select t1.uid,t1.tag,t2.exam_num,ISNULL(t3.question_num,0) from ( select v1.uid,v3.tag from user_info v1 join exam_record v2 on v1.uid=v2.uid left join examination_info v3 on v2.exam_id=v3.exam_id where v1.level=7 group by v1.uid,v3.tag having v3.tag='SQL' ) t1 left join ( select uid,COUNT(exam_id) exam_num from exam_record group by uid ) t2 on t1.uid=t2.uid left join ( select uid,COUNT(question_id) question_num from practice_record group by uid ) t3 on t1.uid=t3.uid 优化

select t1.uid,t1.tag,t2.exam_num,ISNULL(t3.question_num,0) from ( select v1.uid,v3.tag from user_info v1 join exam_record v2 on v1.uid=v2.uid left join examination_info v3 on v2.exam_id=v3.exam_id where v1.level=7 group by v1.uid,v3.tag having v3.tag='SQL' ) t1 left join ( select uid,COUNT(exam_id) exam_num from exam_record group by uid ) t2 on t1.uid=t2.uid left join ( select uid,COUNT(question_id) question_num from practice_record group by uid ) t3 on t1.uid=t3.uid 重写语句优化

SELECT v1.uid, v3.tag, COUNT(DISTINCT v2.exam_id) AS exam_num, COUNT(DISTINCT t3.question_id) AS question_num FROM user_info v1 JOIN exam_record v2 ON v1.uid = v2.uid LEFT JOIN examination_info v3 ON ...

SELECT t1.* FROM awb t1 INNER JOIN awb t2 ON t1.HAWB_NO = t2.HAWB_NO2 OR t1.HAWB_NO = t2.AGENT_NO WHERE t1.HAWB_NO = t2.HAWB_NO and t1.ARR_DATE>getdate()-350语句优化

FROM awb t1 INNER JOIN awb t2 ON t1.HAWB_NO = t2.HAWB_NO2 OR t1.HAWB_NO = t2.AGENT_NO WHERE t1.HAWB_NO = t2.HAWB_NO AND t1.ARR_DATE > (SELECT DATEADD(day, -350, GETDATE())) 这样做可以使 SQL 查询...

WITH RECURSIVE tb_deptrelation (dept_id, dept_f, deptpath) AS ( -- 初始查询 SELECT t1.dept_id, t1.dept_f, t1.dept_id FROM db_attendance.tb_dept AS t1 WHERE dept_f IS NULL UNION ALL -- 递归查询 SELECT t2.dept_id, t2.dept_f, t3.deptpath + '-' + t2.dept_f FROM db_attendance.tb_dept AS t2 JOIN tb_deptrelation t3 ON t2.dept_f = t3.dept_id ) SELECT dept_id, dept_f, deptpath FROM tb_deptrelation;哪有问题

在递归查询的第二个SELECT语句中，连接条件 t2.dept_f = t3.dept_id 可能是错误的。这取决于 tb_dept 表的结构和字段定义。如果 dept_f 字段在 tb_dept 表中表示部门的父级部门ID，那么连接条件应该是 t2...

如何优化如下sql select distinct t1.a from t1,t2 where t1.a=t2.a

FROM t1, t2 WHERE t1.a = t2.a ) subquery - 临时表： sql CREATE TEMPORARY TABLE temp_table AS ( SELECT DISTINCT t1.a FROM t1, t2 WHERE t1.a = t2.a ); SELECT * FROM temp_table; ...

update asks_over_time set is_replied = 1 where exists ( select t1.ask_id from asks t1 inner join asks_over_time t2 on t1.ask_id = t2.ask_id where t2.is_accepted = 1 and t1.handle_time is not null and t1.handle_time != '' and t1.handle_time >= t2.create_time and t2.ask_id = t1.ask_id) sql优化

where t2.is_accepted = 1 and t1.handle_time is not null and t1.handle_time != '' and t1.handle_time >= t2.create_time and t2.ask_id = asks_over_time.ask_id ) 这样的优化可以减少不必要的记录...

