除了SELECT选项，其他选项都不允许使用SELECT中的数据库名称，所以我这样写:

SELECT COUNT(*) AS cnt FROM information_schema.TABLES 
WHERE CONCAT(table_schema,".",table_name)="db_name.table_name";

下面是一个不是SELECT * FROM的表

SHOW TABLES FROM `db` LIKE 'tablename'; //zero rows = not exist

这是从一个数据库专家那里得到的，这是我被告知的:

select 1 from `tablename`; //avoids a function call
select * from INFORMATION_SCHEMA.tables where schema = 'db' and table = 'table' // slow. Field names not accurate
SHOW TABLES FROM `db` LIKE 'tablename'; //zero rows = does not exist

如果想要正确，请使用INFORMATION_SCHEMA。

SELECT * 
FROM information_schema.tables
WHERE table_schema = 'yourdb' 
    AND table_name = 'testtable'
LIMIT 1;

或者，您也可以使用SHOW TABLES

SHOW TABLES LIKE 'yourtable';

如果结果集中有行，则表存在。

你可以这样做:

            string strCheck = "SHOW TABLES LIKE \'tableName\'";
            cmd = new MySqlCommand(strCheck, connection);
            if (connection.State == ConnectionState.Closed)
            {
                connection.Open();
            }
            cmd.Prepare();
            var reader = cmd.ExecuteReader();
            if (reader.HasRows)
            {                             
              Console.WriteLine("Table Exist!");
            }
            else
            {                             
              Console.WriteLine("Table does not Exist!");
            }

如果存在则返回1，不存在则返回0。

set @ret = 0; 
SELECT 1 INTO @ret FROM information_schema.TABLES 
         WHERE TABLE_SCHEMA = DATABASE() AND TABLE_NAME = 'my_table'; 
SELECT @ret;

可以放入mysql函数中

DELIMITER $$
CREATE FUNCTION ExistTable (_tableName varchar(255))
RETURNS tinyint(4)
SQL SECURITY INVOKER
BEGIN
  DECLARE _ret tinyint;
  SET _ret = 0;
  SELECT
    1 INTO _ret
  FROM information_schema.TABLES
  WHERE TABLE_SCHEMA = DATABASE()
  AND TABLE_NAME = _tablename LIMIT 1;
  RETURN _ret;
END
$$
DELIMITER ;

并称之为

Select ExistTable('my_table');

如果存在则返回1，如果不存在则返回0。

上述修改后的解决方案不需要明确了解当前数据库。这样就更灵活了。

SELECT count(*) FROM information_schema.TABLES WHERE TABLE_NAME = 'yourtable' 
AND TABLE_SCHEMA in (SELECT DATABASE());