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: Display Data [Linuxidc @ linuxidc04 ~] $ Mongo localhost: 10000MongoDB shell version: 2.6.4Connecting to: localhost: 10000/test> Db. master. find (){"_ Id": ObjectId ("540942bed89f094a5fbd9b5a"), "uid": 1000}{"_ Id": ObjectId ("540946bcd89f094a5fbd9b5b"), "uid": 1001}{"_ Id": ObjectId ("540956b7789903d8baf6b1b3"), "uid": 1002}> Slave [Linuxidc @ linuxidc04 mongodb-linux-x86_64-2.6.4] $ mongo localhost: 10001MongoDB shell version: 2.6.4Connecting to: localhost: 10001/test> Db. master. find (){"
:1003:75ED74016B24CE3B0D672C5B9753E5FD:915F3C84BF7DEAC005BA89C7BE26F4BC:::IWAM_TCL-HQWEBAPP:1004:558D4251AB2847BE2903B96741F01C95:CEEEE45E12B17A7CAEE2C607BC04907F:::ASPNET:1007:82C09D009D6309CF14C1D8C7226E090D:20AC2CF35AFB2AA03935E74252D53CA4::10.0.0.50 hash:admin:500:AC804745EE68EBEA1AA818381E4E281B:3008C87294511142799DCA1191E69A0F:::Guest:501:NO PASSWORD*********************:NO PASSWORD*********************:::SUPPORT_388945a0:1001:NO PASSWORD*******
注：网上搜的第一篇 1001 这个就不用说了吧1002 简单的大数1003 DP经典问题，最大连续子段和1004 简单题1005 找规律（循环点）1006 感觉有点BT的题，我到现在还没过1007 经典问题，最近点对问题，用分治1008 简单题1009 贪心1010 搜索题，剪枝很关键1011 1012 简单题1013 简单题（有个小陷阱）1014 简单题1015 可以看作搜索题吧1016 经典的搜索1017 简单数学题1018 简单数学题1019 简单数学题1020 简单的字符串处理1021 找规律的数学题1022 数据结构的题（栈的应用）1023 特殊的数（Catalan Number）1024 经典DP，最大M子段和1025 经典DP，最长递增子序列（要用NLogN的方法过）1026 搜索1027 数学题（或用STL）1028 经典问题，整数拆分，用母函数做1029 简单题（一般方法容易超时）1030 简单题，可用模拟过1031 简单题1032 简单题1033 模拟题1034 Candy S
(1003) -- del 1003 and delete 1004 at the same time, because class-use (1005, deep4) exists before (AWT 1004 deep3)│ Skip-class-use (1004)│ AWT (1005)│ Skip-class-use (1006)│ Sampled-color (1007)│ Sampled-new! │ ├ ── Datatransfer (1008)│ Sampled-class-use (1009) =>Efficiency Analysis:In the message table, messageid is not stored in order, but stored in different
select 1, 90Insert into Table2 select 2,100Insert into Table2 select 3, 70 Inline: Select * From Table1 join Table2 on table1.id = table2.id------------- Result -------------ID name ID score------------------------------1 Lee 1 902 Zhang 2 100------------------------------Note: Only the Table1 and Table2 columns that meet the conditions are returned. Left join: Select * From Table1 left join Table2 on table1.id = table2.id------------- Result -------------ID name ID score---------------------
is an easy experiment. Let's take a look: SQL> shutdown abort;ORACLE instance shut down. SQL> startup mount;ORACLE instance started. Total System Global Area 1065353216 bytesFixed Size 2089336 bytesVariable Size 486542984 bytesDatabase Buffers 570425344 bytesRedo Buffers 6295552 bytesDatabase mounted. SQL> alter database open; Crash Recovery will start from alter database open. Let's observe its logs. ========================== Alert. log =Alter database openTue juns 14 18:19:53 2011Beginning c
row has been created.SQL> insert into newperson values (5,100 );One row has been created.SQL> select * from person; PID PAGE--------------------1 202 213 22SQL> select * from newperson;PID PAGE--------------------1 1004 1005 100SQL> merge into person p12 using newperson p23 on (p1.pid = p2.pid)4 when matched then5 update set p1.page = p2.page6 when not matched then7 insert (pid, page) values (p2.pid, p2.page ); The three rows have been merged.-- The
directory of adnroid sdk through the terminal, and then enter ./Android You can open sdkmanager 6. Real machine debugging First, give the system a usb configuration file. Sudo gedit/etc/udev/rules. d/50-android.rules Input SUBSYSTEM = "usb", SYSFS {idVendor} = "0bb4", MODE = "0666" Save and exit. Execute this configuration, Sudo chmod a + rx/etc/udev/rules. d/50-android.rules Then restart adb or eclipse to take effect. If you have more models, add the following content: # ACS 0502 SUBSYSTEM =
Note: Article 1 1001. 1002 simple large number 1003 typical DP problems, maximum continuous sub-segments and 1004 simple questions 1005 find a rule (Cyclic point) 1006 I feel a bit Bt. I haven't done it yet 1007 typical problem: Use the splitting function to resolve the problem recently 1008 simple questions 1009 greedy 1010 search questions, pruning is critical 1011 1012 simple questions 1013 simple questions (there is a small trap) 1014 simple ques
I. background: Although mod_fifo and mod_callcenter can be used as call center applications, it is difficult to customize these two modules in the application, here I use Lua to implement the 5-way Customer Service (1000-1004), a call center number: 9958 ADC implementation solution, which is recorded here. Ii. flowchart: Iii. Specific settings: 1. Add/usr/local/freeswitch/CONF/autoload_confilgs/modules. conf. XML to the mod_lua module, which is loade
