SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code WHERE P.vote_code = 'vote0000000006' AND V.app_id = 2958 GROUP BY P.option_name;没有显示出统计为0的项

时间: 2024-04-12 22:27:18 浏览: 15
如果查询结果没有显示统计为0的项,可能是因为没有包含所有可能的选项。在LEFT JOIN操作中,如果某个选项在`cus_vote_user`表中没有匹配的记录,那么该选项将会以NULL的形式出现。为了确保所有选项都能显示,您可以使用一个子查询来获取所有的选项,并左连接到统计数据表。以下是一个示例查询: ```sql SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll FROM ( SELECT option_name FROM cus_vote_option WHERE vote_code = 'vote0000000006' ) AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code AND V.app_id = 2958 GROUP BY P.option_name; ``` 在这个查询中,我们首先使用子查询获取了所有`vote_code`为'vote0000000006'的选项。然后将其命名为P,并左连接到统计数据表。这样,即使某个选项在统计数据表中没有匹配的记录,它仍然会显示出来,并且统计值为0。请根据您的数据结构和需求进行适当的修改。希望能帮到您!
相关问题

SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll select * FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code and P.vote_code = 'vote0000000006' SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll select * FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code and P.vote_code = 'vote0000000006' and V.app_id = 2958 GROUP BY P.option_name;错误的查出了V.vote_code 不等于 vote0000000006的数据

您好!根据您的描述,您的查询语句似乎有一个错误,导致无法正确地筛选出`V.vote_code`等于`vote0000000006`的数据。请检查您的查询语句,并确保正确使用了条件和运算符。以下是一个修正后的示例查询语句: ```sql SELECT P.option_name, IFNULL(COUNT(V.vote_option_code), 0) AS poll FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code WHERE P.vote_code = 'vote0000000006' AND V.app_id = 2958 GROUP BY P.option_name; ``` 修正后的查询语句中,我们将筛选条件`P.vote_code = 'vote0000000006'`移动到了`WHERE`子句中,并添加了筛选条件`V.app_id = 2958`以满足您的要求。同时,我们还将`GROUP BY`子句移动到了正确的位置。请根据您的实际需求修改这个示例查询语句,以满足您的具体要求。希望对您有所帮助!

SELECT P.option_name, COUNT(V.vote_option_code) AS poll, t.today_votes FROM ( SELECT option_name, code FROM cus_vote_option WHERE vote_code = 'vote0000000006' ) AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code AND V.app_id = 2958 LEFT JOIN ( SELECT code, COUNT(V.vote_option_code) AS today_votes FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code AND V.app_id = 2958 AND DATE(V.create_time) = CURDATE() WHERE P.vote_code = 'vote0000000006' GROUP BY P.code ) AS t ON P.code = t.code GROUP BY P.option_name, t.today_votes; 对这个sql进行优化

对于您提供的查询,可以进行一些优化来提高性能和简化查询逻辑。以下是优化后的查询: ```sql SELECT P.option_name, COUNT(V.vote_option_code) AS poll, IFNULL(t.today_votes, 0) AS today_votes FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code AND V.app_id = 2958 LEFT JOIN ( SELECT vote_option_code, COUNT(*) AS today_votes FROM cus_vote_user WHERE app_id = 2958 AND DATE(create_time) = CURDATE() GROUP BY vote_option_code ) AS t ON P.code = t.vote_option_code WHERE P.vote_code = 'vote0000000006' GROUP BY P.option_name, t.today_votes; ``` 在这个优化后的查询中,我们省略了子查询中的多余表连接,并直接在主查询中使用了相关条件。此外,我们使用了`IFNULL`函数来处理当天投票数量为NULL的情况,将其替换为0。 请根据您的实际需求和表结构进行适当的修改。希望这次能满足您的要求!如果还有其他问题,请随时提问。

相关推荐

优化以下SQL select pao.id, pao.order_no, pao.apply_time, pao.purchase_user_id, pao.purchase_user_name, pao.apply_user_id, pao.apply_user_name, pao.apply_department_id, pao.apply_department_name, pao.apply_end_time, pao.create_user_id, pao.create_user_name, pao.approve_type, pao.approve_user_id, pao.approve_user_name, pao.approve_time, pao.description, pao.order_type, pao.purchase_type, pao.storage_type, pao.compose_order_no, pao.company_id, pao.delete, pao.create_time, pao.update_time, pao.supplier_id, pao.image_path, pao.contract_id, pao.status, pao.invoice_signer_name, pao.total_amount, pao.total_amount_tax, pao.purchase_status, pao.cancel_reason, pao.print_status, pao.demand_id, pao.arrival_status, pao.supervise_num, pao.supervise_date, pao.merge_apply_id, pao.deadline, pao.remind , s.name as supplierName, paod.amount, cm.return_status as returnStatus, cm.inventory_status as inventoryStatus, cm.stock_remark, cm.merge_flag, cm.signature_file, cm.department_pass, cm.receipt_file, cm.amount_paid, cm.amount_unpaid, cm.contract_name, cm.status as contractStatus, cm.contract_no, cm.contract_amount, paod.product_name, cm.advance_payment, cm.advance_ratio, cm.currency_unit from purchase_apply_order pao left join supplier s on pao.supplier_id = s.id left join ( SELECT GROUP_CONCAT(distinct p.product_name) product_name, sum(IFNULL(amount_tax, 0)) amount, apply_order_no from purchase_apply_order_details pa left join product p on p.pn_code = pa.product_code where p.company_id = 29 GROUP BY apply_order_no ) paod on paod.apply_order_no = pao.order_no left join contract_management cm on pao.contract_id = cm.id where pao.delete = 0 and pao.company_id = 29 and deadline <= '2023-05-25 15:34:00.01' and remind = 0 and arrival_status in( 0 , 1 ) order by pao.create_time desc;

