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;
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;
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。
这是我如何使用PHP计算表和所有记录:
$dtb = mysql_query("SHOW TABLES") or die (mysql_error());
$jmltbl = 0;
$jml_record = 0;
$jml_record = 0;
while ($row = mysql_fetch_array($dtb)) {
$sql1 = mysql_query("SELECT * FROM " . $row[0]);
$jml_record = mysql_num_rows($sql1);
echo "Table: " . $row[0] . ": " . $jml_record record . "<br>";
$jmltbl++;
$jml_record += $jml_record;
}
echo "--------------------------------<br>$jmltbl Tables, $jml_record > records.";
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。
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