select a.dt,count(distinct b.account_id)/count(distinct a.account_id)as rate from ( select account_id,dt from src_logoutrole_day where dt between 20180604 and 20180606 )a left join ( select account_id,dt from src_logoutrole_day where dt between 20180605 and 20180607 )b on a.account_id = b.account_id and datediff(from_unixtime(unix_timestamp(b.dt,'yyyyMMdd')),from_unixtime(unix_timestamp(a.dt,'yyyyMMdd'))) = 1 group by a.dt order by a.dt;有错吗

时间: 2024-01-10 10:04:56 浏览: 20
这个 SQL 语句看起来没有语法错误,但是可能存在逻辑或数据问题。这个 SQL 语句是为了计算在一段时间内,前一天和后一天都有登录的玩家占总登录玩家数的比例。需要确认以下几点: 1. 数据表 src_logoutrole_day 中是否包含 account_id 和 dt 两个字段。 2. 时间范围是否正确,确保时间范围内有数据。 3. 是否存在同一天内多次登录的玩家,如果存在,需要去重。 4. 是否存在某个日期的登录数据没有被匹配到对应的前一天或后一天数据,这种情况下会导致计算不准确。 如果以上问题都排除了,那么可以执行该 SQL 语句并检查输出结果来验证逻辑是否正确。
相关问题

select a.dt,count(b.account_id)/count(a.account_id) from ( select distinct account_id from src_loginrole_day where dt <=20180606 and dt>= 20180604 )as a left JOIN ( select distinct account_id from src_loginrole_day where dt <=20180607 and dt>= 20180605 )as b on a.account_id = b.account_id and datediff(from_unixtime(unix_timestamp(b.dt,'yyyyMMdd')),from_unixtime(unix_timestamp(a.dt,'yyyyMMdd'))) = 1 group by a.dt;哪里有错

这个 SQL 语句存在一些问题,下面是对这个 SQL 语句的分析和修改建议: 1. SQL 语句中使用了子查询,但是子查询没有别名,需要为子查询添加别名。 2. 在子查询中使用了 `DISTINCT` 关键字,但是在这个场景下使用 `GROUP BY` 更为合适。 3. 在子查询中筛选出了时间段内的 `account_id`,在外层查询中应该通过 `WHERE` 条件进一步过滤时间段。 4. 在 `SELECT` 语句中使用了聚合函数 `COUNT`,但是在 `GROUP BY` 子句中没有包含所有非聚合列,需要将所有非聚合列都包含在 `GROUP BY` 子句中。 5. 在 `DATEDIFF` 函数中使用了 `FROM_UNIXTIME`,但是这个函数的参数应该是一个时间字符串,而不是时间戳。需要将 `UNIX_TIMESTAMP` 函数放到 `FROM_UNIXTIME` 函数的内部,再将结果作为参数传递给 `DATEDIFF` 函数。 修改后的 SQL 语句如下所示: ``` SELECT a.dt, COUNT(b.account_id) / COUNT(a.account_id) FROM ( SELECT dt, account_id FROM src_loginrole_day WHERE dt BETWEEN 20180604 AND 20180606 GROUP BY dt, account_id ) AS a LEFT JOIN ( SELECT dt, account_id FROM src_loginrole_day WHERE dt BETWEEN 20180605 AND 20180607 GROUP BY dt, account_id ) AS b ON a.account_id = b.account_id AND DATEDIFF(FROM_UNIXTIME(UNIX_TIMESTAMP(b.dt, 'yyyyMMdd')), FROM_UNIXTIME(UNIX_TIMESTAMP(a.dt, 'yyyyMMdd'))) = 1 WHERE a.dt BETWEEN 20180604 AND 20180606 GROUP BY a.dt; ```

select a.dt, count(distinct a.account_id) role, count(distinct b.account_id) ret2 from ( select 'involve' as type,account_id,dt from src_loginrole_day where dt between 20180604 and 20180606 )a left join ( select 'involve' as type,account_id,dt from src_loginrole_day where dt between 20180605 and 20180607 )b on a.account_id = b.account_id and datediff(from_unixtime(unix_timestamp(b.dt,'yyyyMMdd')),from_unixtime(unix_timestamp(a.dt,'yyyyMMdd'))) = 1 group by a.dt order by a.dt;哪里有错

这个 SQL 语句中存在一个错误,因为 `from src_loginrole_day` 后面缺少表的别名。正确的语句如下: ``` select a.dt, count(distinct a.account_id) role, count(distinct b.account_id) ret2 from ( select 'involve' as type, account_id, dt from src_loginrole_day where dt between 20180604 and 20180606 ) a left join ( select 'involve' as type, account_id, dt from src_loginrole_day where dt between 20180605 and 20180607 ) b on a.account_id = b.account_id and datediff(from_unixtime(unix_timestamp(b.dt,'yyyyMMdd')), from_unixtime(unix_timestamp(a.dt,'yyyyMMdd'))) = 1 group by a.dt order by a.dt; ``` 在子查询中,给 `src_loginrole_day` 表起一个别名,例如 `s`,然后在 `select` 子句中使用这个别名来指代该表。

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SELECT t1.supplier_id, t1.ky_count, t1.ky_amount, IFNULL(t2.ky_refund_count,0) as ky_refund_count, IFNULL(t2.ky_refund_amount,0) as ky_refund_amount FROM ( SELECT a.supplier_id, count( DISTINCT c.order_no ) AS ky_count, SUM( IFNULL( c.record_amount, 0 )) / 100 AS ky_amount FROM settle_order a INNER JOIN settle_order_receipt b ON a.order_id = b.order_id INNER JOIN cash_withdrawal_record c ON b.third_trade_no = c.order_no WHERE a.is_del = 0 AND a.order_time >= '2023-05-28 00:00:00' AND a.order_time < '2023-05-29 00:00:00' AND a.order_type in (70,75) AND a.supplier_id IN (78,63,58,57,64,72,71,74,83,77,70,69,67,82,65,87,73,59,66,60,86,85,79,80,84,90) AND b.channel_code = 61 AND c.con_bank_account_no IN ( 247, 325 ) AND c.record_status = 1 AND c.record_time > '2023-05-01 00:00:00' GROUP BY a.supplier_id ) t1 LEFT JOIN ( SELECT a.supplier_id, count( DISTINCT b.order_no ) AS ky_refund_count, SUM( IFNULL( b.record_amount, 0 )) / 100 AS ky_refund_amount FROM settle_order_refund a INNER JOIN cash_withdrawal_record b ON a.third_refund_id = b.order_no WHERE a.is_del = 0 AND a.order_type in (70,75) AND a.apply_time >= '2023-05-28 00:00:00' AND a.apply_time < '2023-05-29 00:00:00' AND a.supplier_id IN (78,63,58,57,64,72,71,74,83,77,70,69,67,82,65,87,73,59,66,60,86,85,79,80,84,90) AND a.channel_code = 61 AND b.con_bank_account_no IN ( 247, 325 ) AND b.record_status = 5 AND b.record_time > '2023-05-01 00:00:00' GROUP BY a.supplier_id ) t2 ON t1.supplier_id = t2.supplier_id order by t1.supplier_id asc

优化这条sql: select distinct (select product_name from t_product from where id = #{productId} and mark = 1 and status = 1) as productName, (select count(0) from t_clue a where a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId} and a.status in(1,2,3,31,32,33)) as clueCount, (select count(0) from t_clue a left join t_clue_appendix b on a.clue_code = b.clue_code where a.distribution_status != 4 and a.mark = 1 and b.file_url is not null and a.product_id = #{productId} and a.status in (3,31,32,33)) as intentionCount, (select count(0) from t_clue a where a.status in (4,5,7,8) and a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId} and a.status = 4) as incomingCount, (select count(0) from t_clue a where a.status in (5,7,8) and a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId} and a.status in (5,7,8)) as approvedCount, (select count(0) from t_clue a where a.status = 6 and a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId} and a.status = 6) as rejectionCount, (select count(0) from t_clue a where a.status in (7,8) and a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId}) as loanCount, (select count(0) from t_clue a where a.status = 8 and a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId}) as swipeCount, (select sum(a.loan_amount) from t_clue a where a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId}) as loanMoney, (select sum(a.use_amount) from t_clue a where a.distribution_status != 4 and a.mark = 1 and a.product_id = #{productId}) as swipeMoney

