Mysql workbench ошибка 1050 - Не ошибается лишь тот, кто ничего не делает!

I’m adding this table:

CREATE TABLE contenttype (
        contenttypeid INT UNSIGNED NOT NULL AUTO_INCREMENT,
        class VARBINARY(50) NOT NULL,
        packageid INT UNSIGNED NOT NULL,
        canplace ENUM('0','1') NOT NULL DEFAULT '0',
        cansearch ENUM('0','1') NOT NULL DEFAULT '0',
        cantag ENUM('0','1') DEFAULT '0',
        canattach ENUM('0','1') DEFAULT '0',
        isaggregator ENUM('0', '1') NOT NULL DEFAULT '0',
        PRIMARY KEY (contenttypeid),
        UNIQUE KEY packageclass (packageid, class)
);

And I get a 1050 «table already exists»

But the table does NOT exist. Any ideas?

EDIT: more details because everyone seems to not believe me

DESCRIBE contenttype

yields:

1146 — Table ‘gunzfact_vbforumdb.contenttype’ doesn’t exist

and

CREATE TABLE gunzfact_vbforumdb.contenttype(
contenttypeid INT UNSIGNED NOT NULL AUTO_INCREMENT ,
class VARBINARY( 50 ) NOT NULL ,
packageid INT UNSIGNED NOT NULL ,
canplace ENUM( '0', '1' ) NOT NULL DEFAULT '0',
cansearch ENUM( '0', '1' ) NOT NULL DEFAULT '0',
cantag ENUM( '0', '1' ) DEFAULT '0',
canattach ENUM( '0', '1' ) DEFAULT '0',
isaggregator ENUM( '0', '1' ) NOT NULL DEFAULT '0',
PRIMARY KEY ( contenttypeid ) ,

Yields:

1050 — Table ‘contenttype’ already exists

Источник

Using the command:

CREATE TABLE IF NOT EXISTS `test`.`t1` (
    `col` VARCHAR(16) NOT NULL
) ENGINE=MEMORY;

Running this twice in the MySQL Query Browser results in:

Table ‘t1’ already exists Error 1050

I would have thought that creating the table «IF NOT EXISTS» would not throw errors. Am I missing something or is this a bug? I am running version 5.1. Thanks.

Adam Wagner

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asked Oct 30, 2009 at 16:39

Works fine for me in 5.0.27

I just get a warning (not an error) that the table exists;

answered Oct 30, 2009 at 16:47

EliEli

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As already stated, it’s a warning not an error, but (if like me) you want things to run without warnings, you can disable that warning, then re-enable it again when you’re done.

SET sql_notes = 0;      -- Temporarily disable the "Table already exists" warning
CREATE TABLE IF NOT EXISTS ...
SET sql_notes = 1;      -- And then re-enable the warning again

answered Aug 8, 2010 at 12:33

gdtgdt

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You can use the following query to create a table to a particular database in MySql.

create database if not exists `test`;

USE `test`;

SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;

/*Table structure for table `test` */

CREATE TABLE IF NOT EXISTS `tblsample` (

  `id` int(11) NOT NULL auto_increment,   
  `recid` int(11) NOT NULL default '0',       
  `cvfilename` varchar(250)  NOT NULL default '',     
  `cvpagenumber`  int(11) NULL,     
  `cilineno` int(11)  NULL,    
  `batchname`  varchar(100) NOT NULL default '',
  `type` varchar(20) NOT NULL default '',    
  `data` varchar(100) NOT NULL default '',
   PRIMARY KEY  (`id`)

);

bummi

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answered Dec 18, 2015 at 7:40

Sachin ParseSachin Parse

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create database if not exists `test`;

USE `test`;

SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;

