Question

我正在使用Advantage sql数据库，并尝试减去2个日期以获取天数。我以为在ADV SQL中，您可以减去日期，就可以计算出天数。

（（已完成）-（开始日期））

1-23-19-1-3-19 = 20天

Answer 1

如果两者均为DATE类型，则可以直接在ADS SQL中减去日期。如果它们都是TIMESTAMP，则可以将TIMESTAMPDIFF()与TSI_SQL_DAYS用作间隔。如果一个是DATE，另一个是TIMESTAMP，则必须将TIMESTAMP转换为DATE，然后才能直接减去。

以下是所有这三种选择的示例。

将两列都设为DATE类型：

create table Test (ID integer, StartDate Date, Completed Date);
insert into Test values (1, '2019-01-01', '2019-01-31');
insert into Test values (2, '2019-01-01', '2018-02-28');
select id, startdate, completed, completed - startdate as days from test;

输出：

id  startdate    completed    days      
--- -----------  -----------  ----
1   01/01/2019   01/31/2019     30
2   01/01/2019   02/28/2019     58

将两列都设为TIMESTAMP类型：

create table Test2 (ID integer, StartDate TimeStamp, Completed TimeStamp);
insert into Test2 (ID, StartDate, Completed)
  values (1, CreateTimeStamp(2019, 1, 1, 0, 0, 0, 1),
             CreateTimeStamp(2019, 1, 31, 0, 0, 0, 1));
insert into Test2 (ID, StartDate, Completed)
  values (2, CreateTimeStamp(2019, 1, 1, 0, 0, 0, 1),
             CreateTimeStamp(2019, 2, 28, 0, 0, 0, 1));
select id, startdate, completed, 
  TimeStampDiff(SQL_TSI_DAY, StartDate, Completed) as days from test2;

输出：

id  startdate    completed    days      
--- -----------  -----------  ----
  1 01/01/2019   01/31/2019     30
  2 01/01/2019   02/28/2019     58

一个DATE和一个TIMESTAMP：

create table Test3 (ID integer, StartDate TimeStamp, Completed Date);
insert into Test3 (ID, StartDate, Completed)
  values (1, CreateTimeStamp(2019, 1, 1, 0, 0, 0, 1), '2019-01-31');
insert into Test3 (ID, StartDate, Completed)
  values (2, CreateTimeStamp(2019, 1, 1, 0, 0, 0, 1), '2019-02-28');
select id, startdate, completed, 
  Completed - Cast(StartDate AS SQL_DATE) as days from test3;

输出：

id  startdate    completed    days      
--- -----------  -----------  ----
  1 01/01/2019   01/31/2019     30
  2 01/01/2019   02/28/2019     58

Answer 2

我认为timestampdiff()应该做您想要做的事情：

timestampdiff(SQL_TSI_DAY, startdate, completed)

Answer 3

这是一个非常不寻常的数据库，可引起您的疑问-文档建议（https://devzone.advantagedatabase.com/dz/webhelp/advantage10/index.html?master_supported_scalar_functions.htm）您需要：

TIMESTAMPDIFF(SQL_TSI_DAY,completed, startdate)

编辑：包括演员

TIMESTAMPDIFF(SQL_TSI_DAY,completed, cast(startdate as SQL_TIMESTAMP))

文档：

TIMESTAMPDIFF（interval，timestamp1，timestamp2）

基于减时间戳1返回整数间隔的数量   从时间戳记2。支持的时间间隔值为：SQL_TSI_FRAC_SECOND，   SQL_TSI_SECOND，SQL_TSI_MINUTE，SQL_TSI_HOUR，SQL_TSI_DAY，   SQL_TSI_WEEK，SQL_TSI_MONTH，SQL_TSI_QUARTER，SQL_TSI_YEAR。

减去2日期以获得天数

3 个答案: