Question

我正在使用sqlite来记录时间戳

INSERT INTO packets VALUES ( strftime('%%J','now') );

然后提取已用时间

SELECT strftime('%%J','now') - first_timestamp FROM packets;

运作良好。如果我等一分钟，结果是0.0007（〜= 1 * 60/24 * 60 * 60）

我希望在数小时，分钟和秒钟内看到这一点，但

sqlite> SELECT time(0.0007);
12:01:00

12来自哪里？

这＆＃39;工作＆＃39;

sqlite> SELECT time(0.0007-0.5);
00:01:00

但似乎太奇怪了，无法使用。

按照CL的说明，我提交了这段代码

std::string TimeSinceFirstPacket()
{
    // open database
    Open();

    // read timestamp of first packet
    // this is stored as a Julian day for convenince in calculating and formatting the elapsed time
    // Note that for Julian days
    //   1.0 is 24 hours
    //    -0.5 represents the previous midnight
    // discussion at http://stackoverflow.com/q/38284268/16582

    int dbret = DB.Query(
             " SELECT "
             "       first_timestamp, "
             "       ( strftime('%%J','now') - first_timestamp ) < 1.0, "
             "       time( strftime('%%J','now') - first_timestamp - 0.5 ) "
             " FROM packets;");

    // check for successful db read
    if( dbret != 1 )
        return "error";

    // check that timestamp has been initialized
    if( DB.myResultA[ 0 ] == "0" )
        return "none";

    // check that elapsed time is less than 24 hours
    if( DB.myResultA[ 1 ] == "0" )
        return ">24hr";

    // return human readable hh::mm::ss elapsed time
    return DB.myResultA[ 2 ];

Answer 1

朱利安的日子不计入午夜，而是从正午：

> select julianday('2000-01-01 00:00:00');
2451544.5
> select julianday('2000-01-01 12:00:00');
2451545.0

所以为了得到午夜以来的时间，你必须计算你的数字和代表午夜的数字之间的差异。

格式化儒略日

1 个答案: