Question

TABLE：

**timeslot**:
----------
id_timeslot   times
1             09:00
2             09:30
3             10:00
4             10:30
5             11:00

**bookslot**
id   id_timeslot     date        b_ref
-------------------------------------------
1    2               2010-02-22  001 
2    3               2010-02-22  001
3    4               2010-02-22  001
4    5               2010-02-22  001
5    2               2010-02-25  002
6    3               2010-02-27  003
7    4               2010-02-27  003
8    5               2010-02-27  003

PHP

$q = $mysqli->query("SELECT * FROM bookslot  
LEFT JOIN timeslot ON bookslot.id_timeslot = timeslot.id_timeslot   
WHERE bookslot.status = 1 
GROUP BY bookslot.b_ref  
ORDER BY bookslot.date ASC, bookslot.id_timeslot ASC LIMIT 20");

HTML结果：

DATE         TIMES  
2010-02-22   10:30
2010-02-25   09:30
2010-02-27   11:00

任何人都会注意到桌面上的结果。时间是不正确的顺序？
我用ASC / DESC改变了另一种方式，但仍显示最后一次id_timeslot？

预期结果：

DATE         TIMES  
2010-02-22   09:30
2010-02-25   09:30
2010-02-27   10:00

Answer 1

您的GROUP BY bookslot.b_ref正在对记录进行分组，因此您只会在每种情况下看到最后一次。

尝试使用

SELECT date, time, MIN(bookslot.id_timeslot)
FROM bookslot  
LEFT JOIN timeslot ON bookslot.id_timeslot = timeslot.id_timeslot   
WHERE bookslot.status = 1 
GROUP BY bookslot.b_ref  
ORDER BY bookslot.date ASC, bookslot.id_timeslot ASC LIMIT 20

Answer 2

虽然你的SQL在语法上是正确的，但它会产生意想不到的结果。

通常，必须在SELECT子句中指定GROUP BY的列，或者将其括在聚合函数中。否则，MySQL将自行决定在GROUP BY操作中消除哪些记录。 ORDER BY并不重要，因为它是在GROUP BY操作之后应用的。您应该更好地修改您的查询：

SELECT b_ref, MIN(ADDTIME(date, times)) AS complete_datetime FROM bookslot LEFT JOIN timeslot ON bookslot.id_timeslot = timeslot.id_timeslot WHERE bookslot.status = 1 GROUP BY bookslot.b_ref ORDER BY bookslot.date, bookslot.id_timeslot

Answer 3

由于目标似乎是收集每个图书位置的最早时段，因此需要使用MIN缩小结果

SELECT b.id, b.id_timeslot, b.date, MIN(`date`) , t.times
FROM bookslot b
LEFT JOIN timeslot t ON b.id_timeslot = t.id_timeslot
GROUP BY b.b_ref

php mysql asc / desc命令

3 个答案: