是否有一种方法可以获得MySQL数据库中所有表的行计数,而无需在每个表上运行SELECT count() ?
当前回答
SELECT TABLE_NAME,SUM(TABLE_ROWS)
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = 'your_db'
GROUP BY TABLE_NAME;
这就是你所需要的。
其他回答
如果你知道表的数量和它们的名称,并且假设它们每个都有主键,你可以使用交叉连接结合COUNT(distinct [column])来获得来自每个表的行:
SELECT
COUNT(distinct t1.id) +
COUNT(distinct t2.id) +
COUNT(distinct t3.id) AS totalRows
FROM firstTable t1, secondTable t2, thirdTable t3;
下面是一个SQL Fiddle的例子。
还有一个选择:对于非InnoDB,它使用information_schema中的数据。TABLES(因为它更快),对于InnoDB -选择count(*)来获得准确的计数。它还会忽略视图。
SET @table_schema = DATABASE();
-- or SET @table_schema = 'my_db_name';
SET GROUP_CONCAT_MAX_LEN=131072;
SET @selects = NULL;
SELECT GROUP_CONCAT(
'SELECT "', table_name,'" as TABLE_NAME, COUNT(*) as TABLE_ROWS FROM `', table_name, '`'
SEPARATOR '\nUNION\n') INTO @selects
FROM information_schema.TABLES
WHERE TABLE_SCHEMA = @table_schema
AND ENGINE = 'InnoDB'
AND TABLE_TYPE = "BASE TABLE";
SELECT CONCAT_WS('\nUNION\n',
CONCAT('SELECT TABLE_NAME, TABLE_ROWS FROM information_schema.TABLES WHERE TABLE_SCHEMA = ? AND ENGINE <> "InnoDB" AND TABLE_TYPE = "BASE TABLE"'),
@selects) INTO @selects;
PREPARE stmt FROM @selects;
EXECUTE stmt USING @table_schema;
DEALLOCATE PREPARE stmt;
如果你的数据库有很多大的InnoDB表,计算所有行会花费更多的时间。
SELECT SUM(TABLE_ROWS)
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = '{your_db}';
从文档中注意到:对于InnoDB表,行数只是用于SQL优化的粗略估计。您需要使用COUNT(*)来获得精确的计数(成本更高)。
你可以试试这个。这对我来说很好。
SELECT IFNULL(table_schema,'Total') "Database",TableCount
FROM (SELECT COUNT(1) TableCount,table_schema
FROM information_schema.tables
WHERE table_schema NOT IN ('information_schema','mysql')
GROUP BY table_schema WITH ROLLUP) A;
像@Venkatramanan和其他人一样,我找到了INFORMATION_SCHEMA。TABLES不可靠(使用InnoDB, MySQL 5.1.44),每次运行时给出不同的行数,即使是在静态表上。这里有一种生成大型SQL语句的相对hack(但是灵活/适应性强)的方法,您可以将其粘贴到新的查询中,而不需要安装Ruby宝石之类的东西。
SELECT CONCAT(
'SELECT "',
table_name,
'" AS table_name, COUNT(*) AS exact_row_count FROM `',
table_schema,
'`.`',
table_name,
'` UNION '
)
FROM INFORMATION_SCHEMA.TABLES
WHERE table_schema = '**my_schema**';
它产生如下输出:
SELECT "func" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.func UNION
SELECT "general_log" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.general_log UNION
SELECT "help_category" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.help_category UNION
SELECT "help_keyword" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.help_keyword UNION
SELECT "help_relation" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.help_relation UNION
SELECT "help_topic" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.help_topic UNION
SELECT "host" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.host UNION
SELECT "ndb_binlog_index" AS table_name, COUNT(*) AS exact_row_count FROM my_schema.ndb_binlog_index UNION
复制粘贴,除了最后一个UNION,可以得到漂亮的输出,
+------------------+-----------------+
| table_name | exact_row_count |
+------------------+-----------------+
| func | 0 |
| general_log | 0 |
| help_category | 37 |
| help_keyword | 450 |
| help_relation | 990 |
| help_topic | 504 |
| host | 0 |
| ndb_binlog_index | 0 |
+------------------+-----------------+
8 rows in set (0.01 sec)
推荐文章
- 如何关闭mysql密码验证?
- 如何在Ruby On Rails中使用NuoDB手动执行SQL命令
- 查询JSON类型内的数组元素
- 确定记录是否存在的最快方法
- MySQL区分大小写查询
- 获得PostgreSQL数据库中当前连接数的正确查询
- 在SQL选择语句Order By 1的目的是什么?
- MySQL数据库表中的最大记录数
- 原则-如何打印出真正的sql,而不仅仅是准备好的语句?
- PHP/MySQL插入一行然后获取id
- 我如何循环通过一组记录在SQL Server?
- 如何从命令行通过mysql运行一个查询?
- 外键约束可能导致循环或多条级联路径?
- 使用LIMIT/OFFSET运行查询,还可以获得总行数
- 当恢复sql时,psql无效命令\N