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code as follows: #重启数据库 Sqlplus '/as sysdba ' Sql>shutdown Immediate Sql>startup #创建一个临时表空间temp02, used as a temporary replacement Sql>create temporary tablespace temp02 tempfile '/u2/oradb/oradata/topprod/temp02.dbf ' size 10M autoextend on next 10M ; #将系统临时表空间指向temp02 Sql>alter database default temporary tablespace temp02; #删除原来的临时表空间temp Sql>drop tablespace temp including contents and datafiles; #创建新的临时表空间temp Sql>cr

The "preface" is explained in the blog post:This article will introduce the User and table space creation related commands through personal tone, in the current time point "May 16, 2017 " under the grasp of the technical level is limited, there may be a lot of knowledge understanding is not enough in-depth or comprehensive, I hope you point out the issue of common exchange, In the follow-up work and study, if found that the content of this article and the actual situation is biased, will improve

=512mbulk_insert_buffer_size =64mmyisam_sort_buffer_size=64mmyisam_max_sort_file_size=10gmyisam_repair_ Threads=1myisam_recoverlong_query_time=2slow_query_logslow_query_log_file=/data/mysql/slow.logskip-lockingskip-name-resolveserver-id= 1innodb_additional_mem_pool_size=16minnodb_buffer_pool_size=512minnodb_data_file _path=ibdata1:10m:autoextendinnodb_file_io_threads=4innodb_thread_concurrency= 8innodb_flush_log_at_trx_commit=1innodb_log_buffer_size=16minnodb_log_file_size =128minnodb_log_files_

When executing a stored procedure, we may encounter permission problems.● Definer permission Stored Procedure● Caller permission Stored ProcedureWhen creating a stored procedure in a database, the User-Defined permission is in the default mode.After the AUTHID CURRENT_USER keyword is specified, it is the stored procedure of caller permission.The most fundamental difference between them is whether role can take effect in the stored procedure.(I) Permission stored procedure of the definerThe role

classHandleajax {@RequestMapping ("/ajax.do") Public voidHandleajax (httpservletresponse response)throwsexception{//Virtual out someDataMapNewHashmap(); UserInfo U1=NewUserInfo (); U1.setage (12); U1.setname ("Congratulations on employment"); UserInfo U2=NewUserInfo (); U2.setage (122); U2.setname ("Smooth Employment"); Map.put ("001", U1); Map.put ("001",

This problem actually seems difficult, but listen to Werkeytom_ftd said can use block chain water, so it is very happy to hit a block chain of the Chairman tree, insert operation directly into a block, note if the size of the block 2*block will block apart, notice that each modification or insert to modify the subsequent state, Post code:#include #include #include #include #include #define FO (i,a,b) for (int i=a;i#define FD (I,A,B) for (int i=a;i>=b;i--)Using namespace Std;constintN =35010; con

; ++tail; Q[tail]=s; B[s]=false; while (Head { ++head; int U=h[q[head]]; while (u!=0) { if (W[u]+dis[q[head]] { Dis[to[u]]=dis[q[head]]+w[u]; Pre[to[u]]=u; if (B[to[u]]) { B[to[u]]=false; ++tail; Q[tail]=to[u];}}U=next[u];}B[q[head]]=true;}if (dis[t]}void Get (){ int now=t; int mx=0x7fffffff; while (Now!=s) { Mx=min (Mx,flow[pre[now]); Now=from[pre[now]];}now=t;while (Now!=s){Ans+=mx*w[pre[now]];FLOW[PRE[NOW]]-=MX; FLOW[PRE[NOW]^1]+=MX;Now=from[pre[now]];}}int main (){Len=1; s=0;N=read (); M=

].v.push_back (p2); Sharp[tot].v.push_back (p3); Sharp[tot].v.push_back (p4); } if(s==" Line") { for(intI=1; i2; i++) {scanf ("(%d,%d)",a,b); Point P (1.0*a,1.0*b); Sharp[tot].v.push_back (P); } } if(s=="Triangle") { for(intI=1; i3; i++) {scanf ("(%d,%d)",a,b); Point P (1.0*a,1.0*b); Sharp[tot].v.push_back (P); } } if(s=="Polygon") { intx; scanf ("%d",x); for(intI=1; i) {scanf ("(%d,%d)",a,b); Point P (1.0*a,1.0*b); S

Intersects an edge of a rectangle, the output tInternally, the output TThe title is not correct, the background data of the rectangular coordinate is not guaranteed to be the upper left and lower right corner, it may be the lower left and upper right corner, so pre-processing a bit.#include #include#include#include#include#include#includeusing namespacestd;intT;DoubleXstart,ystart,xend,yend;DoubleX1,y1,x2,y2;Doublea,b,c,d;Const Doubleeps=1e-8;#defineZero (x) (((x) >0? ( X):(-X)) structpoint{Doub