, so its probability equals 100%.This loop structure allows us to use a small number of user connections to obtain a larger request per second.Once you have a good understanding of them, you can create a convenient alias to quickly observe the Tsung report.Vim ~/.BASHRCAlias treport= "/usr/lib/tsung/bin/tsung_stats.pl; Firefox report.html "SOURCE ~/.BASHRCand start Tsung.[Root@loadnode1 ~] Tsung StartStarting Tsung"Log directory is:/root/.tsung/log/20120421-
// CREATE PROCEDURE Delchannel (in PID int) BEGIN DECLARE PL INT; DECLARE PN INT; DECLARE AFF INT; DECLARE af INT DEFAULT 0; SET @result = null; SET @parentid = null; SET @name = null; SELECT A. ' LfT ', Ifnull (COUNT (b. ' CID '), 0), a. ' ParentID ', A. ' Name ' into the PL,PN, @parentid, @name from ' Pcms_channel ' as a left JOIN ' Pcms_channel ' as B on a. ' CID ' =b ' ParentID ' WHERE A. ' CID ' =pid GROUP by B. ' ParentID '; IF PL!PN THEN IF pl!=1 THEN START TRANSACTION; UPDA
> INSERT into Sales_range values (1004, ' John ', 1000, ' 2011-01-01 '); 1 row created. sql> INSERT into Sales_range values (1005, ' John ', 1000, ' 2011-11-01 '); 1 row created. Sql> SELECT * from Sales_range partition (P1); salesman_id salesman_name Sales_Amount sales_date ----------- --------------- ------------ ------------------- 1001 Tom 1000 2011-02-01 00:00:00 1004 John 1000 2011-01-01 00:00
() { UserInfo userInfo1 = new UserInfo () { UserId = "1001", UserName = "John", PassWord = "123456" }; Mongocollection.save (USERINFO1); UserInfo UserInfo2 = new UserInfo () { UserId = "1002", UserName = "Dick", PassWord = "123456", Detail = new Detail () { Address = "Hubei", Age = 20, Email = "Lisi@163.com" } }; Mongocollection.save (UserInfo2); UserInfo UserInfo3 = new UserInfo () { UserId = "1003", UserName = "Harry", PassWord = "123456", Detail = new Detail ()
Set concurrency mode The Entity Framework implements an optimistic concurrency pattern (optimistic concurrency model). By default, concurrency is not checked when entity update data is submitted to the database. For high frequency concurrency properties, you need to set the concurrency mode of the property to fixed. These properties are added to the WHERE clause section of the T-SQL script to compare the client's values with the values on the database side. Sample code: public void UpdatePro
(1004, ' King II ', ' Male ', ' physics Department ');INSERT into Sporter (sporterid,name,sex,department) VALUES (1005, ' Li Na ', ' female ', ' psychology Department ');INSERT into Sporter (sporterid,name,sex,department) VALUES (1006, ' Sun-hee ', ' female ', ' math Department ');2. Increase Project DataINSERT into Item (itemid,itemname,location) VALUES (' x001 ', ' Man 5-kilometer ', ' one playground ');INSERT into Item (itemid,itemname,location) V
', 50000, NULL, 10);INSERT into EMP VALUES (1004, ' Liu Bei ', ' manager ', 1009, ' 2001-04-02 ', 29750, NULL, 20);INSERT into EMP VALUES (1006, ' Guan Yu ', ' manager ', 1009, ' 2001-05-01 ', 28500, NULL, 30);INSERT into EMP VALUES (1007, ' Zhang Fei ', ' manager ', 1009, ' 2001-09-01 ', 24500, NULL, 10);INSERT into EMP VALUES (1008, ' Zhuge Liang ', ' analyst ', 1004, ' 2007-04-19 ', 30000, NULL, 20);INS
-------------------------------------------------ID Name | ID Score1 Lee | 1 902 Zhang | 2 1004 Wang | 3 70------------------------------------------------- eg. Left Join:select *from table1 left joins table2 on Table1.id=table2.id -------------Results-------------ID Name ID Score------------------------------1 Lee 1 902 Zhang 2 1004 Wang Null NULL------------------------------ Right Join:select *from table
row in the right table, all select list columns in the right table in the associated result set row are null (NULL).(2) SQL statementsSELECT * FROM table1 LEFT join table2 on Table1.id=table2.id-------------Results-------------ID Name ID Score------------------------------1 Lee 1 902 Zhang 2 1004 Wang Null NULL------------------------------NOTE: All clauses that contain table1, return table2 corresponding fields according to specified criteria, non-c
[[emailprotected] ~]# ID tomuid=1004 (Tom) gid=1004 (Tom) groups=1004 (Tom) [[emailprotected] ~ ]# ID maryuid=1006 (Mary) gid=1006 (Mary) groups=1006 (Mary), 1007 (admin) [[emailprotected] ~]# ID beanuid=1005 (bean ) gid=1005 (Bean) groups=1005 (Bean), 1007 (admin) [[emailprotected] ~]# ID jackuid=1002 (Jack) gid=1002 (Jack) groups =1002 (Jack) [[emailprotected]
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