我提出这样的质疑:

SELECT DISTINCT s.*
FROM INFORMATION_SCHEMA.STATISTICS s
LEFT OUTER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS t 
    ON t.TABLE_SCHEMA = s.TABLE_SCHEMA 
       AND t.TABLE_NAME = s.TABLE_NAME
       AND s.INDEX_NAME = t.CONSTRAINT_NAME 
WHERE 0 = 0
      AND t.CONSTRAINT_NAME IS NULL
      AND s.TABLE_SCHEMA = 'YOUR_SCHEMA_SAMPLE';

您已找到所有索引索引。

把。

按顺序获取一列中每个索引的所有索引列。

SELECT table_name AS `Table`,
       index_name AS `Index`,
       GROUP_CONCAT(column_name ORDER BY seq_in_index) AS `Columns`
FROM information_schema.statistics
WHERE table_schema = 'sakila'
GROUP BY 1,2;

裁判:http://blog.9minutesnooze.com/mysql-information-schema-indexes/

查询数据库中所有已禁用的索引

SELECT INDEX_SCHEMA, COLUMN_NAME, COMMENT 
FROM information_schema.statistics
WHERE table_schema = 'mydb'
AND COMMENT = 'disabled'

您可以使用此查询来获取指定数据库中每个表的索引名和索引号。

SELECT TABLE_NAME,
       COUNT(1) index_count,
       GROUP_CONCAT(DISTINCT(index_name) SEPARATOR ',\n ') indexes
FROM INFORMATION_SCHEMA.STATISTICS
WHERE TABLE_SCHEMA = 'mydb'
      AND INDEX_NAME != 'primary'
GROUP BY TABLE_NAME
ORDER BY COUNT(1) DESC;

SHOW INDEX FROM mytable FROM mydb;

SHOW INDEX FROM mydb.mytable;

见文档。

select
    table_name,
    index_name,
    seq_in_index,
    column_name,
    non_unique,
    index_type,
    comment
from
    information_schema.statistics
where 1=1
    and table_schema = 'my_schema'
    and table_name = 'my_table'
order by 1,2,3,4,5,6