? Union U{Char S [9];Int N;Double D;}; S occupies 9 bytes, N occupies 4 bytes, and D occupies 8 bytes. Therefore, it requires at least 9 bytes of space. However, the actual size is not 9, and the size is 16 by the sizeof operator. This is because of the byte alignment problem. 9 cannot be divisible by 4 or 8. Therefore, adding byte to 16 is in line with the alignment of all members. From this we can see that the space occupied by a consortium depends not only on the wi

I. Prepare partitions /Dev/hda1 (ext2/ext3 format) 2. Prepare the user directory Add User u1, U2 # Useradd u1 # Passwd u1 # Useradd U2 # Passwd U2 Iii. Mount (1) manual mounting Mount-O usrquota, maid/dev/hda1/mnt/hda1 (2) automatic mounting # Vi/etc/fstab Add /Dev/hda5/mnt/hda5 ext3 defaults, usrquota, kgquota 0 0 Iv. Directory preparation # Mkdir/mnt/hda5/u1dir

View the sizeof problem of unionCPUPeer Consider the following: (default alignment) Union u{Double;Int B;};UnionU2{Char A [13];Int B;};Union U3{Char A [13];Char B;};Cout Cout Cout We all know that the size of Union depends on the size of one member that occupies the largest space among all its members. So for u, the size is the largest double type member A, so sizeof (u) = sizeof (double) = 8. However, for U2 and U3, the

)/2; if(pmid) Link (T[U].LC,L,MID,P); ElseLink (t[u].rc,mid+1, r,p);}voidMergeintAMP;U1,intU2) { if(u1==0) {u1=u2;return ;} if(u2==0)return ; T[U1].C+=t[u2].c; Merge (T[U1].LC,T[U2].LC); Merge (t[u1].rc,t[u2].rc);}intFindans (intU1,intU2,intLintRintk) { if(t[u1].

To understand YUV422, first look at YUV444 YUV 4:4:4YUV Three channel sampling rate is the same, so in the generated image, the three component information of each pixel is complete, that is, each pixel is represented by three elements (each component is usually 8 bits), after 8 bit quantization, uncompressed each pixel occupies 3 bytes.The following four pixels are: [Y0 U0 V0] [Y1 U1 V1] [Y2 U2 V2] [Y3 U3 V3] A total of 12 bytes, representing only 4

bytes, n occupies 4 bytes, and d occupies 8 bytes. Therefore, it requires at least 9 bytes of space. However, the actual size is not 9, and the size is 16 by the sizeof operator. This is because of the byte alignment problem. 9 cannot be divisible by 4 or 8. Therefore, adding byte to 16 is in line with the alignment of all members. From this we can see that the space occupied by a consortium depends not only on the widest member, but also on all Members, that is, the size of the consortium must

1. Overview A class file is a group of 8-byte binary streams. Each data item is arranged strictly in a sequential and compact manner in the class file without any separators, this makes almost all the content stored in the entire class file necessary for the program to run. When a data item needs to occupy more than 8 bytes of space, it will be separated into multiple 8-bit bytes for storage in the way of the top position. The class file uses a pseudo structure similar to the C-language structur

1. Saving between many-to-many tablesWay One: U2 = User.objects.get (id=2) g2 = Group.objects.get (id=2) U2.group_relation.add (G2) #在关系表中存入关联数据Way two:name = Request. Post.get (' name ', None) Username = Request. Post.get (' username ', None) password = Request. Post.get (' password ', None) gplist = Request. Post.getlist (' groupId ') user = Users.objects.create (name=name,username=username,pa

should be a no-op.*/ - Public voidFollow (intFollowerid,intFolloweeid) { in User U1; the User U2; the if(Users.containskey (Followerid)) { Aboutu1 =Users.get (Followerid); the}Else{ theu1 =NewUser (Followerid); the } + - if(Users.containskey (Followeeid)) { theU2 =Users.get (Followeeid);Bayi}Else{ theU2 =NewUser (Followeeid); the } - U1.followees.add (U2); -

= sessionFactory Based on sqlSessionFactory. openSession (); String statement = "one. to. one. mapper. ordersMapper. selectOrderAndUserByOrderID "; UserMapper userMapper = sqlSession. getMapper (UserMapper. class); // For the first query, issue an SQL statement and put the query result into the cache. User u1 = userMapper. selectUserByUserId (1); System. out. println (u1); // The second query. Because it is the same sqlSession, the query results will be searched in the cache. // If yes, the que

If you are a lazy programmer, you may be annoyed when you see the following code: Abstract class U { } Class u1 extends U { Public static function create (){ Return new u1 (); } } Class u2 extends U { Public static function create (){ Return new u2 (); } } This code works normally, but it is annoying to have a lot of repeated code. I don't want to add the create method to each subclass. If you put the creat