Question

我正在尝试找出这两个日期之间的差异，并将其显示为天数，小时数，分钟数。

以下是它们现在的格式，以下是我提出的但它似乎无法正常工作......

5/18/17 4:57 PM

select *,strftime('%H:%M',CAST ((julianday('Completed Date') - julianday('Created Date')) AS REAL),'12:00') from report

Answer 1

5/18/17 4:57 PM在SQLite中不是有效的时间字符串格式，因此将strftime用于这种格式是没有意义的。您只能将strftime与YYYY-MM-DD或YYYY-MM-DD HH:MM:SS等有效时间字符串格式一起使用。

如果您要计算Created Date和Completed Date之间的差异，则需要计算这两个日期之间的秒数，并使用它来计算天数，小时和分钟。

SELECT strftime('%s',"Completed Date") - strftime('%s', "Created Date") from report;

一天有86400秒，一小时有3600秒，一分钟有60秒。你应该可以从这里计算其余部分。

Answer 2

假设您不能或不想更改数据库结构的日期格式。如果可能这样做，那就更容易了输出（见下文）有不同的日期格式，仅用于显示，当然可以在输出中保留您自己的格式。

我正在使用一些常用的表表达式

首先将日期拆分为可重复使用的部分
修复个别格式
将它们安排到两个适合strftime的日期

有一些沉重的字符串操作和一点点大胆＆＃34; + 12＆＃34; /＆＃34; + AM＆＃34;在它中间。

然后我使用kimbaudis提议进行计算归功于kimbaudi直接秒，。我失去了一些时间试图玩朱利安日。

WITH parts -- splitting for reuse
(   wrongcomonth, 
    codayyear, wrongcohours, wrongcomins, wrongcom, 
    wrongcrmonth,  
    crdayyear, wrongcrhours, wrongcrmins, wrongcrm
) as
(select  
    substr(Completed_Date, instr(Completed_Date, '/'), -2) as wrongcomonth,
    substr(Completed_Date, instr(Completed_Date, ' '), -5) as codayyear,
    substr(Completed_Date, instr(Completed_Date, ':'),-2) as wrongcohours,
    substr(Completed_Date, instr(Completed_Date, ':')+1,2) as wrongcomins,
    substr(Completed_Date, instr(Completed_Date, 'M')+1,-2) as wrongcom,
    substr(Created_Date, instr(Created_Date, '/'), -2) as wrongcrmonth,
    substr(Created_Date, instr(Created_Date, ' '), -5) as crdayyear,
    substr(Created_Date, instr(Created_Date, ':'),-2) as wrongcrhours,
    substr(Created_Date, instr(Created_Date, ':')+1,2) as wrongcrmins,
    substr(Created_Date, instr(Created_Date, 'M')+1,-2) as wrongcrm
 from report),
 -- another "with"
     pretty -- fixing formats
(   coyear, comonth, coday, cohours, comins,
    cryear, crmonth, crday, crhours, crmins
) as
(select
    '20'||substr( codayyear, -2) as coyear,
    substr('0'||wrongcomonth, -2) as comonth,
    substr(replace(replace('_'||codayyear, '_/','_0'), '_', ''), 1,2) as coday,
    substr('0'||(wrongcohours+replace(wrongcom, 'PM', '12')), -2) as cohours,
    substr('0'||replace(wrongcomins, ' ', ''), -2, 2) as comins,
    '20'||substr( crdayyear, -2) as cryear,
    substr('0'||wrongcrmonth, -2) as crmonth,
    substr(replace(replace('_'||crdayyear, '_/','_0'), '_', ''), 1,2) as crday,
    substr('0'||(wrongcrhours+replace(wrongcrm, 'PM', '12')), -2) as crhours,
    substr('0'||replace(wrongcrmins, ' ', ''), -2, 2) as crmins
from parts),
-- another "with"
    times -- two, in correct format and order
(Completion_Time, Creation_Time
) as 
( select
    coyear||'-'||comonth||'-'||coday||' '||cohours||':'||comins as Completion_Time,
    cryear||'-'||crmonth||'-'||crday||' '||crhours||':'||crmins as Creation_Time
from pretty)
-- now the real thing, as proposed by kimbaudi
select  Creation_Time, Completion_Time, 
       ((strftime('%s', Completion_Time)-strftime('%s', Creation_Time))/(24*60*60))||'d '||
       (((strftime('%s', Completion_Time)-strftime('%s', Creation_Time))/(60*60))%24)||'h '||
       (((strftime('%s', Completion_Time)-strftime('%s', Creation_Time))/(60))%60)||'m'
from times;

MCVE .dump，有一些不同的日期和第二行用于测试：

-- cleanup 
drop table if exists report;

-- make playground
BEGIN TRANSACTION;
CREATE TABLE report ("Completed_Date" date, "Created_Date" date);
INSERT INTO report VALUES('5/8/17 10:57 AM','3/8/17 9:33 AM');
INSERT INTO report VALUES('12/31/16 4:1 PM','10/10/16 7:31 PM');
COMMIT;

输出（注意测试的故意不同日期，与OP给出的日期相反）：

2017-03-08 09:33|2017-05-08 10:57|61d 1h 24m
2016-10-10 19:31|2016-12-31 16:01|81d 20h 30m

SQLite日期之间的差异

2 个答案: