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15

You created the table in the SYS schema (which you should never, ever do. Really, never). When you log in as teacher1 any statement looks for objects in that schema. But there is no TEACHER1.INSTRUCTORS table, because the real name is SYS.INSTRUCTORS (did I mention what a bad idea it is to create objects in the SYS schema?). You need to run select * from ...


13

You can cycle the error log by calling sp_cycle_errorlog and then that will close the current error log and cycle the log extensions. Basically, it'll create a new error log file that SQL Server will be hitting. Then the archived error log(s) can be treated accordingly (delete/move with caution). This will not technically "truncate" the log, it'll just ...


13

I hope you are using LOAD DATA INFILE. Try to use LOAD DATA LOCAL INFILE instead of LOAD DATA INFILE. Other issue might be this, please visit the following links : MySQL LOAD DATA. When you login in MySQL do like below, abdul@xmpp3:~/Desktop/Jiva$ mysql -uroot -p --local-infile Enter password: Welcome to the MySQL monitor. Commands end with ; or \g. ...


12

No. This hard-coded restriction exists for good reasons related to possible excessive query plan compilation time. You can workaround it by listing the VALUES clauses in a CTE, then INSERTing from the CTE, but I do not recommend it. Break the INSERT statement up into VALUES clauses of <= 1,000 lines each as a workaround, or use an alternative data ...


10

State codes and their meaning. 1 'Account is locked out' 2 'User id is not valid' 3-4 'Undocumented' 5 'User id is not valid' 6 'Undocumented' 7 'The login being used is disabled' 8 'Incorrect password' 9 'Invalid password' 10 'Related to a SQL login being bound to Windows domain password policy enforcement. ...


9

There are 3 methods that MySQL can use to write to the binary logs: STATEMENT This means that every SQL statement on the master is recorded in the log, and executed on the slave. This can cause problems if the SQL statement contains statements such as "NOW()", "RAND()" and anything non-deterministic. This also requires support from the storage engine in ...


8

There are 3 different items in this question: Incomplete startup packet occuring at server start is inconsequential, you may ignore it. Read Incomplete startup packet help needed (in pgsql-general mailing-list) for more. syntax error at or near "exit" at character 1 means that a client issued exit as if it was an SQL statement. The could not exec error ...


7

Based on a_horse_with_no_name's comment, I started searching around psql and found the solution: \set VERBOSITY verbose SELECT * FROM tgvbn(); ERROR: 42883: function vfjkb() does not exist ... Now that goes into .psqlrc. Details and further options can be found in the psql documentation.


5

I had the same error, trying to debug locally some application that connects to remote DBs. When using php < 5.3, everything worked. When using php 5.3 or greater, error shows up. After hours spent reading and tweaking, i realized this only occurred when connecting as a certain user. All databases i use have new password length (41). I changed the ...


5

You generally don't want to rely on implicit conversions of strings to dates. That only leads to pain and suffering since different users will have different date formats. Either use ANSI date literals or use the to_date function with an explicit format mask UPDATE bb_product SET salestart = date '2012-06-01', saleend = date '2012-06-15', ...


5

Try to repair the table: mysql> LOCK TABLES `fruitful_user_count` WRITE; mysql> CREATE TABLE `fruitful_user_count_new` LIKE `fruitful_user_count`; mysql> INSERT INTO `fruitful_user_count_new` SELECT * FROM `fruitful_user_count`; mysql> RENAME TABLE `fruitful_user_count` TO `fruitful_user_count_old`; mysql> RENAME TABLE `...


5

Your problem has to do with mysql.user and the way you upgraded to MySQL 5.6 If you look my answer to Cannot GRANT privileges as root, I show you the description of mysql.user from MySQL 4.1 to MySQL 5.6. The column plugin is column #41 in mysql.user in MySQL 5.5/5.6 mysql> SELECT column_name,ordinal_position FROM information_schema.columns -> ...


5

Well it's quite simple. You have enabled the safe update mode (as the error stated). Your statement has a where clause, right. But it's interpreted as a JOIN not as a WHERE as you provide a syntax like a INNER JOIN. Your code: UPDATE table_a a, table_b b SET a.update_me = b.update_from_me WHERE a.a_id = b.b_id; Is technically the same (and interpreted ...


4

From the Connection.setAutoCommit docs: NOTE: If this method is called during a transaction and the auto-commit mode is changed, the transaction is committed. If setAutoCommit is called and the auto-commit mode is not changed, the call is a no-op. But I don't think it's very readable/obvious in your code. You should probably simply commit before ...


