pb_trans_id, ROW_NUMBER() OVER( PARTITION BY CASE WHEN trans_id IS NOT NULL THEN trans_id ELSE pb_trans_id END ORDER BY create_ts DESC ) AS rn FROM repldoadm.vw_ire_trade_transactions ) timestamp_table ON ( a.transaction_id = timestamp_table.trans_id OR a.transaction_id = timestamp_table.pb_trans_id ) AND timestamp_table.rn = 1 -- 注意:原SQL缺失h、f、g表的关联逻辑,需要补充后才能正常执行
核心改动说明
- 时间戳子查询中新增
ROW_NUMBER()窗口函数,按照关联ID分组(优先匹配trans_id,为空则取pb_trans_id),每组内按原始创建时间create_ts倒序排序,最新的记录排序序号为1 - 左联条件中新增
timestamp_table.rn = 1,仅关联每组最新的记录,彻底避免单ID匹配多条时间记录导致的重复行问题 - 排序时使用原始时间类型的
create_ts而非转成字符串的timeformated,避免字符串排序出现逻辑错误 - 筛选条件放在关联的ON子句而非WHERE子句中,保留左联特性,无匹配时间记录时也不会过滤左表数据
内容的提问来源于stack exchange,提问作者paddyusa
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