Основная проблема с проверенным ответом состоит в том, что вам придется анализировать выходные данные, чтобы получить информацию. Вот запрос, позволяющий получить их более удобным способом:
SELECT cols.TABLE_NAME, cols.COLUMN_NAME, cols.ORDINAL_POSITION,
cols.COLUMN_DEFAULT, cols.IS_NULLABLE, cols.DATA_TYPE,
cols.CHARACTER_MAXIMUM_LENGTH, cols.CHARACTER_OCTET_LENGTH,
cols.NUMERIC_PRECISION, cols.NUMERIC_SCALE,
cols.COLUMN_TYPE, cols.COLUMN_KEY, cols.EXTRA,
cols.COLUMN_COMMENT, refs.REFERENCED_TABLE_NAME, refs.REFERENCED_COLUMN_NAME,
cRefs.UPDATE_RULE, cRefs.DELETE_RULE,
links.TABLE_NAME, links.COLUMN_NAME,
cLinks.UPDATE_RULE, cLinks.DELETE_RULE
FROM INFORMATION_SCHEMA.`COLUMNS` as cols
LEFT JOIN INFORMATION_SCHEMA.`KEY_COLUMN_USAGE` AS refs
ON refs.TABLE_SCHEMA=cols.TABLE_SCHEMA
AND refs.REFERENCED_TABLE_SCHEMA=cols.TABLE_SCHEMA
AND refs.TABLE_NAME=cols.TABLE_NAME
AND refs.COLUMN_NAME=cols.COLUMN_NAME
LEFT JOIN INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS AS cRefs
ON cRefs.CONSTRAINT_SCHEMA=cols.TABLE_SCHEMA
AND cRefs.CONSTRAINT_NAME=refs.CONSTRAINT_NAME
LEFT JOIN INFORMATION_SCHEMA.`KEY_COLUMN_USAGE` AS links
ON links.TABLE_SCHEMA=cols.TABLE_SCHEMA
AND links.REFERENCED_TABLE_SCHEMA=cols.TABLE_SCHEMA
AND links.REFERENCED_TABLE_NAME=cols.TABLE_NAME
AND links.REFERENCED_COLUMN_NAME=cols.COLUMN_NAME
LEFT JOIN INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS AS cLinks
ON cLinks.CONSTRAINT_SCHEMA=cols.TABLE_SCHEMA
AND cLinks.CONSTRAINT_NAME=links.CONSTRAINT_NAME
WHERE cols.TABLE_SCHEMA=DATABASE()
AND cols.TABLE_NAME="table"