4

I think the permissions are correct because otherwise you wouldn't have gotten here. My guess is that a modification you have made has caused the problem. This is particularly the case given that it is a segmentation fault. What you really need to do is look at the call stack at the time the core was dumped and see if you can isolate where in the code ...


4

Life has many great mysteries. For the MySQL DBA, one of life's greatest mysteries is your question. It's been this way going on 12 years. In fact, I found a bug report that was resolved when the disappearance of mysql.sock was caused by trying to start mysqld with mysqld already running (a.k.a. Shooting mysqld in the Foot). I have personally addressed ...


4

"MSSQL.1" is not the directory for a SQL Server 2008 R2 instance. You should verify the directory you are trying to place the database in. The default data path for a normal installation is: "C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLServer\MSSQL\Data". Edit If this is indeed the path of the databases for your SQL Server 2008 R2 instance ...


4

Suggested solutions: Execute the following command REPAIR TABLE table_name; If that (step one) doesn't help try to change the temp directory in the cnf file to new location with bigger space; If step one and two did not help you need to restore the database to previous stage.


4

'Cannot generate SSPI context' is a generic error. It can be caused by many issues, like an outaded password, clock drift, Active Directory access permissions, failure to register an SPN and so on and so forth. There is no solution to this problem. The only 'solution' is to investigate the cause, as per KB811889 and/or Troubleshooting Kerberos Errors. ...


4

What you see is completely normal and expected. Note: Because smallserial, serial and bigserial are implemented using sequences, there may be "holes" or gaps in the sequence of values which appears in the column, even if no rows are ever deleted. A value allocated from the sequence is still "used up" even if a row containing that value is never ...


4

According to Infinite recompile message in the errorlog on the SQL Programmability & API Development Team Blog, this message is triggered when a statement in a batch recompiles 100 times in a row. This message does not necessarily mean there is a problem; it exists to help troubleshoot statements that might legitimately be recompiling that often (for ...


4

According to the error message, your database contains Enterprise Edition features in the table Country. First of all, check the following on the original (old) server: SELECT SERVERPROPERTY('Edition'); That should tell you if you're running on Standard, Enterprise or Developer Edition. Developer Edition supports all of the Enterprise features (with ...


4

It appears you are putting the foreign key on the wrong table. The column PersonID on the Jobs table should reference the PersonID on the Employee table. If you are assigning persons to jobs, then your original foreign key should be on JobId. This would require adding the JobId to the Employees table and removing the PersonId from the Jobs table. If ...


4

You have several SQL Server errors in the posted message: String or binary data would be truncated. The INSERT statement conflicted with the FOREIGN KEY constraint "FK_UPC_Item". The conflict occurred in database "PosBe", table "dbo.Item", column 'Item_ID'. The INSERT statement conflicted with the FOREIGN KEY constraint "FK_VendorPartNo_Item". ...


3

The Check Database Integrity Task provided in the maintenance plan issue DBCC CHECKDB WITH NO_INFOMSGS on the database selected. You can view its command by clicking the view-SQL in the task setup. If you doubt the generated SQL command, you can use SQL profiler to see its SQL command. If corruption was found, the agent job with this maintenance task will ...


3

Try like below, it will help you... CREATE TABLE `read` (`order` MEDIUMINT NOT NULL AUTO_INCREMENT, `title` VARCHAR(50) NOT NULL, `url` VARCHAR(50) NOT NULL, PRIMARY KEY(`order`) )DEFAULT CHARSET=UTF8;


3

I'm sure there are dba.se and StackOverflow duplicates, but it was far faster for me to post this link: http://connect.microsoft.com/SQLServer/feedback/details/537419/sql-server-should-not-raise-illogical-errors In short, the Query Optimizer is really free to rewrite the plan as it sees best. Sometimes, it means a column is transformed (CAST -> int) close ...


3

Since you did this changes to the MySQL side, your only other option is to downgrade PHP. Other links support this: StackOverflow : MySQL PHP incompatibility ServerFault : mysql_connect(): The server requested authentication method unknown to the client [mysql_old_password] in Another Blog : http://datatables.net/forums/discussion/12188/notice-array-to-...


3

The solution of the problem is that table MCOUNTRY did not have the primary key. It was only configured with unique constraint that was called is the same way that primary keys. Even when the unique column was used in query Oracle could not provide expected result. Something obvious but still lesson to learn, unique constraint is not the same as primary ...


3

Synonyms have nothing to do with privileges. They are simply a way to simplify naming. The error you are getting appears to indicate that user_app does not have privileges on the user.clients table. You'd need to grant that GRANT SELECT, INSERT, UPDATE, DELETE ON user.clients TO user_app; Of course, in your actual system, I'm guessing that there ...



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