我有三個表SQL查詢的三個表
表1 color_sets_info
與set_id
(PK),set_name
列
表2 colors
與set_id
(FK),color_id
(PK),color_name
,color_formula
列
表3. mixed_colors
與color_id1
(FK),color_id2
(FK),color_formula
列
外鍵colors.set_id
引用color_sets_info.set_id
外鍵mixed_colors.color_id1
引用colors.color_id
外鍵mixed_colors.color_id2
引用colors.color_id
如何從mixed_colors具體set_name
或color_formula
和所有相關的列得到:
colors.color_name
(爲mixed_colors.color_id1
),
colors.color_name
(爲mixed_colors.color_id2
),
color_sets_info.set_name
(爲第一colors.color_name
),
color_sets_info.set_name
(對於第二個colors.color_name
)
mixed_colors.color_formula
?
例如:
color_sets_info colors
+--------+-----------+ +--------+----------+------------+---------------+
| set_id | set_name | | set_id | color_id | color_name | color_formula |
+--------+-----------+ +--------+----------+------------+---------------+
| 1 | somename1 | | 1 | 1 | black | R0G0B0 |
| 2 | somename2 | | 1 | 2 | yellow | R255G255B0 |
| 3 | somename3 | | 2 | 3 | green | R0G255B255 |
+--------+-----------+ | 3 | 4 | red | R255G0B0 |
+--------+----------+------------+---------------+
mixed_colors
+-----------+-----------+---------------+
| color_id1 | color_id2 | color_formula |
+-----------+-----------+---------------+
| 1 | 4 | R127G0B0 |
| 2 | 3 | R127G255B127 |
| 3 | 1 | R0G127B127 |
+-----------+-----------+---------------+
我需要從mixed_colorscolor_formula
和兩個set_names
和兩個color_names
每個混合色,其中1)僅使用somename1和somename2顏色組以獲得2)R127G0B0公式
哪些是你使用,MySQL或SQLServer數據庫?你已經嘗試過哪些SQL? – talegna
抱歉,sqlite ...... – Pylyp
我知道如何爲兒童構建SQL,例如'SELECT color_sets_info.set_name,color_name,color_formula FROM colors JOIN color_sets_info ON colors.set_id = color_sets_info.set_id',但被子clidrens卡住,特別是在** mixed_colors **中有兩個'color_id' ** – Pylyp