/*Table structure for table `test` */

***CREATE TABLE IF NOT EXISTS `tblsample` (
  `id` int(11) NOT NULL auto_increment,   
  `recid` int(11) NOT NULL default '0',       
  `cvfilename` varchar(250)  NOT NULL default '',     
  `cvpagenumber`  int(11) NULL,     
  `cilineno` int(11)  NULL,    
  `batchname`  varchar(100) NOT NULL default '',
  `type` varchar(20) NOT NULL default '',    
  `data` varchar(100) NOT NULL default '',
   PRIMARY KEY  (`id`)
);***

tedder42

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answered Nov 16, 2017 at 9:28

BalkishanBalkishan

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I have a solution to a problem that may also apply to you. My database was in a state where a DROP TABLE failed because it couldn’t find the table… but a CREATE TABLE also failed because MySQL thought the table existed. (This state could easily mess with your IF NOT EXISTS clause).

I eventually found this solution:

sudo mysqladmin flush-tables

For me, without the sudo, I got the following error:

mysqladmin: refresh failed; error: 'Access denied; you need the RELOAD privilege for this operation'

(Running on OS X 10.6)

answered Nov 23, 2010 at 0:05

Craig WalkerCraig Walker

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Create mysql connection with following parameter. «‘raise_on_warnings’: False». It will ignore the warning. e.g.

config = {'user': 'user','password': 'passwd','host': 'localhost','database': 'db',   'raise_on_warnings': False,}
cnx = mysql.connector.connect(**config)

answered Feb 8, 2014 at 10:25

VinayVinay

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I had a similar Problem as @CraigWalker on debian: My database was in a state where a DROP TABLE failed because it couldn’t find the table, but a CREATE TABLE also failed because MySQL thought the table still existed. So the broken table still existed somewhere although it wasn’t there when I looked in phpmyadmin.

I created this state by just copying the whole folder that contained a database with some MyISAM and some InnoDB tables

cp -a /var/lib/mysql/sometable /var/lib/mysql/test

(this is not recommended!)

All InnoDB tables where not visible in the new database test in phpmyadmin.

sudo mysqladmin flush-tables didn’t help either.

My solution: I had to delete the new test database with drop database test and copy it with mysqldump instead:

mysqldump somedatabase -u username -p -r export.sql
mysql test -u username -p < export.sql

answered Jun 22, 2017 at 4:13

rubo77rubo77

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Well there are lot of answeres already provided and lot are making sense too.

Some mentioned it is just warning and some giving a temp way to disable warnings. All that will work but add risk when number of transactions in your DB is high.

I came across similar situation today and here is the query I came up with…

declare
begin
  execute immediate '
    create table "TBL" ("ID" number not null)';
  exception when others then
    if SQLCODE = -955 then null; else raise; end if;
end;
/

This is simple, if exception come while running query it will be suppressed. and you can use same for SQL or Oracle.

answered Apr 23, 2020 at 11:16

Kunal VohraKunal Vohra

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If anyone is getting this error after a Phpmyadmin export, using the custom options and adding the «drop tables» statements cleared this right up.

answered Dec 2, 2018 at 17:44

nizz0knizz0k

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Источник

Выдаёт ошибку — #1050 — Table ‘users’ already exists.

CREATE TABLE users (
    usersId int(11) PRIMARY KEY AUTO_INCREMENT NOT NULL,
    usersEmail varchar(128) NOT NULL,
    usersUid varchar(128) NOT NULL,
    usersPwd varchar(128) NOT NULL
);

Вопрос задан

04 авг. 2022
1649 просмотров

CREATE TABLE IF NOT EXISTS users (

Потому что нельзя создать таблицу с именем существующей.

DROP TABLE users;
CREATE TABLE users (
    usersId int(11) PRIMARY KEY AUTO_INCREMENT NOT NULL,
    usersEmail varchar(128) NOT NULL,
    usersUid varchar(128) NOT NULL,
    usersPwd varchar(128) NOT NULL
);

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Минуточку внимания

Источник

If you’re getting an error that reads something like “ERROR 1050 (42S01): Table ‘customers’ already exists” when trying to create a table in MySQL, it’s probably because there’s already a table in the database with the same name.

