error 1062

Modify the mysql configuration file, etcmy. cnf. Add a line of slave_skip_errors1062 under [mysqld] to save and restart the mysql service. Check master-slave replication again. Modify the mysql configuration file/etc/my. cnf. Add a line of slave_skip_errors = 1062 under [mysqld] to save and restart the mysql service. Check master-slave replication again. Environment Description: Mysql> show slave stsatus \ G; The

Tags: last_sql_errno:1062I. Description of the Environment1. CentOS 6.62.mysql 5.6.253. Gtid-based replicationIi. Description of the exceptionMistakenly inserts a data from the node as the master node, the same data is inserted in the primary, from the node insert, causing the primary key conflict, slave thread exception.Third, the processing process1. View the error message([email protected]) [(none)]> show Slave status\g;1. Row ********************

Last_Errno: 1062, Last_Error, it will take time to check the problem. 1. Slave Database error message: Mysql> show slave status/G* *************************** 1. row ***************************Slave_IO_State: Waiting for master to send eventMaster_Host: xxxx0402.china.online.ea.comMaster_User: replicationMaster_Port: 3306Connect_Retry: 60Master_Log_File: mysql-bin.000154Read_Master_Log_Pos: 56680675Relay_

, INSERT INTO An error occurs when executing a statement when trying to insert an item name that is the same as shown in the two identical (case-only difference)!One of the workarounds is to change the two field types to binary:ALTER TABLE book_entity_matches MODIFY COLUMN rg_project_name VARCHAR(100) BINARY NOT NULL;ALTER TABLE book_entity_matches MODIFY COLUMN rf_project_name VARCHAR(100) BINARY NOT NULL;Other methods include specifying that the fie

Online environment I have never encountered 1062 of the problem, the test environment development environment is constantly similar problems, in the past in order to catch the time is skip or directly redo, this will have time, just to check the problem 1 error message from library: Mysqlgt; Show Slave Status\g1. Row ***************************Slave_io_state:waiting for Master to send eventMaster_Host:x

Tags: space write air var traffic work care part forThe site suddenly appears with the following error: MySQL Error: Duplicate entry ' 1 ' for key ' views 'MySQL Errno: 1062Message: Duplicate entry ' 1 ' for key ' views ' I use Myisamchk Cdb_posts.myi-r Fixed the table, but the contents of the table views were emptied, fortunately, my side is the article Access scale, not very important, before also made a

Title Address: http://poj.org/problem?id=1062Beginning to understand the error, thought that as long as the level of the adjacent node is not greater than the level limit on OK, it is necessary in this scenario, the node level difference must not be greater than the level limit.Problem-solving ideas: Because must contain the number 1th node (Chiefs), so all about the 1th nodes within the scope of the list, each time is a different diagram, take the sm

Nonsense not to say, the first question:Text ReverseTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 17744 Accepted Submission (s): 6730Problem Descriptionignatius likes to the write words in reverse. Given a single line of text which are written by Ignatius, you should reverse all the words and then output them.Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T tes

1062. Talent and Virtue (25)About years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people ' s talent and VI Rtue. According to He theory, a man being outstanding in both talent and virtue must is a "sage (sage)"; Being less excellent but with one's virtue outweighs talent can be called a "nobleman (gentleman)"; Being good in neither are a "fool Man (Fool)"; Yet a fool man was better than a "small man (villain)"

Topic 1062: Segmented functions time limit:1 seconds Memory limit:32 MB Special question: No submitted:2877 Resolution:1671 Title Description: Write a program that calculates the value of the following segment function, Y=f (x).y=-x+2.5; 0y=2-1.5 (x-3) (x-3); 2y=x/2-1.5; 4 Input: A floating-point number n Output:

Http://poj.org/problem? Id = 1062 This question was initially determined to be violent .. But n Click Submit! The shortest path... The diagram is clever. The required item corresponding to each item is equivalent to a constraint. we can connect the required item to a "discount" side to this item, and the cost is the preferential price. Then, a fee is charged for each item from a source. In this way, many paths of the source-the first item are formed.

