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问题

数据库查询11秒 好慢 肿么回事

SQL: SELECT Vod.vod_id AS vod_id,Vod.vod_cid AS vod_cid,Vod.vod_name AS vod_name,Vod.vod_ename AS vod_ename,Vod.vod_titl...
缝合额 2019-12-01 19:29:23 809 浏览量 回答数 2

问题

数据库查询好慢哦 19秒

[font=Arial, "]SQL: SELECT Vod.vod_id AS vod_id,Vod.vod_cid AS vod_cid,Vod.vod_name AS vod_name,Vod.vod_ename AS...
缝合额 2019-12-01 21:55:35 1697 浏览量 回答数 0

问题

hibernate 的问题啊

用DetachedCriteria类 组装查询条件,咋个生成这样的sql语句了捏? 如果没没条件的话还行 ,一加上where条件查询,就会出 where y1_.... 然后就报错&#...
长安归故里. 2020-01-31 13:38:29 0 浏览量 回答数 1

问题

Entity Framework使用Where方法查询单行记录,但生成的SQL语句没有Where子句是为什么?

在使用ENtityFramework查询数据的时候,其中有一个表,即使在Where方法中传入了查询条件,生成的SQL语句中始终不带Where子句,请问这是为什么?调用栈: public IList<Device> GetDev...
a123456678 2019-12-01 20:16:51 1039 浏览量 回答数 1

问题

如何用上一行的值填充空列?

我正在尝试执行一些操作。我想让所有颜色填充值。但是,当我有空列时,我想用前一个非空列中的值填充它。 with cte as ( select '2019-11-12 16:01:55' as t...
保持可爱mmm 2019-12-01 21:58:43 2 浏览量 回答数 1

回答

关于效率,我不会说太多,因为我没有针对其他方法进行过测试,但是如果没有临时表,这似乎是一个不错的选择。 SELECT COUNT(u.userID) AS total, m.month FROM ( SELECT 'Jan' AS MONTH UNION SELECT 'Feb' AS MONTH UNION SELECT 'Mar' AS MONTH UNION SELECT 'Apr' AS MONTH UNION SELECT 'May' AS MONTH UNION SELECT 'Jun' AS MONTH UNION SELECT 'Jul' AS MONTH UNION SELECT 'Aug' AS MONTH UNION SELECT 'Sep' AS MONTH UNION SELECT 'Oct' AS MONTH UNION SELECT 'Nov' AS MONTH UNION SELECT 'Dec' AS MONTH ) AS m LEFT JOIN users u ON MONTH(STR_TO_DATE(CONCAT(m.month, ' 2013'),'%M %Y')) = MONTH(u.userRegistredDate) AND YEAR(u.userRegistredDate) = '2013' GROUP BY m.month ORDER BY 1+1; 如果您基于日期格式进行并集,甚至可以减少工作量和查询负担。 SELECT COUNT(u.userID) AS total, DATE_FORMAT(merge_date,'%b') AS month, YEAR(m.merge_date) AS year FROM ( SELECT '2013-01-01' AS merge_date UNION SELECT '2013-02-01' AS merge_date UNION SELECT '2013-03-01' AS merge_date UNION SELECT '2013-04-01' AS merge_date UNION SELECT '2013-05-01' AS merge_date UNION SELECT '2013-06-01' AS merge_date UNION SELECT '2013-07-01' AS merge_date UNION SELECT '2013-08-01' AS merge_date UNION SELECT '2013-09-01' AS merge_date UNION SELECT '2013-10-01' AS merge_date UNION SELECT '2013-11-01' AS merge_date UNION SELECT '2013-12-01' AS merge_date ) AS m LEFT JOIN users u ON MONTH(m.merge_date) = MONTH(u.userRegistredDate) AND YEAR(m.merge_date) = YEAR(u.userRegistredDate) GROUP BY m.merge_date ORDER BY 1+1;来源:stack overflow
保持可爱mmm 2020-05-17 20:30:58 0 浏览量 回答数 0

