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.";

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.";

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)。

基于上面@Nathan的回答，但不需要“删除最终的联合”，并带有对输出进行排序的选项，我使用以下SQL。它生成另一个SQL语句，然后运行:

select CONCAT( 'select * from (\n', group_concat( single_select SEPARATOR ' UNION\n'), '\n ) Q order by Q.exact_row_count desc') as sql_query
from (
    SELECT CONCAT(
        'SELECT "', 
        table_name, 
        '" AS table_name, COUNT(1) AS exact_row_count
        FROM `', 
        table_schema,
        '`.`',
        table_name, 
        '`'
    ) as single_select
    FROM INFORMATION_SCHEMA.TABLES 
    WHERE table_schema = 'YOUR_SCHEMA_NAME'
      and table_type = 'BASE TABLE'
) Q 

您确实需要一个足够大的group_concat_max_len服务器变量的值，但是从MariaDb 10.2.4开始，它应该默认为1M。

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

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 |
+----------------+-----------------+