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1062 Solution Report: note that the end of a row may be a space, and remember to use a carriage return character for getchar. 1 #includeView code HDU 1062 text reverse (water question, string processing)

9-degree OJ 1062 piecewise functions (simulation) Question 1062: piecewise Functions Time Limit: 1 second Memory limit: 32 MB Special question: No Submit: 2877 Solution: 1671 Description: Compile a program and calculate the value of the following piecewise Function y = f (x.Y =-x + 2.5; 0 Y = 2-1.5 (X-3) (X-3); 2 Y = x/2-1.5; 4 Input: A floating poin

Topic 1062: Segmented functions time limit:1 seconds Memory limit:32 MB Special question: No submitted:2892 Resolution:1682 Title Description: Write a program that calculates the value of the following segment function, Y=f (x).y=-x+2.5; 0y=2-1.5 (x-3) (x-3); 2y=x/2-1.5; 4 Input: A floating-point number n Output:

1062 maximum number of questions in a sequence Source: Ural 1079 reference time limit: 1 second space limit: 131072 kb score: 10 difficulty: 2 level algorithm question description There is a sequence:A [0] = 0A [1] = 1A [2I] = A [I]A [2I + 1] = A [I] + A [I + 1]Enter n to calculate the maximum number in a [0]-A [n.A [0] = 0, a [1] = 1, a [2] = 1, a [3] = 2, a [4] = 1, A [5] = 3, a [6] = 2, a [7] = 3, a [8] = 1, a [9] = 4, A [10] = 3.For example, n = 5

No way. If you want to check the number of questions, you have to go crazy about a simple question ~~~ Hdoj 1062 text reverse http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1062 Simple String Control seems to have been done before, but it hasn't been checked in the question bank ..... A was wrong several times during training before it was very difficult. Today, I am confident that I will lose my skills in

Apple TV and iOS Local Security Restriction Bypass Vulnerability (CVE-2015-1062) Release date:Updated on: Affected Systems:Apple TV Apple iOS Description:Bugtraq id: 73003CVE (CAN) ID: CVE-2015-1062 IOS is an operating system developed by Apple for mobile devices. It supports iPhone, iPod touch, iPad, and Apple TV. In versions earlier than Apple TV 7.1 and earlier than iOS 8.2, MobileStorageMounter does

1062 Route Selectiontime limit: 1 sspace limit: 128000 KBtitle level: Diamonds DiamondTitle DescriptionDescriptionIn network communication, the shortest path is often required. However, there is a problem with the shortest path transfer: If there is only one shortest path given between the two end nodes. The signal transmission is at risk of interruption in the event of any node or link failure in the path. As a result, the following improvements have

Expensive offer time limit: 1000 msmemory limit: 10000 kbthis problem will be judged on PKU. Original ID: 1062 64-bit integer Io format: % LLD Java class name: Main the young explorer came to an Indian tribe. There he fell in love with the daughter of the chief, so he went to ask for advice from the Chief. The chief promised to marry his daughter only when he used 10000 gold coins as a dowry. The explorer asked the chief to lower his requirements if h

= 0;for (int i=1; i{Dist[i] = Price[0][i];}for (int i=1; i{int mmin = INF;int k = 0;for (int j=1; j{if (!vis[j] dist[j] {Mmin = Dist[j];K = J;}}if (k = = 0) break;Vis[k] = true;Temp_p + = Mmin;for (int j=1; j{if (!vis[j] price[k][j] > 0 dist[j] > Dist[k] + price[k][j])DIST[J] = Dist[k] + price[k][j];}}return dist[1];}int main (){while (Cin >> m >> N){Init ();for (int i=1; i{CIN >> Price[0][i] >> lv[i] >> x[i];for (int j=0; j{int T, V;Cin >> T >> v;Price[t][i] = v;}}for (int i=1; i{for (int j