यहां नेस्टेड सबक्वेरी पर आधारित एक समाधान है। सबसे पहले, मैंने कुछ और मामलों को पकड़ने के लिए कुछ पंक्तियाँ जोड़ीं। लेन-देन 10, उदाहरण के लिए, लेन-देन 12 द्वारा रद्द नहीं किया जाना चाहिए, क्योंकि लेन-देन 11 बीच में आता है।
> select * from transactions order by date_time;
+----+---------+------+---------------------+--------+
| id | account | type | date_time | amount |
+----+---------+------+---------------------+--------+
| 1 | 1 | R | 2012-01-01 10:01:00 | 1000 |
| 2 | 3 | R | 2012-01-02 12:53:10 | 1500 |
| 3 | 3 | A | 2012-01-03 13:10:01 | -1500 |
| 4 | 2 | R | 2012-01-03 17:56:00 | 2000 |
| 5 | 1 | R | 2012-01-04 12:30:01 | 1000 |
| 6 | 2 | A | 2012-01-04 13:23:01 | -2000 |
| 7 | 3 | R | 2012-01-04 15:13:10 | 3000 |
| 8 | 3 | R | 2012-01-05 12:12:00 | 1250 |
| 9 | 3 | A | 2012-01-06 17:24:01 | -1250 |
| 10 | 3 | R | 2012-01-07 00:00:00 | 1250 |
| 11 | 3 | R | 2012-01-07 05:00:00 | 4000 |
| 12 | 3 | A | 2012-01-08 00:00:00 | -1250 |
| 14 | 2 | R | 2012-01-09 00:00:00 | 2000 |
| 13 | 3 | A | 2012-01-10 00:00:00 | -1500 |
| 15 | 2 | A | 2012-01-11 04:00:00 | -2000 |
| 16 | 2 | R | 2012-01-12 00:00:00 | 5000 |
+----+---------+------+---------------------+--------+
16 rows in set (0.00 sec)
सबसे पहले, प्रत्येक लेन-देन के लिए "उसी खाते में सबसे हाल के लेन-देन की तारीख" को पकड़ने के लिए एक क्वेरी बनाएं:
SELECT t2.*,
MAX(t1.date_time) AS prev_date
FROM transactions t1
JOIN transactions t2
ON (t1.account = t2.account
AND t2.date_time > t1.date_time)
GROUP BY t2.account,t2.date_time
ORDER BY t2.date_time;
+----+---------+------+---------------------+--------+---------------------+
| id | account | type | date_time | amount | prev_date |
+----+---------+------+---------------------+--------+---------------------+
| 3 | 3 | A | 2012-01-03 13:10:01 | -1500 | 2012-01-02 12:53:10 |
| 5 | 1 | R | 2012-01-04 12:30:01 | 1000 | 2012-01-01 10:01:00 |
| 6 | 2 | A | 2012-01-04 13:23:01 | -2000 | 2012-01-03 17:56:00 |
| 7 | 3 | R | 2012-01-04 15:13:10 | 3000 | 2012-01-03 13:10:01 |
| 8 | 3 | R | 2012-01-05 12:12:00 | 1250 | 2012-01-04 15:13:10 |
| 9 | 3 | A | 2012-01-06 17:24:01 | -1250 | 2012-01-05 12:12:00 |
| 10 | 3 | R | 2012-01-07 00:00:00 | 1250 | 2012-01-06 17:24:01 |
| 11 | 3 | R | 2012-01-07 05:00:00 | 4000 | 2012-01-07 00:00:00 |
| 12 | 3 | A | 2012-01-08 00:00:00 | -1250 | 2012-01-07 05:00:00 |
| 14 | 2 | R | 2012-01-09 00:00:00 | 2000 | 2012-01-04 13:23:01 |
| 13 | 3 | A | 2012-01-10 00:00:00 | -1500 | 2012-01-08 00:00:00 |
| 15 | 2 | A | 2012-01-11 04:00:00 | -2000 | 2012-01-09 00:00:00 |
| 16 | 2 | R | 2012-01-12 00:00:00 | 5000 | 2012-01-11 04:00:00 |
+----+---------+------+---------------------+--------+---------------------+
13 rows in set (0.00 sec)
प्रत्येक लेन-देन और उसके पूर्ववर्ती को एक ही पंक्ति में प्राप्त करने के लिए एक सबक्वेरी के रूप में इसका उपयोग करें। जिन लेन-देन में हम रुचि रखते हैं उन्हें बाहर निकालने के लिए कुछ फ़िल्टरिंग का उपयोग करें - अर्थात्, 'ए' लेनदेन जिनके पूर्ववर्ती 'आर' लेनदेन हैं जिन्हें वे बिल्कुल रद्द कर देते हैं -
SELECT
t3.*,transactions.*
FROM
transactions
JOIN
(SELECT t2.*,
MAX(t1.date_time) AS prev_date
FROM transactions t1
JOIN transactions t2
ON (t1.account = t2.account
AND t2.date_time > t1.date_time)
GROUP BY t2.account,t2.date_time) t3
ON t3.account = transactions.account
AND t3.prev_date = transactions.date_time
AND t3.type='A'
AND transactions.type='R'
AND t3.amount + transactions.amount = 0
ORDER BY t3.date_time;
+----+---------+------+---------------------+--------+---------------------+----+---------+------+---------------------+--------+
| id | account | type | date_time | amount | prev_date | id | account | type | date_time | amount |
+----+---------+------+---------------------+--------+---------------------+----+---------+------+---------------------+--------+
| 3 | 3 | A | 2012-01-03 13:10:01 | -1500 | 2012-01-02 12:53:10 | 2 | 3 | R | 2012-01-02 12:53:10 | 1500 |
| 6 | 2 | A | 2012-01-04 13:23:01 | -2000 | 2012-01-03 17:56:00 | 4 | 2 | R | 2012-01-03 17:56:00 | 2000 |
| 9 | 3 | A | 2012-01-06 17:24:01 | -1250 | 2012-01-05 12:12:00 | 8 | 3 | R | 2012-01-05 12:12:00 | 1250 |
| 15 | 2 | A | 2012-01-11 04:00:00 | -2000 | 2012-01-09 00:00:00 | 14 | 2 | R | 2012-01-09 00:00:00 | 2000 |
+----+---------+------+---------------------+--------+---------------------+----+---------+------+---------------------+--------+
4 rows in set (0.00 sec)
उपरोक्त परिणाम से यह स्पष्ट है कि हम लगभग वहां हैं - हमने अवांछित लेनदेन की पहचान कर ली है। LEFT JOIN का उपयोग करना हम इन्हें पूरे लेन-देन सेट में से फ़िल्टर कर सकते हैं:
SELECT
transactions.*
FROM
transactions
LEFT JOIN
(SELECT
transactions.id
FROM
transactions
JOIN
(SELECT t2.*,
MAX(t1.date_time) AS prev_date
FROM transactions t1
JOIN transactions t2
ON (t1.account = t2.account
AND t2.date_time > t1.date_time)
GROUP BY t2.account,t2.date_time) t3
ON t3.account = transactions.account
AND t3.prev_date = transactions.date_time
AND t3.type='A'
AND transactions.type='R'
AND t3.amount + transactions.amount = 0) t4
USING(id)
WHERE t4.id IS NULL
AND transactions.type = 'R'
ORDER BY transactions.date_time;
+----+---------+------+---------------------+--------+
| id | account | type | date_time | amount |
+----+---------+------+---------------------+--------+
| 1 | 1 | R | 2012-01-01 10:01:00 | 1000 |
| 5 | 1 | R | 2012-01-04 12:30:01 | 1000 |
| 7 | 3 | R | 2012-01-04 15:13:10 | 3000 |
| 10 | 3 | R | 2012-01-07 00:00:00 | 1250 |
| 11 | 3 | R | 2012-01-07 05:00:00 | 4000 |
| 16 | 2 | R | 2012-01-12 00:00:00 | 5000 |
+----+---------+------+---------------------+--------+
