Question

像这样的东西

Gap : 30 min
Time A : 12:00
Time B : 15:00

结果

12:00-12:30
12:30-13:00
13:00-13:30
13:30-14:00
13:00-14:30
13:30-15:00

同样的问题在question中被问到但是我需要在mysql中使用它

Answer 1

您可以使用计数表并添加多个30分钟INTERVAL：

SELECT DATE_FORMAT(timeA + INTERVAL n * 30 MINUTE, '%k:%i') as timeA,
       DATE_FORMAT(timeA + INTERVAL (n+1) * 30 MINUTE , '%k:%i') as timeB,
       CONCAT(DATE_FORMAT(timeA + INTERVAL n * 30 MINUTE, '%k:%i'),'-',
       DATE_FORMAT(timeA + INTERVAL (n+1) * 30 MINUTE , '%k:%i')) as time_range
FROM tab
CROSS JOIN 
(
   SELECT a.N + b.N * 10 n
     FROM 
    (SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
   ,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
) n
WHERE timeA + INTERVAL (n+1) * 30 MINUTE <= timeB

的 SqlFiddleDemo

输出：

╔════════╦════════╦═════════════╗ ║ timeA ║ timeB ║ time_range ║ ╠════════╬════════╬═════════════╣ ║ 12:00 ║ 12:30 ║ 12:00-12:30 ║ ║ 12:30 ║ 13:00 ║ 12:30-13:00 ║ ║ 13:00 ║ 13:30 ║ 13:00-13:30 ║ ║ 13:30 ║ 14:00 ║ 13:30-14:00 ║ ║ 14:00 ║ 14:30 ║ 14:00-14:30 ║ ║ 14:30 ║ 15:00 ║ 14:30-15:00 ║ ╚════════╩════════╩═════════════╝

有变量并支持非统一差距：

SET @gap = 45; SET @start = '11:00:00'; SET @end = '15:00:00'; SELECT CONCAT(DATE_FORMAT(timeA + INTERVAL n * @gap MINUTE, '%k:%i'),'-', DATE_FORMAT(IF(timeA + INTERVAL (n+1) * @gap MINUTE < timeB, timeA + INTERVAL (n+1) * @gap MINUTE, timeB) , '%k:%i')) as time_range FROM (SELECT CAST(@start AS TIME) AS timeA, CAST(@end AS TIME) AS timeB) AS tab CROSS JOIN ( SELECT a.N + b.N * 10 n FROM (SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a ,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b ) n WHERE timeA + INTERVAL (n) * @gap MINUTE <= timeB

的 SqlFiddleDemo2

输出：

╔═════════════╗ ║ time_range ║ ╠═════════════╣ ║ 11:00-11:45 ║ ║ 11:45-12:30 ║ ║ 12:30-13:15 ║ ║ 13:15-14:00 ║ ║ 14:00-14:45 ║ ║ 14:45-15:00 ║ -- not 15.30 ╚═════════════╝

Answer 2

如果您可以使用SEQUENCE Engine，您可以使用：

SELECT
  CONCAT(
    "SELECT CONCAT (
    TIME('11:00') + interval ((seq-1) * 30) Minute,
    ' - ',
    TIME('11:00') + interval (seq * 30) Minute) AS time_range
    FROM seq_1_to_",
    CAST( (TIME_TO_SEC( timediff('13:00','11:00')) /60/30) AS UNSIGNED)
) INTO @sql;
PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

<强>输出

+---------------------+
| time_range          |
+---------------------+
| 11:00:00 - 11:30:00 |
| 11:30:00 - 12:00:00 |
| 12:00:00 - 12:30:00 |
| 12:30:00 - 13:00:00 |
+---------------------+
4 rows in set (0.00 sec)

使用参数显示两个不同时间之间的所有时间

2 个答案: