是不是这样查就可以了
SELECT status, count(distinct p.region_id), count(distinct p.peer_id) FROM information_schema.TIKV_REGION_STATUS s JOIN information_schema.TIKV_REGION_PEERS p ON s.region_id = p.region_id JOIN information_schema.TIKV_STORE_STATUS m ON m.store_id = p.store_id group by status;
SELECT
‘TOTAL_REGIONS’ AS METRIC,
COUNT() AS COUNT
FROM INFORMATION_SCHEMA.TIKV_REGION_STATUS
WHERE REGION_ID IN (
SELECT REGION_ID FROM INFORMATION_SCHEMA.TIKV_STORE_STATUS
WHERE STORE_ID IN (SELECT STORE_ID FROM INFORMATION_SCHEMA.TIKV_STORE_STATUS)
)
UNION ALL
SELECT
‘REPLICA_COUNT_’ || REPLICA_COUNT,
COUNT()
FROM INFORMATION_SCHEMA.TIKV_REGION_STATUS
WHERE REPLICA_COUNT IS NOT NULL
GROUP BY REPLICA_COUNT
UNION ALL
SELECT
‘ABNORMAL_REPLICA_COUNT’,
COUNT(*)
FROM INFORMATION_SCHEMA.TIKV_REGION_STATUS
WHERE REPLICA_COUNT != 3; – 假设你期望是 3 副本