回答

17题 select company,avg(salary) as salary from ( select * from ( select * from ( select x.sort,x.company,x.salary,y.total_num from ( select row_number() over(partition by company order by salary) as sort, * from employee ) as x inner join (select count(1) as total_num, company from employee group by company) as y on x.company = y.company ) as z where z.total_num % 2 = 0 ) as b where b.sort = b.total_num / 2 or b.sort = b.total_num / 2 + 1 ) as c group by company union all select company,salary from ( select * from ( select x.sort,x.company,x.salary,y.total_num from ( select row_number() over(partition by company order by salary) as sort, * from employee ) as x inner join (select count(1) as total_num, company from employee group by company) as y on x.company = y.company ) as z where z.total_num % 2 = 1 ) as b where b.sort = (b.total_num + 1) / 2; or select case when total_num_of_each_company % 2 = 1 then salary else (salary + next_row_salary) / 2 end as mid_salary, * from ( select case when total_num_of_each_company % 2 = 1 then (total_num_of_each_company + 1) / 2 else total_num_of_each_company / 2 end as mid_row_num, * from ( select last_value(sort_by_salary_num) over(partition by company order by company) as total_num_of_each_company, * from( select row_number() over(partition by company order by salary) as sort_by_salary_num, lead(salary) over(partition by company order by salary) as next_row_salary, * from employee ) as x ) as y ) as z where z.mid_row_num = z.sort_by_salary_num; 18题 select product_id,year as first_year,quantity, price from ( select row_number() over(partition by product_id order by year) as sort, * from sales ) as x where sort = 1 order by product_id; 如果结果中不展示product name,那么第二张表,product可以不用,直接一张sales就可以了。 19题 select one.stuid from (select * from sc as x where x.cid = '001') as one INNER JOIN (select * from sc as x where x.cid = '002') as two on one.stuid = two.stuid where one.course > two.course; 20题 select salary as secondHighSalary from ( select dense_rank() over(partition by 1 order by salary desc) as dense_rank_sort, salary from employee_bak ) as x where x.dense_rank_sort = 2 limit 1;
o_coby 2019-12-18 12:07:09 0 浏览量 回答数 0

问题

greenplum 大量sum语句多个临时表关联,如何优化

如题:在 greenplum 大量sum语句多个临时表关联,如何优化。 sql 示例: SELECT tk.column1 AS column1, tk.column2 AS column2, t...
亲亲橘子香 2020-05-09 18:46:48 0 浏览量 回答数 1

问题

查询报错ERROR [SqlExceptionHelper] 无法转换为内部表示?报错

@Koala_考拉 你好,想跟你请教个问题:Hibernate: select count(*) as col_0_0_ from AMST_USER_INFO user0_ where 1=1 ...
爱吃鱼的程序员 2020-06-20 16:56:17 0 浏览量 回答数 1

问题

计算每行的唯一值(在索引轴上,而不是列方向)

我有下表: WITH data AS ( SELECT 10 AS A, 10 AS B, 10 AS C UNION ALL SELECT 20 AS A, 10 AS B, 20 AS C UNION ALL SELEC...
祖安文状元 2020-01-03 18:59:37 0 浏览量 回答数 1

问题

hibernate 左连接 如何加连接条件?报错

HQL : FROM  FileCheckOrg as fco right outer join fco.org as o WITH o.semesterId =? where fco.type=? 生成sql如...
爱吃鱼的程序员 2020-06-22 16:57:53 1 浏览量 回答数 1

问题

SQL Server中的时间范围仅适用于日期

我有一个名为Name的表Transactions,它包含这样的数据: transaction_id cust_id tran_date 80712190438 270351 28-...
祖安文状元 2020-01-03 16:35:22 0 浏览量 回答数 1

