use_libiconv, change the value 1 to 0 ### then make make install again
# Configuration test, test if you can run $/usr/local/coreseek/bin/indexer-C/usr/local/coreseek/etc/sphinx-min.conf.dist # The following prompts for a normal test: coreseek Fulltext 3.2 [sph00000.9.9-release (r2117)] Copyright (c) 2007-2010, Beijing choice Software Technologies Inc (http://www.coreseek.com) using Config File '/usr/local/coreseek/etc/sphinx-min.conf.dist '... total 0 reads, 0.000 sec, 0.0 KB/call
can be understood as follows:
Rollup (a, B) Statistical columns include: (a, B), (),()Rollup (a, B, c) Statistical columns include: (a, B, c), (a, B), (),()
Unless it is not the beginning, there will be a foundation. Otherwise, you will be confused about some conceptual things, so please read the following data and use the data to speak.
The above table uses group by rollup (grade, clazz) to calculate the total score and average score. First, let's look at the results and then analyze them.
, ndbcluster, federated} [innodb]-Mysql-engine-trx = STRING whether storage engine used is transactional or not {yes, no, auto} [auto]-Mysql-ssl = [on | off] use SSL connections, if available in the client library [off]-Myisam-max-rows = N max-rows parameter for MyISAM tables [0, 1000000]-Mysql-create-options = STRING additional options passed to create table []Test
1. CPU Test
[Root @ localhost bin] #./sysbench-test = cpu-max-prime = 20000 run
Sysbench 0.4.12: multi-threaded system evaluation b
SQL database Interview Questions and answers
(22:52:54)
ReprintedBytes
Summary of course number of three c courses by Ye Ping
Category: Program World
Student (S #, Sname, Sage, Ssex) Student table S #: Student ID; Sname: Student name; Sage: Student age; Ssex: Student genderCourse (C #, Cname, T #) curriculum C #, Course No.; Cname: Course name; T #: Instructor No.SC (S #, C #, score) Student table S #: Student ID; C #, course number; score: scoreTeacher (T #, Tname) Instructor table T #: Instruc
50 Common SQL statements
Student (s#,sname,sage,ssex) Student table
Course (c#,cname,t#) timetable
SC (s#,c#,score) score sheet
Teacher (t#,tname) Teacher's Table
Problem:
1, Query the "001" Course than "002" course scores of all students of the school number;
Select a.s# from (select S#,score from SC where c#= ' 001 ') A, (select S#,score
From SC where c#= ' 002 ') b
where A.score>b.score and a.s#=b.s#;
2, the average score is greater than 60 students of the school number and average score;
S
Student (s#,sname,sage,ssex) Student tableCourse (c#,cname,t#) timetableSC (s#,c#,score) score tableTeacher (t#,tname) Teacher tableProblem:1, the inquiry "001" course is higher than "002" of all students of the school number;Select a.s# from (select S#,score from SC where c#= ' 001 ') A, (select S#,scoreFrom SC where c#= ' 002 ') bwhere A.score>b.score and a.s#=b.s#;2, the query average score is more than 60 points of the student's number and average score;Select S#,
without using magic methodsSequential execution of script n times under Php5.6.26-1Average/minimum/maximum of statistical execution timeSequential execution of script n times under Php7.0.12-2Average/minimum/maximum of statistical execution timeAt present, personal ability is limited, only in this way, if you have a better plan or advice can tell me, thank you, haha~
Test
__construct
First, let's take a look at the experiment of the constructor __construct, the PHP script is as follows:
PHP5.6
network connectivity is good, the lab is successful, and the network can communicate, to achieve higher security, you must use other protocols and other devices. R1 # ping 192.168.3.2Type escape sequence to abort.Sending 5, 100-byte ICMP Echos to 192.168.3.2, timeout is 2 seconds:!!!!!Success rate is 100 percent (5/5), round-trip min/avg/max = 63/85/94 MSR1 # ping 192.168.3.1Type escape sequence to abort.Sending 5, 100-byte ICMP Echos to 192.168.3.1,
study the page exchange activity further. You may need to increase the memory to reduce the page feed requirement (you can multiply this number by 4 K to get the hard disk data traffic caused by this ). A large value of pages/sec does not necessarily indicate memory problems, but may be caused by running programs that use memory ing files.