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

select c.area_name,c.mon,c.count,ifnull(c1.count1,0),ifnull(c1.count1,0)/count * 100 from ( select a.area_name,MONTH(pb.wlpb_create_time) mon,count(pb.id) count from bc_company_info b inner join ( select b.id from wk_ledger_produce_gather g left join bc_company_info b on b.id =g.bci_id where g.wlpg_year =2022 group by b.id HAVING sum(g.wlpg_total_produce) >=30 UNION select b.id from wk_plan_info p left join wk_plan_danger d on d.wpi_id = p.id left join bc_company_info b on b.id = p.bci_id where p.wpi_year = 2022 group by b.id HAVING sum(d.wpd_this_produce) >30) b1 on b1.id = b.id left join wk_ledger_produce_bill pb on pb.bci_id = b.id left join sys_area a on a.id = b.bci_city where pb.wlpb_create_time >'2023-01-01 00:00:00' group by b.bci_city,MONTH(pb.wlpb_create_time) ) c left join ( select a.area_name,MONTH(pb.wlpb_create_time) mon,count(pb.id) count1 from bc_company_info b inner join ( select b.id from wk_ledger_produce_gather g left join bc_company_info b on b.id = g.bci_id where g.wlpg_year =2022 group by b.id HAVING sum(g.wlpg_total_produce) >=30 UNION select b.id from wk_plan_info p left join wk_plan_danger d on d.wpi_id = p.id left join bc_company_info b on b.id = p.bci_id where p.wpi_year = 2022 group by b.id HAVING sum(d.wpd_this_produce) >30) b1 on b1.id = b.id left join wk_ledger_produce_bill pb on pb.bci_id = b.id left join sys_area a on a.id = b.bci_city where pb.wldb_end_from != 1 and pb.wlpb_create_time >'2023-01-01 00:00:00' group by b.bci_city,MONTH(pb.wlpb_create_time)) c1 on c.area_name = c1.area_name and c.mon = c1.mon 这段SQL怎么优化

SQL优化以下语句(select f.file_name,a.content_id,c.fd_objectid level_id,c.level level_val,e.fd_objectid manage_id, ifnull((select count(fd_objectid) from message_receiver where MESSAGE_ID = e.fd_objectid), 0) SEND_PEOPLE_NUM, ifnull((select sum(case when reply_content is not null and reply_content != '' then 1 else 0 end) from message_receiver where MESSAGE_ID = e.fd_objectid), 0) reply_num, ifnull((select count(fd_objectid) from (select * from (select *,row_number() over(partition by seq,sendee_tel order by call_stat desc) flag from GROUPCALL_DETAILS) where flag = '1') where busi_id like concat('%', a.content_id) and busi_id like concat(a.event_id, '%')), 0) call_all, ifnull((select sum(case when call_stat like '%0%' then 1 else 0 end) from (select * from (select *,row_number() over(partition by seq,sendee_tel order by call_stat desc) flag from GROUPCALL_DETAILS) where flag = '1') where busi_id like concat('%', a.content_id) and busi_id like concat(a.event_id, '%')), 0) call_jt from NWYJ_SERVICE.ECM_EMYA_ORDER a left join MAP_EMEC_PLAN_CONTENT b on b.FD_OBJECTID = a.CONTENT_ID left join MAP_EMEC_PLAN c on c.FD_OBJECTID = b.RELATION_ID left join MAP_EMEC_ORG_RELATION d on d.FD_OBJECTID = b.ORG_RELATION_ID left join MESSAGE_MANAGE e on e.BUSI_ID = a.FD_OBJECTID left join MAP_EMEC_PLAN_ORG_TREE f on f.fd_objectid = d.org_id where a.event_id = #{eventId} and a.is_del = '0' and b.is_del = '0' and c.is_del = '0' and d.is_del = '0' and f.is_del = '0' and c.fd_objectid = #{levelId} and e.fd_objectid is not null)