回答

您需要进行三向联接: CREATE TABLE new_table AS SELECT p.*, d.content AS age FROM people AS p JOIN details AS d ON d.person_id = p.id JOIN taxonomy AS t ON t.id = d.detail_id WHERE t.taxonomy = 'age' 演示 或者,如果您已经创建了表,则可以执行以下操作: INSERT INTO new_table (id, last_name, first_name, email, age) SELECT p.id, p.last_name, p.first_name, p.email, d.content AS age FROM people AS p JOIN details AS d ON d.person_id = p.id JOIN taxonomy AS t ON t.id = d.detail_id WHERE t.taxonomy = 'age' 要获得多个属性,您必须分别为每个属性将详细信息和分类表联接在一起: CREATE TABLE new_table AS SELECT p.*, d1.content AS age, d2.content AS gender, d3.content AS height FROM people AS p JOIN details AS d1 ON d1.person_id = p.id JOIN taxonomy AS t1 ON t1.id = d1.detail_id JOIN details AS d2 ON d2.person_id = p.id JOIN taxonomy AS t2 ON t2.id = d2.detail_id JOIN details AS d3 ON d3.person_id = p.id JOIN taxonomy AS t3 ON t3.id = d3.detail_id WHERE t1.taxonomy = 'age' AND t2.taxonomy = 'gender' AND t3.taxonomy = 'height'来源:stack overflow
保持可爱mmm 2020-05-17 16:55:54 0 浏览量 回答数 0

回答

这样的话SQL有两种写法:法一SELECT e.EMPLOYEE_NAME AS `姓名`, d.DEPARTMENT_NAME AS `部门号`, e.SALARY AS `工资` FROM DEPARTMENTS AS d, EMPLOYEES AS e WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID ORDER BY d.DEPARTMENT_ID DESC, SALARY ASC法二SELECT e.EMPLOYEE_NAME AS `姓名`, d.DEPARTMENT_NAME AS `部门号`, e.SALARY AS `工资` FROM EMPLOYEES AS e LEFT JOIN DEPARTMENTS AS d ON e.DEPARTMENT_ID = d.DEPARTMENT_ID ORDER BY d.DEPARTMENT_ID DESC, SALARY ASC
落地花开啦 2019-12-02 01:46:36 0 浏览量 回答数 0

回答

MstrSelectSql = "SELECT StuInfo.NO AS 学号,StuInfo.Name AS 姓名,StuInfo.Sex AS 性别,StuInfo.BirDate AS 出生日期,StuInfo.PolType AS 政治面貌,StuScore.English AS 大学英语,StuScore.Computer AS 计算机,StuScore.Math AS 大学高等数学,StuScore.PE AS 大学体育 " & _" FROM StuInfo INNER JOIN StuScore " & _" ON StuInfo.NO=StuScore.NO " & _" ORDER BY StuInfo.NO ASC"
吴孟桥 2019-12-02 02:41:08 0 浏览量 回答数 0

