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
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**';
其他回答
简单的方法:
SELECT
TABLE_NAME, SUM(TABLE_ROWS)
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = '{Your_DB}'
GROUP BY TABLE_NAME;
结果示例:
+----------------+-----------------+
| TABLE_NAME | SUM(TABLE_ROWS) |
+----------------+-----------------+
| calls | 7533 |
| courses | 179 |
| course_modules | 298 |
| departments | 58 |
| faculties | 236 |
| modules | 169 |
| searches | 25423 |
| sections | 532 |
| universities | 57 |
| users | 10293 |
+----------------+-----------------+
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)
我不知道为什么这么难,但这就是生活。 下面是执行实际计数的bash脚本。只需将其保存为(例如count_rows.sh),使其可执行(例如chmod 755 count_rows.sh),并运行它(例如。/count_rows.sh)
#!/bin/bash
readarray -t TABLES < <(mysql --skip-column-names -u myuser -pmypassword mydbname -e "show tables")
# now we have an array like:
# TABLES='([0]="customer" [1]="order" [2]="product")'
# You can print out the array with:
#declare -p TABLES
for i in "${TABLES[@]}"
do
#echo $i
COUNT=$(mysql --skip-column-names -u username -pmypassword mydbname -e "select count(*) from $i")
echo $i : $COUNT
done
这是我获得实际计数的方法(不使用模式)
它更慢,但更准确。
这个过程有两步
获取数据库的表列表。你可以使用它 Mysql -uroot -p mydb -e“显示表” 在这个bash脚本中创建表列表并将其分配给数组变量(与下面的代码一样,用一个空格分隔) 数组=(table1 table2 table3) ${array[@]}中的I 做 echo $我 Mysql -uroot mydb -e "select count(*) from $i" 完成 运行该程序: Chmod +x script.sh;。/ script.sh
如果你使用数据库information_schema,你可以使用下面的mysql代码(where部分使查询不显示行为空值的表):
SELECT TABLE_NAME, TABLE_ROWS
FROM `TABLES`
WHERE `TABLE_ROWS` >=0