To fix this issue, either change the name of the table you’re trying to create, or check the existing table to see if it’s the one you actually need.

Example of Error

Here’s an example of code that produces the error:

CREATE TABLE Customers (
  CustomerId int NOT NULL PRIMARY KEY,
  CustomerName varchar(60) NOT NULL
  );

Result:

ERROR 1050 (42S01): Table 'customers' already exists

In this case, I’m trying to create a table called Customers, but it already exists in the database.

Solution 1

The most obvious solution is to change the name of the table we’re creating:

CREATE TABLE Customers2 (
  CustomerId int NOT NULL PRIMARY KEY,
  CustomerName varchar(60) NOT NULL
  );

Result:

Query OK, 0 rows affected (0.02 sec)

Here, I simply renamed the table to Customers2. In practice, we would probably give it a more appropriate name.

We should also be mindful that if there’s already a table with the same name as the one we’re trying to create, there’s a possibility that our desired table has already been created. In this case we wouldn’t need to recreate it (unless we had good reason). We could either just go ahead and use it, or we could alter it to suit our new requirements.

Solution 2

Another way to deal with this error is to suppress it. We can modify our CREATE TABLE statement to only create the table if it doesn’t already exist:

CREATE TABLE IF NOT EXISTS Customers (
  CustomerId int NOT NULL PRIMARY KEY,
  CustomerName varchar(60) NOT NULL
  );

Result:

Query OK, 0 rows affected, 1 warning (0.00 sec)

In this case, we got a warning. Let’s check the warning:

SHOW WARNINGS;

Result:

+-------+------+----------------------------------+
| Level | Code | Message                          |
+-------+------+----------------------------------+
| Note  | 1050 | Table 'Customers' already exists |
+-------+------+----------------------------------+
1 row in set (0.00 sec)

The warning explicitly tells us that the table already exists.

The Table REALLY Doesn’t Exist?

If you believe that the table really doesn’t exist, perhaps there’s something else going on. See this article on Stack Overflow for a discussion on possible solutions.

Источник

Я добавляю эту таблицу:

CREATE TABLE contenttype (
        contenttypeid INT UNSIGNED NOT NULL AUTO_INCREMENT,
        class VARBINARY(50) NOT NULL,
        packageid INT UNSIGNED NOT NULL,
        canplace ENUM('0','1') NOT NULL DEFAULT '0',
        cansearch ENUM('0','1') NOT NULL DEFAULT '0',
        cantag ENUM('0','1') DEFAULT '0',
        canattach ENUM('0','1') DEFAULT '0',
        isaggregator ENUM('0', '1') NOT NULL DEFAULT '0',
        PRIMARY KEY (contenttypeid),
        UNIQUE KEY packageclass (packageid, class)
);

И я получаю таблицу 1050 «уже существует»

Но таблица НЕ существует. Любые идеи?

EDIT: подробности, потому что все, кажется, не верят мне:)

DESCRIBE contenttype

дает:

1146 — Таблица ‘gunzfact_vbforumdb.contenttype’ не существует

CREATE TABLE gunzfact_vbforumdb.contenttype(
contenttypeid INT UNSIGNED NOT NULL AUTO_INCREMENT ,
class VARBINARY( 50 ) NOT NULL ,
packageid INT UNSIGNED NOT NULL ,
canplace ENUM( '0', '1' ) NOT NULL DEFAULT '0',
cansearch ENUM( '0', '1' ) NOT NULL DEFAULT '0',
cantag ENUM( '0', '1' ) DEFAULT '0',
canattach ENUM( '0', '1' ) DEFAULT '0',
isaggregator ENUM( '0', '1' ) NOT NULL DEFAULT '0',
PRIMARY KEY ( contenttypeid ) ,

Урожайность:

1050 — Табл. ‘contenttype’ уже существует

Источник