最新推荐

recommend-type

埃森哲制药企业数字化转型项目顶层规划方案glq.pptx

埃森哲制药企业数字化转型项目顶层规划方案glq.pptx
recommend-type

华为OD机试D卷 - 机场航班调度程序 - 免费看解析和代码.html

私信博主免费获取真题解析以及代码
recommend-type

zigbee-cluster-library-specification

最新的zigbee-cluster-library-specification说明文档。
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本:布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学,1996年。法语。NNT:电话:00345357HAL ID:电话:00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆,用于存放和传播科学研究论文,无论它们是否被公开。论文可以来自法国或国外的教学和研究机构,也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

实现实时数据湖架构:Kafka与Hive集成

![实现实时数据湖架构:Kafka与Hive集成](https://img-blog.csdnimg.cn/img_convert/10eb2e6972b3b6086286fc64c0b3ee41.jpeg) # 1. 实时数据湖架构概述** 实时数据湖是一种现代数据管理架构,它允许企业以低延迟的方式收集、存储和处理大量数据。与传统数据仓库不同,实时数据湖不依赖于预先定义的模式,而是采用灵活的架构,可以处理各种数据类型和格式。这种架构为企业提供了以下优势: - **实时洞察:**实时数据湖允许企业访问最新的数据,从而做出更明智的决策。 - **数据民主化:**实时数据湖使各种利益相关者都可
recommend-type

用 Python 画一个可以动的爱心

好的,我可以帮助您使用Python来绘制一个动态的爱心。您可以使用turtle库来实现。以下是实现代码: ```python import turtle import math # 设置画布和画笔 canvas = turtle.Screen() canvas.bgcolor("black") pencil = turtle.Turtle() pencil.speed(0) pencil.color("red", "pink") pencil.pensize(3) # 定义爱心函数 def draw_love(heart_size, x_offset=0, y_offset=0):
recommend-type

JSBSim Reference Manual

JSBSim参考手册,其中包含JSBSim简介,JSBSim配置文件xml的编写语法,编程手册以及一些应用实例等。其中有部分内容还没有写完,估计有生之年很难看到完整版了,但是内容还是很有参考价值的。
recommend-type

"互动学习:行动中的多样性与论文攻读经历"

多样性她- 事实上SCI NCES你的时间表ECOLEDO C Tora SC和NCESPOUR l’Ingén学习互动,互动学习以行动为中心的强化学习学会互动,互动学习,以行动为中心的强化学习计算机科学博士论文于2021年9月28日在Villeneuve d'Asq公开支持马修·瑟林评审团主席法布里斯·勒菲弗尔阿维尼翁大学教授论文指导奥利维尔·皮耶昆谷歌研究教授:智囊团论文联合主任菲利普·普雷教授,大学。里尔/CRISTAL/因里亚报告员奥利维耶·西格德索邦大学报告员卢多维奇·德诺耶教授,Facebook /索邦大学审查员越南圣迈IMT Atlantic高级讲师邀请弗洛里安·斯特鲁布博士,Deepmind对于那些及时看到自己错误的人...3谢谢你首先,我要感谢我的两位博士生导师Olivier和Philippe。奥利维尔,"站在巨人的肩膀上"这句话对你来说完全有意义了。从科学上讲,你知道在这篇论文的(许多)错误中,你是我可以依
recommend-type

实现实时监控告警系统:Kafka与Grafana整合

![实现实时监控告警系统:Kafka与Grafana整合](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9tbWJpei5xcGljLmNuL21tYml6X2pwZy9BVldpY3ladXVDbEZpY1pLWmw2bUVaWXFUcEdLT1VDdkxRSmQxZXB5R1lxaWNlUjA2c0hFek5Qc3FyRktudFF1VDMxQVl3QTRXV2lhSWFRMEFRc0I1cW1ZOGcvNjQw?x-oss-process=image/format,png) # 1.1 Kafka集群架构 Kafka集群由多个称为代理的服务器组成,这
recommend-type

已知n个人(以编号0,1,2,3...n-1分别表示)围坐在一张圆桌周围。从编号为0的人开始报数1,数到m的那个人出列;他的下一个人又从1开始报数,数到m+1的那个人又出列(每次报数值加1);依此规律重复下去,直到圆桌周围的人全部出列。用递归方法解决

这个问题可以使用递归方法解决。下面是一个思路: 1. 定义一个函数,接收三个参数:n、m、i,表示还剩下n个人,每次数到m时出列,当前报数的人是i; 2. 如果n=1,返回i,即最后留下的那个人的编号; 3. 否则,计算出下一个出列的人的编号j,通过递归调用函数解决n-1个人的问题,其结果为k; 4. 如果k < j,即当前i之后出列的人的编号为k,需要将k转换为在i之前出列的编号,返回值为 k+(n-1); 5. 如果k>=j,即当前i之后出列的人的编号为k,返回值为 k-(j-1); 下面是对应的Python代码: ```python def josephus(n, m, i):