Question

我想获得我在DB中的每个月的统计信息

SELECT      monthname(created_at) AS month, YEAR(created_at) AS year, count(*) AS number
FROM        tableName
WHERE       type_of_user = "someType"
GROUP BY    year, month(created_at)
ORDER BY    created_at DESC

现在它只给我一个月，但是即使该月我没有任何存储数据，我也需要获取每个月的统计信息

Answer 1

创建一个calendar表。您打算使用的每一年，每个月需要输入一次。

然后从日历表中选择，并加入从当前查询中获得的值。使用COALESCE()将零值放在条目为NULL的位置（例如，tableName中没有该月份和年份的记录）。

SELECT MONTHNAME(date) as month, 
       YEAR(date) as year,
       COALESCE(number, 0) as number
FROM calendar AS C
LEFT JOIN (
  SELECT created_at, COUNT(*) as number
  FROM tableName AS T
  WHERE T.type_of_user = 'someType'
  GROUP BY YEAR(created_at), MONTH(created_at)
) AS T 
ON MONTH(T.created_at) = MONTH(C.date) AND YEAR(T.created_at) = YEAR(C.date)
GROUP BY month, YEAR(created_at)
ORDER BY MONTH(date), YEAR(date)

SQL摆弄http://sqlfiddle.com/#!9/e0a4dc/

Answer 2

package mitJava2010;

public class As3_Marathon {
    public static void main(String[] arguments) {
        String[] names = { "Elena", "Thomas", "Hamilton", "Suzie", "Phil", "Matt", "Alex", "Emma", "John", "James",
                "Jane", "Emily", "Daniel", "Neda", "Aaron", "Kate" };
        int[] times = { 70, 273, 278, 329, 445, 402, 388, 275, 243, 334, 412, 393, 299, 343, 317, 265 };

        for (int i = 0; i < names.length; i++) {
            System.out.println(names[i] + ": " + times[i]);
        }

        int f = findTheFastest(times);
        int sf = find2fastest(times);
        System.out.println("The fastest is " + names[f] + " with " + times[f] + " minutes");

        System.out.println("The second fastest is " + names[sf] + " with " + times[sf] + " minutes");
    }

    private static int find2fastest(int[] times) {
        int f = findTheFastest(times);
        int sf = (f != 0) ? 0 : 1;
        for (int i = sf + 1; i < times.length; i++) {
            if (times[i] < times[sf] && i != f)
                sf = i;

        }
        return sf;
    }

    private static int findTheFastest(int[] times) {
        int minIn = 0;
        for (int i = 1; i < times.length; i++) {
            if (times[i] < times[minIn])
                minIn = i;
        }
        return minIn;

    }
}

加入具有所有记录的表将每月给您

。

即使数据库中没有该月的数据，我也可以获取一年中的每个月

2 个答案: