这是我目前的代码:
选择e.AUDIT_ID,s.USER_ID,s.AUD_ID
来自USERS_123 e,用户的
其中e.USER_ID = s.USER_ID
结果:
e.AUDIT_ID|s.USER_ID,\x{e76f} s.AUD_ID
2222389
2222400中转abcdef0203中转机2222400
2222399中转abcdef0202中转机2222399
2222398中转abcdef0201中转机2222398
2222397中转abcdef0200中转机2222397
我想知道的是:
上述结果的Col e.AUDIT_ID中的Dupe值(如果有的话)
会显示出这种类型的欺骗的东西:
2222389 abcdef019 2222389
2222389 abcdef019 2222389
2222389 abcdef019 2222389
2222388
2222387
发布于 2017-09-01 20:59:29
更新:
SELECT e.AUDIT_ID, s.USER_ID, s.AUD_ID
FROM USERS_123 e, USERS s
WHERE e.USER_ID = s.USER_ID
GROUP BY e.AUDIT_ID, s.USER_ID, s.AUD_ID
HAVING COUNT(*) > 1 若要获得存在的任何重复AUDIT_ID (旧响应),请执行以下操作:
SELECT audit_id
FROM users_123
GROUP BY audit_id
HAVING COUNT(*) > 1;发布于 2017-09-01 20:51:49
复制AUDIT_ID?好吧,按ID和计数分组:
SELECT audit_id
FROM users_123
HAVING COUNT(*) > 1;https://stackoverflow.com/questions/46007787
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