回答

这是我为几乎相同的堆栈编写的(我们需要标准化硬件的制造商名称,并且有各种各样的变体)。不过,这是客户端(准确地说是VB.Net)-并使用Levenshtein距离算法(已修改,以获得更好的结果): Public Shared Function FindMostSimilarString(ByVal toFind As String, ByVal ParamArray stringList() As String) As String Dim bestMatch As String = "" Dim bestDistance As Integer = 1000 'Almost anything should be better than that! For Each matchCandidate As String In stringList Dim candidateDistance As Integer = LevenshteinDistance(toFind, matchCandidate) If candidateDistance < bestDistance Then bestMatch = matchCandidate bestDistance = candidateDistance End If Next Return bestMatch End Function 'This will be used to determine how similar strings are. Modified from the link below... 'Fxn from: http://ca0v.terapad.com/index.cfm?fa=contentNews.newsDetails&newsID=37030&from=list Public Shared Function LevenshteinDistance(ByVal s As String, ByVal t As String) As Integer Dim sLength As Integer = s.Length ' length of s Dim tLength As Integer = t.Length ' length of t Dim lvCost As Integer ' cost Dim lvDistance As Integer = 0 Dim zeroCostCount As Integer = 0 Try ' Step 1 If tLength = 0 Then Return sLength ElseIf sLength = 0 Then Return tLength End If Dim lvMatrixSize As Integer = (1 + sLength) * (1 + tLength) Dim poBuffer() As Integer = New Integer(0 To lvMatrixSize - 1) {} ' fill first row For lvIndex As Integer = 0 To sLength poBuffer(lvIndex) = lvIndex Next 'fill first column For lvIndex As Integer = 1 To tLength poBuffer(lvIndex * (sLength + 1)) = lvIndex Next For lvRowIndex As Integer = 0 To sLength - 1 Dim s_i As Char = s(lvRowIndex) For lvColIndex As Integer = 0 To tLength - 1 If s_i = t(lvColIndex) Then lvCost = 0 zeroCostCount += 1 Else lvCost = 1 End If ' Step 6 Dim lvTopLeftIndex As Integer = lvColIndex * (sLength + 1) + lvRowIndex Dim lvTopLeft As Integer = poBuffer(lvTopLeftIndex) Dim lvTop As Integer = poBuffer(lvTopLeftIndex + 1) Dim lvLeft As Integer = poBuffer(lvTopLeftIndex + (sLength + 1)) lvDistance = Math.Min(lvTopLeft + lvCost, Math.Min(lvLeft, lvTop) + 1) poBuffer(lvTopLeftIndex + sLength + 2) = lvDistance Next Next Catch ex As ThreadAbortException Err.Clear() Catch ex As Exception WriteDebugMessage(Application.StartupPath , [Assembly].GetExecutingAssembly().GetName.Name.ToString, MethodBase.GetCurrentMethod.Name, Err) End Try Return lvDistance - zeroCostCount End Function
心有灵_夕 2019-12-29 12:50:04 0 浏览量 回答数 0

问题

映射到hbm.xml文件时无法提取ResultSet

我有一个联合查询 SELECT al.C_PERSIST_ID AS id, tkl.C_PERSIST_VERSION AS version, tkl.C_TENANT_ID AS tenantId, tkl.C_MESSAGE AS ...
小六码奴 2019-12-01 19:59:52 9 浏览量 回答数 0

问题

MySQL错误1449:指定为定义者的用户不存在?mysql

当我运行以下查询时,我得到一个错误: SELECT a.sl_id AS sl_id, a.quote_id AS quote_id, a.sl_date AS sl_date, a.sl_type AS sl...
保持可爱mmm 2020-05-13 14:43:34 1 浏览量 回答数 1

回答

尝试这个: ;with Ranges as ( select 1 as NumberStart, 10 as NumberEnd ), InputSet as ( select 2 as NumberStart, 3 as NumberEnd union all select 5, 7 ), cte as ( select (select min(NumberStart) from Ranges) as rn, (select max(NumberEnd) from Ranges) as rn_end union all select rn+1, t.rn_end from cte t where t.rn < t.rn_end ) select min(isnull(t2.NumberStart,t1.rn)) as NumberStart, coalesce(t2.NumberEnd,v3.NumberEnd-1,v4.NumberEnd) as NumberEnd, count(*) as Cnt from cte t1 left join InputSet t2 on t1.rn between t2.NumberStart and t2.NumberEnd outer apply ( select min(NumberStart) as NumberEnd from InputSet tt where tt.NumberStart > t1.rn ) v3 outer apply ( select max(NumberEnd) as NumberEnd from Ranges ) v4 group by coalesce(t2.NumberEnd,v3.NumberEnd-1,v4.NumberEnd) order by 1 option (maxrecursion 32000) 但!如果间隔设置不正确,它将不起作用:您必须确保间隔(InputSet)不相交,并且Start小于End。
祖安文状元 2020-01-03 16:16:19 0 浏览量 回答数 0

问题

SQL Server如何比较2个表以提取差异

到目前为止,我所做的识别不正确付款的操作是运行不正确的查询并检索从12/01/2019开始的所有记录,然后将这些记录保存在临时表中。 使用不正确的表进行查询: SELECT DISTINCT ...
祖安文状元 2020-01-03 18:50:30 1 浏览量 回答数 1

回答

每行计算出3个辅助列,距离最早那天的天数,比本行时间早的行数,缺少的天数(距离最早那天的天数-比本行时间早的行数)。 连续登陆的那几天缺少的天数是相同的,按用户和缺少的天数统计一下就是连续登陆的天数。 按这个思路简单写下SQL: SELECT uid, min_date_diff - before_date_count miss_date_count, count(*) FROM ( SELECT user_time.*, datediff( user_time.time, user_min_time.min_time ) + 1 min_date_diff, ( SELECT count(*) FROM user_time i WHERE i.uid = user_time.uid AND i.time <= user_time.time ) before_date_count FROM user_time INNER JOIN ( SELECT uid, min(time) min_time FROM user_time GROUP BY uid ) user_min_time ON user_min_time.uid = user_time.uid ) user_time_stat GROUP BY uid, miss_date_count 结果: uid miss_date_count count(*) 1 0 4 1 5 2 2 0 3 ######我觉得这种记录之所以有需求是设计问题,连续登陆应该在每天都计算,如果发现最近一次登陆不连续,马上覆盖重算,比这么计算简单的多######我之前就是这样搞得,每天第一次登陆+1,如果最近的一个登陆不是昨天,从0开始计算,方便的很。根本不需要像LZ那样!######我擦,感觉太难了,我想想###### 已找到答、案 http://www.oschina.net/question/573517_118821 谢谢上面回答,吐槽一下,发完贴然后居然找不到该贴了,隔了一天,今天才显示出来,不知道怎么回事。 ######CREATE TABLE countLine AS SELECT 1 AS id, 1 AS uid, 20150101 AS TIME UNION ALL SELECT 2 AS id, 1 AS uid, 20150102 AS TIME UNION ALL SELECT 3 AS id, 2 AS uid, 20150101 AS TIME UNION ALL SELECT 4 AS id, 1 AS uid, 20150103 AS TIME UNION ALL SELECT 5 AS id, 2 AS uid, 20150102 AS TIME UNION ALL SELECT 6 AS id, 2 AS uid, 20150103 AS TIME UNION ALL SELECT 7 AS id, 1 AS uid, 20150104 AS TIME UNION ALL SELECT 8 AS id, 1 AS uid, 20150110 AS TIME UNION ALL SELECT 9 AS id, 1 AS uid, 20150111 AS TIME; --查询语句 SELECT uid,COUNT(1) AS count_time FROM (SELECT id,uid,TIME,TIME-IF(@uid=uid,(@rn:=@rn+1),@rn:=1) AS diff ,@uid :=uid FROM countLine e,(SELECT @rn:=0,@uid='',@rnx:=0) c ORDER BY uid,TIME) a GROUP BY uid,diff; 结果    uid  count_time   ------  ------------      1             4      1             2      2             3
kun坤 2020-06-09 11:44:12 0 浏览量 回答数 0

回答

SELECTSUM(d.read_num) AS READ,SUM(d.click_num) AS click,SUM(d.read_num - t.read_num) AS diffREAD,SUM(d.click_num - t.click_num) AS diffclick,u.aidFROMtc_url AS uLEFT JOIN tc_data AS dON d.uid = u.idLEFT JOIN(SELECT_d.read_num,_d.click_num,_u.aidFROMtc_url AS _uLEFT JOIN tc_data AS _dON _d.uid = _u.idWHERE _u.createtime = '昨天') AS tON t.aid = u.aidWHERE u.createtime = '今天'GROUP BY u.aid你自己试试
吴孟桥 2019-12-02 02:49:16 0 浏览量 回答数 0

回答

自己找到解决办法了, let x = CGFloat(((attributeDict["x"]! as String )as NSString).floatValue) let y = CGFloat(((attributeDict["y"]! as String ) as NSString).floatValue) let Width = CGFloat(((attributeDict["Width"]! as String )as NSString).floatValue) let Height = CGFloat(((attributeDict["Height"]! as String ) as NSString).floatValue) m_backgroundRect = CGRectMake(x,y,Width,Height)
爵霸 2019-12-02 02:08:13 0 浏览量 回答数 0

回答

以下是一种非常直接的方法:spark.sql(""" select 1 as id1,2 as id2 union select 3 as id1,4 as id2 """).createOrReplaceTempView("table1")spark.sql(""" select 1 as c1, 'london' as city union select 4 as c1, 'texas' as city """).createOrReplaceTempView("table2")spark.sql(""" select 2 as c1, 'paris' as city union select 3 as c1, 'arizona' as city """).createOrReplaceTempView("table3")spark.table("table1").show()spark.table("table2").show()spark.table("table3").show()for simplicity, union table2 and table 3spark.sql(""" select from table2 union all select from table3 """).createOrReplaceTempView("city_mappings")spark.table("city_mappings").show()now join to the ids:spark.sql(""" select id1, id2, city from table1 join city_mappings on c1 = id1 or c1 = id2""").createOrReplaceTempView("id_to_city")
社区小助手 2019-12-02 01:52:19 0 浏览量 回答数 0

回答

楼上误人子弟 啊,不懂就不要说了,把多个结果结果集生成 一个表,多表关联更新,还可以用case where 方法1(我更倾向于方法1).UPDATE table1, (SELECT 1 AS id, 1 AS scoreA, 2 AS scoreB UNION ALL SELECT 1 AS id,91 AS scoreA,98 AS scoreB UNION ALL SELECT 2 AS id,92 AS scoreA,97 AS scoreB) temp SET table1.scoreA = temp.scoreA, table1.scoreB = temp.scoreB WHERE table1.id = temp.id ;方法2.UPDATE table1 SET scoreA= CASE id WHEN 10001 THEN 98 WHEN 10002 THEN 97 WHEN 10004 THEN 92 WHEN 10006 THEN 92 WHEN 10039 THEN 93 END WHERE id IN ( 10001, 10002, 10004, 10006, 10039 )
30676716 2019-12-02 01:44:04 0 浏览量 回答数 0

回答

正如浮点类型的加法不准确一样,如果超出精度,则十进制类型的乘积可能不准确(或导致不准确)。请参阅数据类型转换以及十进制和数字。 由于您将NUMERIC(24,8)和相乘NUMERIC(24,8),并且SQL Server将仅检查类型而不是内容,因此当它无法保存所有48位精度(最大为38)时,它可能会尝试保存潜在的16位非十进制数字(24-8) )。将它们中的两个结合起来,您将获得32个非十进制数字,仅剩下6个十进制数字(38-32)。 因此原始查询 SELECT A, B, C, A + B * C FROM ( SELECT CAST(0.12345678 AS NUMERIC(24,8)) AS A, CAST(0 AS NUMERIC(24,8)) AS B, CAST(500 AS NUMERIC(24,8)) AS C ) T 减少到 SELECT A, B, C, A + D FROM ( SELECT CAST(0.12345678 AS NUMERIC(24,8)) AS A, CAST(0 AS NUMERIC(24,8)) AS B, CAST(500 AS NUMERIC(24,8)) AS C, CAST(0 AS NUMERIC(38,6)) AS D ) T 同样,在NUMERIC(24,8)和之间NUMERIC(38,6),SQL Server将尝试保存潜在的32位非十进制数字,因此A + D减少为 SELECT CAST(0.12345678 AS NUMERIC(38,6)) 它给你0.123457四舍五入后。
心有灵_夕 2019-12-29 12:54:59 0 浏览量 回答数 0

