Question

我在此处使用以下查询来获取我的数据并每周显示该值，但是，当有新年且结果将显示为周50, 51, 52 of 2016和周1, 2, 3 of 2017时，它将会导致order by的{{1}}周数表示2017年的结果出现在2016年之前，我该如何更改？

1,2,3,50,51,52

Answer 1

也许不是最优雅的解决方案，但你可以在你所分组的领域中包括年份：

SELECT 
    CONVERT(VARCHAR,DATEPART(year, date)) + '_' + CONVERT(VARCHAR,DATEPART(wk, date)) AS Year_Week,
    SUM((kg * rep * sett)) AS weight,
    SUM(kg / max * rep * sett) AS avg,
    SUM((rep * sett)) AS reps,
    MAX(kg / max) AS peak
FROM
    Test
WHERE
    date BETWEEN @0 AND @1 AND exercise < 4
GROUP BY CONVERT(VARCHAR,DATEPART(year, date)) + '_' + CONVERT(VARCHAR,DATEPART(wk, date))
ORDER BY CONVERT(VARCHAR,DATEPART(year, date)) + '_' + CONVERT(VARCHAR,DATEPART(wk, date))

或（可能更好），只需将它们视为完全独立的字段：

SELECT 
    DATEPART(year, date) AS Year,
    DATEPART(wk, date) AS Week,
    SUM((kg * rep * sett)) AS weight,
    SUM(kg / max * rep * sett) AS avg,
    SUM((rep * sett)) AS reps,
    MAX(kg / max) AS peak
FROM
    Test
WHERE
    date BETWEEN @0 AND @1 AND exercise < 4
GROUP BY DATEPART(year, date), DATEPART(wk, date)
ORDER BY DATEPART(year, date), DATEPART(wk, date)

Answer 2

您可以按年份部分的汇总添加订单：

SELECT 
    DATEPART(wk, date) AS number,
    SUM((kg * rep * sett)) AS weight,
    SUM(kg / max * rep * sett) AS avg,
    SUM((rep * sett)) AS reps,
    MAX(kg / max) AS peak
FROM
    Test
WHERE
    date BETWEEN @0 AND @1 AND exercise < 4
GROUP BY DATEPART(wk, date)
order by max(datepart(year,date)), number;

然而，上面的数据汇集在同一周的数据中，即使它们属于不同的年份（可能不是你想要的）。如果要将不同年份的数据分开，可以在输出中添加年份列。

SELECT 
    datepart(year,date) as year,
    DATEPART(wk, date) AS number,
    SUM((kg * rep * sett)) AS weight,
    SUM(kg / max * rep * sett) AS avg,
    SUM((rep * sett)) AS reps,
    MAX(kg / max) AS peak
FROM
    Test
WHERE
    date BETWEEN @0 AND @1 AND exercise < 4
GROUP BY datepart(year,date), DATEPART(wk, date)
order by year, number;

SQL Datepart（周），ORDER BY年？

2 个答案: