作者

digoal

日期

2020-08-12

標籤

PostgreSQL , 計算時間間隔 , 數值

背景

計算兩個時間戳的間隔, 然後轉化為秒或者轉化為天為單位的數值.

怎麼算才是正確的?

1、錯誤: 時間相減, 然後轉化為epoch (秒數)

因為interval型別轉換為epoch時, 演算法可能和預期不符.

``` 
postgres=# select extract('epoch' from interval '0.01 year')/3600/24.0; 
?column?

    0

(1 row)

postgres=# select extract('epoch' from interval '1 year')/3600/24.0; 
?column?

365.25 
(1 row)

postgres=# select extract('epoch' from interval '0.5 year')/3600/24.0; 
?column?

  180

(1 row)

postgres=# select extract('epoch' from interval '0.583 year')/3600/24.0; 
?column?

  180

(1 row)

postgres=# select extract('epoch' from interval '0.584 year')/3600/24.0; 
?column?

  210

(1 row) 
```

0.01年的epoch是0 ?

1年的epoch是365.25天?

0.5年的epoch是180天?

0.583年的epoch是180天?

0.584年的epoch是210天?

為什麼?

原因要從make interval說起, 程式碼如下:

src/backend/utils/adt/timestamp.c

``` 
/ 
* make_interval - numeric Interval constructor 
/ 
Datum 
make_interval(PG_FUNCTION_ARGS) 
{ 
int32 years = PG_GETARG_INT32(0); 
int32 months = PG_GETARG_INT32(1); 
int32 weeks = PG_GETARG_INT32(2); 
int32 days = PG_GETARG_INT32(3); 
int32 hours = PG_GETARG_INT32(4); 
int32 mins = PG_GETARG_INT32(5); 
double secs = PG_GETARG_FLOAT8(6); 
Interval *result;

    /*    
     * Reject out-of-range inputs.  We really ought to check the integer    
     * inputs as well, but it's not entirely clear what limits to apply.    
     */    
    if (isinf(secs) || isnan(secs))    
            ereport(ERROR,    
                            (errcode(ERRCODE_DATETIME_VALUE_OUT_OF_RANGE),    
                             errmsg("interval out of range")));

    result = (Interval *) palloc(sizeof(Interval));    
    result->month = years * MONTHS_PER_YEAR + months;    
    result->day = weeks * 7 + days;

    secs = rint(secs * USECS_PER_SEC);    
    result->time = hours * ((int64) SECS_PER_HOUR * USECS_PER_SEC) +    
            mins * ((int64) SECS_PER_MINUTE * USECS_PER_SEC) +    
            (int64) secs;

    PG_RETURN_INTERVAL_P(result);

} 
```

MONTHS_PER_YEAR 
USECS_PER_SEC 
SECS_PER_HOUR 
SECS_PER_MINUTE

每個單位都是整數, 如果不是整數, 則需要轉換為下一級的整數

整數再乘以這個級別轉換為下一級別的常數係數

例如

0.583年的epoch是180天? 
0.584年的epoch是210天?

``` 
postgres=# select 0.584*12; 
?column?

7.008

(1 row)

postgres=# select 0.583*12; 
?column?

6.996

(1 row) 
```

抹掉小數後得到6個月,7個月.

``` 
postgres=# select interval '0.583 year'; 
interval

6 mons 
(1 row)

postgres=# select interval '0.584 year'; 
interval

7 mons 
(1 row)

postgres=# select interval '0.11 month'; 
interval

3 days 07:12:00 
(1 row) 
```

這樣的演算法, 造成結果與預期不符.

2、正確: 時間轉化為epoch後, 兩個epoch值再相減.

``` 
postgres=# select extract('epoch' from now()) - extract('epoch' from timestamp '2018-10-01'); 
?column?

58863397.59471512 
(1 row)

postgres=# select (extract('epoch' from now()) - extract('epoch' from timestamp '2018-10-01'))/3600.0/24.0; 
?column?

681.289469844514 
(1 row) 
```

參考自：https://cdn.modb.pro/db/91966