回答

这可以通过许多CASE表达式来完成: SELECT project, size, startdate, CASE WHEN mon < 2 THEN greatest(size - 50 * (1 - mon), 0) ELSE 0 END AS jan, CASE WHEN mon < 3 THEN greatest(size - 50 * (2 - mon), 0) ELSE 0 END AS feb, CASE WHEN mon < 4 THEN greatest(size - 50 * (3 - mon), 0) ELSE 0 END AS mar, CASE WHEN mon < 5 THEN greatest(size - 50 * (4 - mon), 0) ELSE 0 END AS apr, CASE WHEN mon < 6 THEN greatest(size - 50 * (5 - mon), 0) ELSE 0 END AS may, CASE WHEN mon < 7 THEN greatest(size - 50 * (6 - mon), 0) ELSE 0 END AS jun, CASE WHEN mon < 8 THEN greatest(size - 50 * (7 - mon), 0) ELSE 0 END AS jul, CASE WHEN mon < 9 THEN greatest(size - 50 * (8 - mon), 0) ELSE 0 END AS aug, CASE WHEN mon < 10 THEN greatest(size - 50 * (9 - mon), 0) ELSE 0 END AS sep, CASE WHEN mon < 11 THEN greatest(size - 50 * (10 - mon), 0) ELSE 0 END AS oct, CASE WHEN mon < 12 THEN greatest(size - 50 * (11 - mon), 0) ELSE 0 END AS nov, CASE WHEN mon < 13 THEN greatest(size - 50 * (12 - mon), 0) ELSE 0 END AS dec FROM (SELECT project, size, startdate, extract(month FROM startdate) AS mon FROM mytable) AS q; 回答来源:stackoverflow
is大龙 2020-03-21 18:10:19 0 浏览量 回答数 0

回答

找到方法了, 可以直接在客户端管理这类异常/超时的查询。 1、直接从客户端获取当前连接的进程id客户端新建jdbc连接conn时,通过执行以下sql获取该链接的后台进程pid:select procpid, start, S.client_host, S.client_port, now() - start as lap, current_queryfrom ( select backendid, pg_stat_get_backend_pid(S.backendid) as procpid, pg_stat_get_backend_activity_start(S.backendid) as start, pg_stat_get_backend_client_addr(S.backendid) as client_host, pg_stat_get_backend_client_port(S.backendid) as client_port, pg_stat_get_backend_activity(S.backendid) as current_query from (select pg_stat_get_backend_idset() as backendid) as S ) as S,( select inet_client_addr() as client_host, inet_client_port() as client_port ) as client where client.client_host = S.client_host and client.client_port = S.client_port; 2、客户端连接池管理线程记录conn<->pid的对应关系 3、当客户端发现该连接conn异常/超时后,结合digoal回答的关闭查询pid的方法,将该pid上的查询关闭。pg_cancel_backend(pid)
pg-learner 2019-12-02 01:52:14 0 浏览量 回答数 0

回答

这样的sql语句,可不可行`SELECT tmp.name,COUNT(DISTINCT tmp.shop_id) AS 'locat_sum',SUM(tmp.shop_no) AS 'shop_sum' FROM (SELECT si.name AS 'name' ,(CASE WHEN sl.parent_id=0 THEN sl.location_id ELSE sl.parent_id END) AS 'shop_id' ,(CASE WHEN sl.parent_id=0 THEN 0 ELSE 1 END) AS 'shop_no' FROM shop_info AS si LEFT JOIN shop_location AS sl ON si.location=sl.location_id ) AS tmp GROUP BY tmp.name ; `
落地花开啦 2019-12-02 01:45:23 0 浏览量 回答数 0

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除了secretGeek的答案之外,要从HSL值获得颜色(反之亦然),还可以使用以下本机函数调用(Visual Basic中的示例代码): Public Declare Sub ColorRGBToHLS Lib "shlwapi.dll" _ (ByVal clrRGB As UInteger, _ ByRef pwHue As Short, _ ByRef pwLuminance As Short, _ ByRef pwSaturation As Short) Public Declare Function ColorHLSToRGB Lib "shlwapi.dll" _ (ByVal wHue As Short, _ ByVal wLuminance As Short, _ ByVal wSaturation As Short) As UInteger
游客ufivfoddcd53c 2020-01-03 16:01:58 0 浏览量 回答数 0
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