对于这个估算问题，有一点hack/workaround。

Auto_Increment -由于某些原因，如果您在表上设置了自动增量，则此函数将为数据库返回更准确的行数。

在探索为什么显示表信息与实际数据不匹配时发现了这一点。

SELECT
table_schema 'Database',
SUM(data_length + index_length) AS 'DBSize',
SUM(TABLE_ROWS) AS DBRows,
SUM(AUTO_INCREMENT) AS DBAutoIncCount
FROM information_schema.tables
GROUP BY table_schema;


+--------------------+-----------+---------+----------------+
| Database           | DBSize    | DBRows  | DBAutoIncCount |
+--------------------+-----------+---------+----------------+
| Core               |  35241984 |   76057 |           8341 |
| information_schema |    163840 |    NULL |           NULL |
| jspServ            |     49152 |      11 |            856 |
| mysql              |   7069265 |   30023 |              1 |
| net_snmp           |  47415296 |   95123 |            324 |
| performance_schema |         0 | 1395326 |           NULL |
| sys                |     16384 |       6 |           NULL |
| WebCal             |    655360 |    2809 |           NULL |
| WxObs              | 494256128 |  530533 |        3066752 |
+--------------------+-----------+---------+----------------+
9 rows in set (0.40 sec)

然后，您可以轻松地使用PHP或其他工具返回2个数据列的最大值，以给出行数的“最佳估计”。

即。

SELECT
table_schema 'Database',
SUM(data_length + index_length) AS 'DBSize',
GREATEST(SUM(TABLE_ROWS), SUM(AUTO_INCREMENT)) AS DBRows
FROM information_schema.tables
GROUP BY table_schema;

Auto Increment将始终是+1 *(表数)行，但即使有4000个表和300万行，这也是99.9%的准确性。比估计的行数好多了。

这样做的好处是，performance_schema中返回的行计数也会被擦除，因为greatest对null无效。但是，如果没有带有自动递增功能的表，这可能是个问题。

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。

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)

海报想要行计数，但没有指定哪个表引擎。对于InnoDB，我只知道一种方法，那就是计数。

我是这样摘土豆的:

# Put this function in your bash and call with:
# rowpicker DBUSER DBPASS DBNAME [TABLEPATTERN]
function rowpicker() {
    UN=$1
    PW=$2
    DB=$3
    if [ ! -z "$4" ]; then
        PAT="LIKE '$4'"
        tot=-2
    else
        PAT=""
        tot=-1
    fi
    for t in `mysql -u "$UN" -p"$PW" "$DB" -e "SHOW TABLES $PAT"`;do
        if [ $tot -lt 0 ]; then
            echo "Skipping $t";
            let "tot += 1";
        else
            c=`mysql -u "$UN" -p"$PW" "$DB" -e "SELECT count(*) FROM $t"`;
            c=`echo $c | cut -d " " -f 2`;
            echo "$t: $c";
            let "tot += c";
        fi;
    done;
    echo "total rows: $tot"
}

我对此没有任何断言，只是说这是一种非常丑陋但有效的方法，可以获得数据库中每个表中存在多少行，而不需要使用表引擎，也不需要拥有安装存储过程的权限，也不需要安装ruby或php。是的，生锈了。是的，这很重要。Count(*)是准确的。

像许多其他人一样，我很难用InnoDB在INFORMATION_SCHEMA表上获得准确的值，并且能够通过count()进行查询将无限受益，并且希望在一次查询中完成它。

首先，确保启用大规模group_concats:

SET SESSION group_concat_max_len = 1000000;

然后运行此查询以获得将为数据库运行的结果查询。

SELECT CONCAT('SELECT ', GROUP_CONCAT(table1.count SEPARATOR ',\n')) FROM (
    SELECT concat('(SELECT count(id) AS \'',table_name,' Count\' ','FROM ',table_name,') AS ',table_name,'_Count') AS 'count'
    FROM information_schema.tables 
    WHERE table_schema = '**YOUR_DATABASE_HERE**'
) AS table1

这将生成诸如…

SELECT (SELECT count(id) AS 'table1 Count' FROM table1) AS table1_Count,
   (SELECT count(id) AS 'table2 Count' FROM table2) AS table2_Count,
   (SELECT count(id) AS 'table3 Count' FROM table3) AS table3_Count;

这反过来又产生了以下结果:

*************************** 1. row ***************************
table1_Count: 1
table2_Count: 1
table3_Count: 0

下面的代码为所有故事生成选择查询。只需删除最后的“UNION ALL”选择所有结果，并粘贴一个新的查询窗口运行。

SELECT 
concat('select ''', table_name ,''' as TableName, COUNT(*) as RowCount from ' , table_name , ' UNION ALL ')  as TR FROM
information_schema.tables where 
table_schema = 'Database Name'