好的,
我试图让这一整天都完成,但我对MySQL的了解似乎太有限了。
我有以下表格:
time_entries
|id|comment|ticket_id|
| 1|foo | 1|
| 2|bar | 1|
| 3|baz | 2|
| 4|lorem | 3|
| 5|ipsum | 4|
ticket
|id|name |
| 1|ticket1|
| 2|ticket2|
| 3|ticket3|
| 4|ticket4|
custom_fields
|id|name |
| 1|custom1|
| 2|custom2|
| 3|custom3|
custom_values
|id|custom_field_id|ticket_id|value|
| 1| 1| 1| 22|
| 2| 2| 1| 33|
| 3| 3| 1| 44|
| 4| 1| 2| 55|
| 5| 3| 2| 66|
| 6| 2| 3| 77|
| 7| 1| 4| 88|
我想要获得的是每个time_entry一行,包含故障单信息和自定义值。如果没有为此票证设置自定义值,则结果选择中的值必须为空或为空:
select
|time_entries_comment|ticket_name|custom1_value|custom2_value|custom3_value|
| foo| ticket1| 22| 33| 44|
| bar| ticket1| 22| 33| 44|
| baz| ticket2| 55| NULL| 66|
| lorem| ticket3| NULL| 77| NULL|
| ipsum| ticket4| 88| NULL| NULL|
到目前为止我得到的是:
select
te.comment,
t.name,
cv1.value,
cv2.value,
cv3.value
from time_entries te
LEFT JOIN ticket t ON te.ticket_id = i.id
LEFT JOIN custom_values cv1 ON t.id = cv1.ticket_id
LEFT JOIN custom_fields cf1 ON cv1.custom_field_id = cf1.id AND cf1.id = 1
LEFT JOIN custom_values cv2 ON t.id = cv2.ticket_id
LEFT JOIN custom_fields cf2 ON cv2.custom_field_id = cf2.id AND cf2.id = 2
LEFT JOIN custom_values cv3 ON t.id = cv3.ticket_id
LEFT JOIN custom_fields cf3 ON cv3.custom_field_id = cf3.id AND cf3.id = 3
WHERE t.id = 1;
但是这给了我所有的比赛。
我尝试了内部联接,但是如果此票证没有custom_value,则我没有结果。我还尝试了UNION Left and Right的外部加入但没有成功。
有什么建议可以寻找吗?这里有什么关键词?
感谢您的帮助
答案 0 :(得分:3)
您想要的是流程调用枢轴。 这主要通过GROUP BY和MAX函数完成。
SELECT
time_entries.comment AS time_entries_comment
, ticket.name
, MAX(CASE WHEN custom_values.custom_field_id = 1 THEN custom_values.value ELSE NULL END) AS custom1_value
, MAX(CASE WHEN custom_values.custom_field_id = 2 THEN custom_values.value ELSE NULL END) AS custom2_value
, MAX(CASE WHEN custom_values.custom_field_id = 3 THEN custom_values.value ELSE NULL END) AS custom3_value
FROM
time_entries
INNER JOIN
ticket
ON
time_entries.ticket_id = ticket.id
INNER JOIN
custom_values
ON
time_entries.ticket_id = custom_values.ticket_id
GROUP BY
time_entries.comment
, ticket.name
ORDER BY
time_entries.id ASC
<强>结果强>
time_entries_comment name custom1_value custom2_value custom3_value
-------------------- ------- ------------- ------------- ---------------
foo ticket1 22 33 44
bar ticket1 22 33 44
baz ticket2 55 (NULL) 66
lorem ticket3 (NULL) 77 (NULL)
ipsum ticket4 88 (NULL) (NULL)
答案 1 :(得分:0)
我认为LEFT JOIN是要走的路,因为它允许你获得NULL值。您只需在末尾添加GROUP BY语句
GROUP BY t.id