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

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。

其他回答

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;

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'

你可以用表格把一些东西组合在一起。我从来没有这样做过,但它看起来有一个列用于TABLE_ROWS和一个列用于TABLE NAME。

要获取每个表的行,你可以使用这样的查询:

SELECT table_name, table_rows
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = '**YOUR SCHEMA**';

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)
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)。