SELECT * from T1 where Exists select * from T2 where T1.id= T2.id

您的 SQL 查询语句存在语法错误，应该将 select * from T2 where T1.id= T2.id 用括号括起来，表示这是一个子查询，如下所示： SELECT * FROM T1 WHERE EXISTS (SELECT * FROM T2 WHERE T1.id = T2.id); ...

update asks_over_time set is_replied = 1 where ask_id in ( select t1.ask_id from asks t1 left join asks_over_time t2 on t1.ask_id = t2.ask_id where t2.is_accepted = 1 and t1.handle_time is not null and t1.handle_time != '' and t1.handle_time >= t2.create_time )优化sql

where t2.is_accepted = 1 and t1.handle_time is not null and t1.handle_time != '' and t1.handle_time >= t2.create_time and asks_over_time.ask_id = t1.ask_id ) 这样的优化可以减少不必要的记录...

优化这个sql语句：UPDATE GAMS_ASSETCARD t2 SET ( t2.SHIYRID ) = ( SELECT t1.RECID FROM GAMS_JC_PERSONNEL t1 WHERE t1.STDNAME = t2.SHIYRID_OLD and t1.REMARK IS NULL and t1.UNITID='43B8BFD400000081BC4FCC244F2CD1FB' )where t2.shiyrid is null ;

t1.REMARK IS NULL AND t1.UNITID = '43B8BFD400000081BC4FCC244F2CD1FB' AND t2.SHIYRID IS NULL ) src ON (t2.SHIYRID_OLD = src.SHIYRID_OLD) WHEN MATCHED THEN UPDATE SET t2.SHIYRID = src.RECID; ...

SELECT t3.id, t3.no, t3.name, t3.orgnization, sex, icno, address, birth, register_date, phone, img_status, finger_print_num, card_num, t3.status,t4.img_template FROM ( SELECT t1.*,t2.name AS 'orgnization' FROM tb_employee t1 INNER JOIN tb_department t2 ON t1.orgnization_id = t2.id )t3 INNER JOIN tb_face t4 ON t3.no = t4.no mybatis如何写

SELECT t3.id, t3.no, t3.name, t3.orgnization, sex, icno, address, birth, register_date, phone, img_status, finger_print_num, card_num, t3.status, t4.img_template FROM ( SELECT t1.*,t2.name AS '...

帮我优化如下sql：SELECT T1.quyu, T2.lastNum, T3.jdNum, T4.xcNum FROM (SELECT STREET AS quyu FROM STREET_YXW WHERE 1=1 GROUP BY STREET ) T1 LEFT JOIN ( SELECT T2.STREET_NAME AS quyu,COUNT() AS lastNum FROM V_FIRESAFETYREGISTER_DYC T1 LEFT JOIN V_HSE_COMB_BUILDING_INFO_PA T2 ON LEFT(T1.HOUSEID,19) = T2.BUILDING_CODE WHERE T1.LASTLOGOUTTIME IS NOT NULL and date_format(T1.LASTLOGOUTTIME,'%Y') = ? GROUP BY T2.STREET_NAME ) T2 ON T1.quyu = T2.quyu LEFT JOIN ( SELECT T3.STREET_NAME AS quyu,COUNT(DISTINCT T1.FIREID) AS jdNum FROM V_SX_FIRESAFETYPLAN T1 LEFT JOIN V_FIRESAFETYREGISTER_DYC T2 ON T1.FIREID = T2.ID LEFT JOIN V_HSE_COMB_BUILDING_INFO_PA T3 ON LEFT(T2.HOUSEID,19) = T3.BUILDING_CODE WHERE T1.DATE_NEW IS NOT NULL and date_format(T1.DATE_NEW,'%Y') = ? GROUP BY T3.STREET_NAME ) T3 ON T1.quyu = T3.quyu LEFT JOIN ( SELECT T3.STREET_NAME AS quyu,COUNT() as xcNum FROM V_SX_FIRESAFETYTRAINING T1 LEFT JOIN V_FIRESAFETYREGISTER_DYC T2 ON T1.FIREID = T2.ID LEFT JOIN V_HSE_COMB_BUILDING_INFO_PA T3 ON LEFT(T2.HOUSEID,19) = T3.BUILDING_CODE WHERE T1.TRAININGTIME IS NOT NULL and date_format(T1.TRAININGTIME,'%Y') = ? GROUP BY T3.STREET_NAME ) T4 ON T1.quyu = T4.quyu ORDER BY T2.lastNum limit ?,?

SELECT T1.quyu, T2.lastNum, T3.jdNum, T4.xcNum FROM tmp_street T1 LEFT JOIN tmp_lastNum T2 ON T1.quyu = T2.quyu LEFT JOIN tmp_jdNum T3 ON T1.quyu = T3.quyu LEFT JOIN tmp_xcNum T4 ON T1.quyu = T4.quyu ...

select t1.c_id c_id1, t2.c_id c_id2 from pro t1, pro t2 where t1.c_id < t2.c_id and t1.f_id = t2.f_id;

这是一个 MySQL 查询语句，用于在 pro 表中找出持有相同基金的用户对，并按照用户 ID 排序。查询结果包括两个用户 ID，其中 c_id1 < c_id2。查询中使用了自连接和比较运算符。首先，在 WHERE 子句中，通过自连接将 ...

SELECT t1.organization_id, t1.organization_name, SUM( t2.jl_amount ) yfSjAmount, SUM( t2.sj_amount ) yfJlAmount, SUM( t3.money ) yfSectionAmount, SUM( t3.mileage ) yfSectionLength FROM main_project t1 LEFT JOIN mat_approval_project t2 ON t2.main_id = t1.id LEFT JOIN mat_project_pavement_approval t3 ON t3.project_id = t2.id WHERE t1.project_general_name = '预防性养护' AND t1.project_category_name = '路面工程' AND t1.process_status = '100' AND t1.organization_id = 'b3f029fa-62ab-476e-9876-b87ce69c41c5' AND t1.project_year = '2023' SQL统计错误，请修改

SELECT t1.organization_id, t1.organization_name, SUM(t2.jl_amount) yfSjAmount, SUM(t2.sj_amount) yfJlAmount, SUM(t3.money) yfSectionAmount, SUM(t3.mileage) yfSectionLength FROM main_project t1 LEFT ...

代码如何优化：select t1.* from t1 where NOT EXISTS ( select 1 from t2 where t1.ip=t2.id );

可以考虑以下优化措施： 1. 添加索引：对t1表中的ip列和t2表中的id列添加索引，可以大大提高查询效率。 2. 改写成LEFT JOIN语句：将NOT EXISTS改写成LEFT JOIN，这样可以避免子查询的多次执行...WHERE t2.id IS NULL;

SELECT a.id FROM ex_exam a LEFT JOIN ex_exam_statistic_enp_evaluate b ON a.id = b.qlh_exam_id WHERE a.type IN ("EXAM_EX_NORMAL_TYPE_NORMAL", "EXAM_TYPE_RZ") AND b.qlh_exam_id IS NULL 替换IS NULL

可以使用 "NOT EXISTS" 来替换 "b.qlh_exam_id IS NULL"，得到以下 SQL 查询语句： SELECT a.id FROM ex_exam a WHERE a.type IN ("EXAM_EX_NORMAL_TYPE_NORMAL", "EXAM_TYPE_RZ") AND NOT EXISTS ( SELECT 1 ...

修改这段代码避免死锁 MERGE INTO GAMS_ASSETCARD t2 USING ( SELECT t1.RECID, t1.stdname FROM GAMS_JC_PERSONNEL t1 WHERE t1.REMARK IS NULL AND t1.UNITID = '43B8BFD400000081BC4FCC244F2CD1FB' ) src ON (t2.SHIYRID_OLD = src.stdname ) WHEN MATCHED THEN UPDATE SET t2.SHIYRID = src.RECID where t2.SHIYRID IS NULL and t2.qingcbh='S20221660';

UPDATE SET t2.SHIYRID = src.RECID where t2.SHIYRID IS NULL and t2.qingcbh='S20221660'; COMMIT; 通过在代码开始处添加BEGIN TRANSACTION，并在代码结束处添加COMMIT，你可以将整个代码块包裹在一个事务中。...

