federated authentication

.... Cool And then follow the same line of thought to some of the other common SQL has been optimized, the effect immediately effective After 30 minutes to check the slow SQL records file, not good, found that a good SQL became gray often slow, God horse situation? After analyzing and testing the reason for adding a federated index, and having an or in the SQL statement, the problem is excluded when this or union is used. This time to come to a l

(+)----results equivalent to outer joinsUse of data dictionariesSelf-connect (cheat yourself): select * from Z_course C1, Z_course C2 where C1.cour_code=c2.p_cour_codeHierarchical queryConnect by PRIOR Connection condition start with start condition　　　　　　　　　　　　SELECT * FROM Z_course t connect by prior T.cour_code=t.p_cour_code start with t.cour_name like '% International 'Pseudo-column:Level: plus a column of display levels　　　　　　　　　　　　　　Select t.*, level from Z_course t connect by prior T.cour_

151510 1450 Select Se.last_name,se.salary,se.dept_id,temp.avgsalFrom S_emp SE, (Select Dept_id,avg (Salary) AvgsalFrom S_empGROUP BY dept_id) Tempwhere se.dept_id = temp.dept_idand Se.salary > temp.avgsal; 1. Query for employee information2. For example, to find out the staff information in Sector 41st, which is higher than the average salary of department 41st (this department) Select AVG (Salary), dept_idFrom S_empGroup BY DEPT_ID; Select last_name,salary,dept_idFrom S_emp; Select S1.last_na

join ... On and Traditional Association query queries, the SID appears two times. 4.LEFT OUTER JOIN ... On "SELECT * from students left OUTER JOIN transcript USING (SID) where transcript.date=2015-6"; Select the table students all the data and add the transcript data to the students table, and if the transcript does not meet the criteria, it is not added to the result table, and NULL indicates 5.RIGHT OUTER JOIN ... On "SELECT * from students right OUTER JOIN transcript USING (SID) where transc

Tags: problem SP c new SQL table BS r ROMIn a subquery, if you want to implement the following functions:Select Lib,count (*), select SUM (NEWSNO) from Table1 GROUP by LibFrom Tabel1 t1,table2 T2 where T1.newsno =t2.newsnoGroup by LibYou will be prompted that the subquery returns more than one value. "Error, meaning that a subquery cannot return multiple results, only one result can be returned.Therefore, you can use the following method instead:Select Lib,count (*), select SUM (newsno) from Tab

Briefly in the previous article about the ES synchronization data script roughly, but the actual situation is certainly not as simple as the script in the previous article. For example, I now have three tables, two entity tables, and an association table. The approximate implementation is as follows:The bin directory establishes a Statefile.json file:{ ' type ': ' JDBC ', "Statefile": "Statefile.json", "JDBC": { "url": "jdbc:mysql://", "User": "", "Password": "",

, you cannot use the prefix index for both ORDER by and GROUP by, and you cannot use the prefix index for overwrite scans. # 语法 ALTER TABLE table_name ADD KEY(column_name(prefix_length)); # 示例 ALTER TABLE city ADD KEY(cityname(7)) 1 2 3 4 Overwrite IndexA bit similar to the federated Index, which is to fetch the data only by reading the index when querying T, without having to query the related table two times. The leaf no

INNER JOIN: (INNER join)Inner joins use comparison operators to match rows in two tables based on values in the universal columns of each table(1) SELECT ... From table 1INNER JOIN table 2ON ...SELECT S.sname,c.courseid,c.scoreFrom score as CINNER JOIN Students as SOn c.studentid = S.scode(2) SELECT ... From table 1, table 2WHERE ...SELECT Students.sname, Score.courseid, Score.scoreFrom Students,scoreWHERE Students.scode = Score.studentidOuter joins:(1) LEFT OUTER join: The data of the left tabl

Tags: info create tab creat ONS int const NEW Create1. Create PRIMARY KEY syntaxALTER TABLE table_name ADD CONSTRAINT pk_name PRIMARY KEY (column name);2. Creating FOREIGN KEY syntaxALTER Table news_info[child table name] ADD CONSTRAINT fk_news_info_news_type[constraint name] FOREIGN KEY (info_id) [Child table column] REFERENCES news_type[ primary table name] (ID) [Main table column];3. Using the combined primary keyIf a column cannot uniquely distinguish a table of records, consider multiple co

Label:Group:SELECT * FROM (SELECT row_number () over (ORDER by ProjId) as row_number,* from (Select Projapplydate,projproperty,projid,projstatuscurrent,projguid,projetpid,projregionid,projname,projaddress , Projareasize,projcost,projcreatetime,projchargeperson,projchargepersonphone,projapplyetp,projapplyperson, Projapplypersonphone,projconstructetp,projconstructpm,projconstructpmphone,projdirector,projdirectorphone, Projsafeorgz,projcode,projisenable, Min (consigndate) as Consigndate,max (conbal

1. Simple manual correlationFirst, the results are queried and put into a variable, and then queryu = Db.user.findOne ({author: "Wangwenlong"});For (var p = db.postings.find ({author:u.author});p. Hasnext;) {Printjson (P.next (). title); }2, Dbref Way AssociationExample:Step 1 Get current user informationDb.user.insert ({name: "Wangwenlong"})u1 = Db.user.find ({name: "Wangwenlong"}) [0]Step 2 post and associateDb.postings.insert ({"title": "Hello mongodb!"},users:[new dbref (' users ', u1._id)])

First, the demandUser answer, a total of 3, must be 3 questions are completed to submit.Objective: To query the user answered several questions correctly, answered several wrong questions. (Of course, you can only find the correct number of questions, with 3 minus the number of answers to the wrong question)Second, SQLSelect * from(Select Count(1) Rightcount from(Select * from(Select * 　　　　 fromMic_td_exercise_push_problem MPwheremp.push_id= 347 　　　　 O

Label:Remember a very long SQL, in order to facilitate reading, I have added the paragraph: SELECT * from ( (SELECT 1 asAcheck, S1.id, S1.plan_name, S1.cost_mode, S1.sta_type, S1.cost_type, S1.cost_period_time, S1.cost_period_value, S1. ' Carrier_operator ', S1.cost_settle_type fromWo_serviceplan S1JOINWo_serviceplan_hotspot SH onS1.id=sh.serviceplan_idWHERES1.is_enable= 1 andsh.hotspot_id= 109 ORDER byS1.plan_nameASC ) UNION All (SELECT 2 asAcheck, S2.id, S2.plan_name, S2.cost_m

Original source: http://blog.csdn.net/dba_huangzj/article/details/38657111, featured catalogue:http://blog.csdn.net/dba_huangzj/article/details/37906349 No person shall, without the consent of the author, be published in the form of "original" or used for commercial purposes, and I am not responsible for any legal liability. Previous article: http://blog.csdn.net/dba_huangzj/article/details/38656615 Objective: SQL Server has two types of authentication