优化这条sql 解决bug select a.*,rownum num from ( SELECT t.PROJ_ID,t.PROJ_CODE,t.PROJ_NAME,t.CLIENT_CODE,t.CLIENT_NAME,t.SPEC_CODE,t.SPEC_NAME,t.BUS_UNIT_CODE,t.BUS_UNIT,t.PROJ_DEP_CODE,t.PROJ_DEP,t.PROJECT_MANAGER_CODE,t.PROJECT_MANAGER,t.PROJECT_DEP_MANAGER_CODE,t.PROJECT_DEP_MANAGER,t.IS_SUB_PROJ,t.SUB_PROJ_TYPE_CODE,t.SUB_PROJ_TYPE,t.PARENT_CODE,t.PROJ_GROSS,t.CLIENT_AREA_CODE,t.CLIENT_AREA,t.CLIENT_TYPE_FULL_PATH_CODE,t.CLIENT_TYPE_PULL_PATH,t.BUSINESS_TYPE_CODE,t.BUSINESS_TYPE,t.BUSINESS_LEVEL_CODE,t.BUSINESS_LEVEL,t.BUSINESS_AREA_CODE,t.BUSINESS_AREA_NAME,t.IS_CLOSE,t.IS_IN_COO,t.TAX_RATE,t.IS_AUTHORIZED,t.AUTHORIZED_AMOUNT,t.IS_VIRTUAL,t.INCOME_BUDGET,t.EXPENDITURE_BUDGET,t.P_VALUE,t.CREATE_TIME,t.P_BUD_VALUE,t.P1_BUD_VALUE,t.P2_BUD_VALUE,t.ORG_CODE,t.ORG_NAME,t.PROD_RES_TYPE,t.IS_TECH_COO,t.COO_UNIT_RATIO,t.PROJ_ACHIEVEMENTS_BUD,t.REIMBURSEMENT_COST_BUD,t.COO_COST_BUD,t.MATERIAL_COST_BUD,t.PERFORMANCE_PERCENT,t.SCHE_START_TIME,t.SCHE_END_TIME,t.PROJECT_ACCOUNT_CODE,t.CUSTOMER_TYPE_CODE,t.CUSTOMER_TYPE,t.IS_PURE_OUT_PROJ,t.PROJECT_CREATE_TIME,t.IS_RELATE,t.IS_QUOTA,t.MAIN_PROJECT_CODE,t.PROJ_STATUS,t.IS_LARGE_PROJECT,t.MARKET_DIS_COUNT_RATE,t.PROJECT_CAT,t.MGR_PER_FORMANCE_RATIO,t.P1_VALUE,t.S_VALUE,t.COOP_VALUE,t.H_VALUE,t.DEVICE_BUDGET_COST,t.SUR_FEE_DIS_COUNT_RATE,t.DES_FEE_DIS_COUNT_RATE, (select listagg(p.coo_unit_code, ',') within group(order by p.coo_unit_code) from ( select distinct coo_unit_code from t_spdi_proj where is_sub_proj = 'Y' and sub_proj_type_code = 'wbhz' and PROJ_STATUS != 'P_5' AND PROJ_STATUS != 'P_4' and parent_code = t.proj_code )p ) coo_unit_code, (select listagg(to_char(p.coo_unit), ',') within group(order by p.coo_unit) from ( select distinct coo_unit from t_spdi_proj where is_sub_proj = 'Y' and sub_proj_type_code = 'wbhz' and PROJ_STATUS != 'P_5' AND PROJ_STATUS != 'P_4' and parent_code = t.proj_code )p ) coo_unit from T_SPDI_PROJ t where t.PROJ_STATUS NOT IN ('E','H','W') order by t.proj_id )a

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