SELECT t1.organization_id, t1.organization_name, t1.road_name, SUM( t2.sj_amount ) xfSjAmount, SUM( t2.jl_amount ) xfJlAmount, SUM( t3.xfGeneralSectionAmount ) xfGeneralSectionAmount, SUM( t3.xfGeneralSectionLength ) xfGeneralSectionLength FROM main_project t1 LEFT JOIN mat_approval_project t2 ON t2.main_id = t1.id LEFT JOIN ( SELECT project_id, data_type, SUM( money ) xfGeneralSectionAmount, SUM( mileage ) xfGeneralSectionLength FROM mat_project_pavement_approval GROUP BY project_id ) t3 ON t3.project_id = t2.id WHERE t1.project_general_name = '修复性养护' AND t1.project_category_name = '路面工程' AND t1.process_status = '100' AND t3.data_type = '1' AND t1.road_name = '常祁高速' AND t1.project_year = '2023' 这条sql中t1.road_name多条数据只会展示一个，导致根据t1.road_name查询不到数据

如果t1表中有多个记录的t1.road_name值相同，那么在SELECT语句中使用SUM函数对t2和t3表中的记录进行了合并，使得最终查询结果中只有一个记录。这可能导致你根据t1.road_name无法查到完整的数据。为了解决...

t1.user_id = t2.user_id t1.user_id为数值型，t2.user_id为字符串型则t1和t2怎么连接

由于t1.user_id为数值型，t2.user_id为字符串型，它们的数据类型不同，因此无法直接进行连接。在这种情况下，需要通过类型转换将其中一个字段的数据类型转换为与另一个字段相同的数据类型，然后再进行连接。具体实现...

相关推荐

优化SQL1

font_select.zip_font_select.zip

demo.tar.gz_C demo.tar_DEMO_select.c

SELECT t1.* FROM awb t1 INNER JOIN awb t2 ON t1.HAWB_NO = t2.HAWB_NO2 OR t1.HAWB_NO = t2.AGENT_NO WHERE t1.HAWB_NO = t2.HAWB_NO and t1.ARR_DATE>getdate()-350语句优化

如何优化如下sql select distinct t1.a from t1,t2 where t1.a=t2.a

update asks_over_time set is_replied = 1 where exists ( select t1.ask_id from asks t1 inner join asks_over_time t2 on t1.ask_id = t2.ask_id where t2.is_accepted = 1 and t1.handle_time is not null and t1.handle_time != '' and t1.handle_time >= t2.create_time and t2.ask_id = t1.ask_id) sql优化

SELECT * from T1 where Exists select * from T2 where T1.id= T2.id

update asks_over_time set is_replied = 1 where ask_id in ( select t1.ask_id from asks t1 left join asks_over_time t2 on t1.ask_id = t2.ask_id where t2.is_accepted = 1 and t1.handle_time is not null and t1.handle_time != '' and t1.handle_time >= t2.create_time )优化sql

优化这个sql语句：UPDATE GAMS_ASSETCARD t2 SET ( t2.SHIYRID ) = ( SELECT t1.RECID FROM GAMS_JC_PERSONNEL t1 WHERE t1.STDNAME = t2.SHIYRID_OLD and t1.REMARK IS NULL and t1.UNITID='43B8BFD400000081BC4FCC244F2CD1FB' )where t2.shiyrid is null ;

select t1.c_id c_id1, t2.c_id c_id2 from pro t1, pro t2 where t1.c_id < t2.c_id and t1.f_id = t2.f_id;

代码如何优化 ：select t1.* from t1 where NOT EXISTS ( select 1 from t2 where t1.ip=t2.id );

SELECT a.id FROM ex_exam a LEFT JOIN ex_exam_statistic_enp_evaluate b ON a.id = b.qlh_exam_id WHERE a.type IN ("EXAM_EX_NORMAL_TYPE_NORMAL", "EXAM_TYPE_RZ") AND b.qlh_exam_id IS NULL 替换IS NULL

t1.user_id = t2.user_id t1.user_id为数值型，t2.user_id为字符串型则t1和t2怎么连接

最新推荐

[Oracle] dbms_metadata.get_ddl 的使用方法总结

Vue.js仿Select下拉框效果

JQuery三级无刷新联动_select(C#版本).doc

浪潮服务器NF8460M3安装windows_Server2008.doc

用SELECT... INTO OUTFILE语句导出MySQL数据的教程

基于嵌入式ARMLinux的播放器的设计与实现 word格式.doc

管理建模和仿真的文件

Python字符串为空判断的动手实践：通过示例掌握技巧

box-sizing: border-box;作用是？

经典：大学答辩通过_基于ARM微处理器的嵌入式指纹识别系统设计.pdf

代码如何优化：select t1.* from t1 where NOT EXISTS ( select 1 from t2 where t1.ip=t2.id );