PostgreSQL的時間/日期函數使用

PostgreSQL的經常使用時間函數使用整理以下:運維

1、獲取系統時間函數

1.1 獲取當前完整時間函數

select now();spa

david=# select now();
              now              
-------------------------------
 2013-04-12 15:39:40.399711+08
(1 row)

david=#

current_timestamp 同 now() 函數等效。對象

david=# select current_timestamp;
              now              
-------------------------------
 2013-04-12 15:40:22.398709+08
(1 row)

david=#

 

1.2 獲取當前日期開發

select current_date;input

david=# select current_date;
    date    
------------
 2013-04-12
(1 row)

david=#

 

1.3 獲取當前時間io

select current_time;table

david=# select current_time;
       timetz       
--------------------
 15:43:31.101726+08
(1 row)

david=#

 

2、時間的計算

 

david=# select now();
              now              
-------------------------------
 2013-04-12 15:47:13.244721+08
(1 row)

david=#

 

2.1 兩年後date

david=# select now() + interval '2 years';
           ?column?            
-------------------------------
 2015-04-12 15:49:03.168851+08
(1 row)

david=# select now() + interval '2 year'; 
           ?column?            
-------------------------------
 2015-04-12 15:49:12.378727+08
(1 row)

david=# select now() + interval '2 y';   
           ?column?           
------------------------------
 2015-04-12 15:49:25.46986+08
(1 row)

david=# select now() + interval '2 Y';
           ?column?            
-------------------------------
 2015-04-12 15:49:28.410853+08
(1 row)

david=# select now() + interval '2Y'; 
           ?column?            
-------------------------------
 2015-04-12 15:49:31.122831+08
(1 row)

david=#

 

2.2 一個月後select

david=# select now() + interval '1 month';  
           ?column?           
------------------------------
 2013-05-12 15:51:22.24373+08
(1 row)

david=# select now() + interval 'one month';
ERROR:  invalid input syntax for type interval: "one month"
LINE 1: select now() + interval 'one month';
                                ^
david=#

 

2.3 三週前

david=# select now() - interval '3 week';
           ?column?            
-------------------------------
 2013-03-22 16:00:04.203735+08
(1 row)

david=#

 

2.4 十分鐘後

david=# select now() + '10 min';                 
           ?column?            
-------------------------------
 2013-04-12 16:12:47.445744+08
(1 row)

david=#

 

說明:

interval 能夠不寫,其值能夠是:

Abbreviation Meaning
Y Years
M Months (in the date part)
W Weeks
D Days
H Hours
M Minutes (in the time part)
S Seconds

 

2.5 計算兩個時間差

使用 age(timestamp, timestamp)

david=# select age(now(), timestamp '1989-02-05');
                  age                   
----------------------------------------
 24 years 2 mons 7 days 17:05:49.119848
(1 row)

david=#
david=# select age(timestamp '2007-09-15');       
          age           
------------------------
 5 years 6 mons 27 days
(1 row)

david=#

 

3、時間字段的截取

在開發過程當中,常常要取日期的年,月,日,小時等值,PostgreSQL 提供一個很是便利的EXTRACT函數。

EXTRACT(field FROM source)

field 表示取的時間對象,source 表示取的日期來源,類型爲 timestamp、time 或 interval。

3.1 取年份

david=# select extract(year from now());
 date_part 
-----------
      2013
(1 row)

david=#

 

3.2 取月份

david=# select extract(month from now());    
 date_part 
-----------
         4
(1 row)

david=#
david=# select extract(day from timestamp '2013-04-13');
 date_part 
-----------
        13
(1 row)

david=#
david=# SELECT EXTRACT(DAY FROM INTERVAL '40 days 1 minute');
 date_part 
-----------
        40
(1 row)

david=#

 

3.3 查看今天是一年中的第幾天

david=# select extract(doy from now());
 date_part 
-----------
       102
(1 row)

david=#

 

3.4 查看如今距1970-01-01 00:00:00 UTC 的秒數

david=# select extract(epoch from now());
    date_part     
------------------
 1365755907.94474
(1 row)

david=#

 

3.5 把epoch 值轉換回時間戳

david=# SELECT TIMESTAMP WITH TIME ZONE 'epoch' + 1369755555 * INTERVAL '1 second'; 
        ?column?        
------------------------
 2013-05-28 23:39:15+08
(1 row)

david=#

 

以上是基本的PG時間/日期函數使用,可知足通常的開發運維應用。

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