MySQL根据值并在每个客户的时间戳后删除行

Question

I am trying to delete rows after a timestamp for each customer who have value 1 in MySQL. Example table:

id | timestamp | cust_ID | value 899900 | 2016-04-11 12:00:00 | 500219 | 0 899901 | 2016-04-12 16:00:00 | 500219 | 0 899902 | 2016-04-14 11:00:00 | 500219 | 1 899903 | 2016-04-15 12:00:00 | 500219 | 1 899904 | 2016-04-23 09:00:00 | 500219 | 0 899905 | 2016-05-02 19:00:00 | 500219 | 0 909901 | 2016-04-12 16:00:00 | 500230 | 0 909902 | 2016-04-14 11:00:00 | 500230 | 1 909903 | 2016-04-15 12:00:00 | 500230 | 1 909904 | 2016-04-23 09:00:00 | 500230 | 0 909905 | 2016-05-02 19:00:00 | 500230 | 0 939905 | 2016-05-02 19:00:00 | 500240 | 0

Trying to achieve:

id | timestamp | cust_ID | value 899900 | 2016-04-11 12:00:00 | 500219 | 0 899901 | 2016-04-12 16:00:00 | 500219 | 0 899902 | 2016-04-14 11:00:00 | 500219 | 1 899903 | 2016-04-15 12:00:00 | 500219 | 1 909901 | 2016-04-12 16:00:00 | 500230 | 0 909902 | 2016-04-14 11:00:00 | 500230 | 1 909903 | 2016-04-15 12:00:00 | 500230 | 1 939905 | 2016-05-02 19:00:00 | 500240 | 0

So far I have the following which throws error 1242 'subquery returns more than one row':

CREATE VIEW max_id AS SELECT id , cust_ID , MAX( timestamp ) FROM table WHERE value = 1 GROUP BY cust_ID ; DELETE FROM max_id WHERE id > (SELECT id FROM max_id GROUP BY cust_ID );

Any help would be greatly appreciated!

Answer 1

I'm not 100% what the exactly logic you want is. But, you should be able to use a join . The following deletes all rows for a customer whose timestamp is greater than the maximum timestamp with a "1" for that customer:

delete t
    from table t join
         (select t.cust_Id, max(timestamp) as maxts
          from table t
          group by t.cust_id
         ) tt
         on t.cust_id = tt.cust_id and tt.timestamp > t.timestamp