Mysql 跳过具有相同值的第一个连续记录,然后停止
这是我的桌面场景Mysql 跳过具有相同值的第一个连续记录,然后停止,mysql,sql,Mysql,Sql,这是我的桌面场景 +----------+------------+---------------------+------------------+ | userd_id | user_value | user_datetime | i_want_to_select | +----------+------------+---------------------+------------------+ | 1 | 1 | 2020-01-31 12:
+----------+------------+---------------------+------------------+
| userd_id | user_value | user_datetime | i_want_to_select |
+----------+------------+---------------------+------------------+
| 1 | 1 | 2020-01-31 12:13:14 | |
| 2 | 1 | 2020-01-30 12:13:14 | |
| 3 | 1 | 2020-01-29 12:13:14 | |
| 4 | 2 | 2020-01-28 12:13:14 | |
| 5 | 2 | 2020-01-27 12:13:14 | |
| 6 | 3 | 2020-01-20 12:13:14 | |
| 7 | 1 | 2020-01-19 12:13:14 | this |
| 8 | 1 | 2020-01-18 12:13:14 | this |
| 9 | 1 | 2020-01-17 12:13:14 | this |
+----------+------------+---------------------+------------------+
仅当第一个元素为1时,才需要跳过这些连续值,并在值不同时立即停止
我是用PHP代码实现的,但我希望避免加重PHP处理器和垃圾收集器的负担
foreach ($rows as &$row){
if($row['user_value'] === 1){
unset($row); // remove my row
} else {
break; // at the first different value, stop
}
}
unset($row);
$rows = array_values($rows) //reset the array index/key
谢谢一种方法是:
select t.*
from t
where t.user_datetime >= (select min(t2.user_datetime)
from t t2
where t2.user_id = t.user_id and
t2.user_value <> 1
);
您还可以使用窗口功能,例如:
select t.*
from (select t.*,
min( case when t.user_value <> 1 then t user_datetime) over (partition by t.user_id) as min_not1_datetime
from t
) t
where user_datetime >= min_not1_datetime;
我试试第一个好的,非常感谢。是第一个,但用>号代替