[英]MYSQL - Select Only Records Where Previous Record Column Data Differs
希望這可以通過 sql 實現。 我有一個查詢,它返回一個帶有 time_stamp 和 hash_index 列的數據集。 基本上有以下作用:
1 1583365548 6ff11ad5536f28d66098f6d74f97d877
2 1583365554 6ff11ad5536f28d66098f6d74f97d877
3 1583365556 6ff11ad5536f28d66098f6d74f97d877
4 1583365562 a2e99acb2540d49955ef93fb2684ac25
5 1583365571 a2e99acb2540d49955ef93fb2684ac25
6 1583365572 a2e99acb2540d49955ef93fb2684ac25
7 1583365574 a2e99acb2540d49955ef93fb2684ac25
8 1583365578 a2e99acb2540d49955ef93fb2684ac25
9 1583365580 a2e99acb2540d49955ef93fb2684ac25
我想要做的是進一步過濾此查詢以僅包含 hash_index 與前一條記錄不同的記錄。 這是我可以做的事情而不必將其轉儲到 PHP 中並循環遍歷它嗎?
我目前的查詢如下:
SELECT
(@cnt:=@cnt + 1) AS row_number,
time_stamp,
MD5(GROUP_CONCAT(CONCAT(user_state_name,
option_id,
option_code,
item_id,
item_code))) AS hash_index
FROM
user_state
WHERE
user_id = 2
GROUP BY
time_stamp;
什么是這樣的查詢:
SELECT t1.*
FROM user_state t1
LEFT JOIN user_state t2 ON t1.id-1 = t2.id
WHERE t1.hash_index <> t2.hash_index;
如果您想過濾掉相鄰的重復項,我將只使用lag()
並免除散列和聚合:
SELECT us.*
FROM (SELECT us.*,
LAG(time_stamp) OVER (PARTITION BY user_id ORDER BY time_stamp) as prev_ts,
LAG(time_stamp) OVER (PARTITION BY user_id, user_state_nae, option_id, option_code, item_id, item_code ORDER BY time_stamp) as prev_ts_values
FROM user_state us
WHERE user_id = 2
) t
WHERE prev_ts_values is null or prev_ts_values <> prev_ts;
您可以選擇所需的任何行。
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