Page read/sec: Page hardware fault, a subset of page/sec. In order to parse the reference to memory, the number of times page files must be read. The thresho
from student 1, with an average score of 73.67. Note that transaction 1 is not committed at this time.
private static double ReadAverageMarksByTransaction1(){ return (double)ExecuteScalar("SELECT AVG(mark) AS AvgMark FROM SC WHERE (id = 1)");}private static int ReadTotalCoursesByTransaction1(){ return (int)ExecuteScalar("SELECT COUNT(*) AS num FROM SC WHERE (id = 1)");}private static object ExecuteScalar(string command){ Console.WriteLine("-- E
salary of employees with salaries greater than 500 in each department */Select AVG (SAL) from Scott. EMP where Sal gt; 500 group by deptno/* 13. collect statistics on departments with an average salary greater than 500 */Select deptno from Scott. EMP group by deptno having AVG (SAL) gt; 500/* 14. Calculate the employee bonus for which the Department 30 receives the maximum bonus */Select max (Comm) from
* FROM scott.emp where sal>2500 union select * from Scott.emp where job= ' ANALYST '10. Through the Intersect set operation, query the employee information for which the salary is greater than 2500 and work as analyst.SELECT * from scott.emp where sal>2500 intersectselect * from scott.emp where job= ' ANALYST11. Use the minus collection to query for wages greater than 2500, but work is not employee information for analyst.SELECT * FROM Scott.emp where Sal>2500minusselect * from scott.emp where
.
private static void Setup(){conn1 = new SqlConnection(connectionString);conn1.Open();tx1 = conn1.BeginTransaction(IsolationLevel.RepeatableRead);conn2 = new SqlConnection(connectionString);conn2.Open();tx2 = conn2.BeginTransaction(IsolationLevel.RepeatableRead);}
Step 2: Transaction 1 reads the average score of students whose student ID is 1 and the number of courses they have learned. At this time, I learned three courses from student 1, with an average score of 73.67. Note that transaction 1
Aggregate functions, multiline functions, and grouping functions are all types of functions.Group by and having
Group function: AVG \ sum \ min \ MAX \ count \ stddev \ varianceUse distinct in combination with the group FunctionProcessing of null values in the group Function
Nested group Function
Syntax of the group function:Select group_function (column ),...From table[Where condition][Order by column];
--- Example 1:
learned three courses from student 1, with an average score of 73.67. Note that transaction 1 is not committed at this time.
private static double ReadAverageMarksByTransaction1(){ return (double)ExecuteScalar("SELECT AVG(mark) AS AvgMark FROM SC WHERE (id = 1)");}private static int ReadTotalCoursesByTransaction1(){ return (int)ExecuteScalar("SELECT COUNT(*) AS num FROM SC WHERE (id = 1)");}private static object ExecuteScalar(string command){ C
the search criteria of transaction 1, when transaction 1 reads data again based on the same conditions, it finds that there are more records. Let's use the following cases to reproduce the phantom read problem:
Step 1: set both transaction 1 and transaction 2 to the repeatableread isolation level and start the transaction at the same time.
private static void Setup(){conn1 = new SqlConnection(connectionString);conn1.Open();tx1 = conn1.BeginTransaction(IsolationLevel.RepeatableRead);conn2 = new
file: no such file or directory solution: # In-S/usr/local/mysql-5.5.8/lib/libmysqlclient. so.16/lib64/libmysqlclient. so.16Iii. test options
[Root @ localhost bin] # sysbench test case: sysbench [General-options]... -- test = Iv. Test the instance test environment:
CPU: Name: Intel (r) Xeon (r) CPU e5606 @ 2.13 GHzMemory: 4 GBSystem: rhel5.4 x86DB: percona-5.5.18
1. Test CPU
# sysbench --test=cpu --cpu-max-prime=2000 runMaximum prime number checked in CPU test: 2000Test execution summary: tota
Avg. Disk Queue LengthAvg. avg.Queue queuesLength lengthIt refers to the current disk queue length, and the popular point is that it reflects the time it takes for the disk to complete the request. A higher value indicates that the disk controller has repeatedly retried the disk because of a failure. These failures increase the average disk transfer time.======================================Add, OK, say the exact point.Simplicity can be understood as
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