postgresql Postgres where 子句比较时间戳
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Postgres where clause compare timestamp
提问by a_horse_with_no_name
I have a table where column is of datatype timestamp
我有一个表，其中列是数据类型 timestamp
Which contains records multiple records for a day I want to select all rows corresponding to day
其中包含一天的多条记录我想选择与一天对应的所有行
How do I do it?
我该怎么做？
回答by a_horse_with_no_name
Assuming you actually mean timestampbecause there is no datetimein Postgres
假设你的意思是timestamp因为datetimePostgres 中没有
Cast the timestamp column to a date, that will remove the time part:
将时间戳列转换为日期，这将删除时间部分：
select *
from the_table
where the_timestamp_column::date = date '2015-07-15';
This will return all rows from July, 15th.
这将返回从 7 月 15 日开始的所有行。
Note that the above will notuse an index on the_timestamp_column. If performance is critical, you need to either create an index on that expression or use a range condition:
请注意，以上不会在 上使用索引the_timestamp_column。如果性能至关重要，则需要在该表达式上创建索引或使用范围条件：
select *
from the_table
where the_timestamp_column >= timestamp '2015-07-15 00:00:00'
and the_timestamp_column < timestamp '2015-07-16 00:00:00';
