我在oracles HR模式上执行了以下正确的自联接查询,但我无法真正理解它返回的内容。当我执行完全相同的查询但使用LEFT JOIN
时我理解它返回所有员工,无论他们是否有主管。
经理ID有点令人困惑,例如156, King
- 但King的ID为100.
SELECT emps.employee_id as "Employee", emps.last_name, mgr.employee_id as "Manager", mgr.last_name
FROM employees emps
RIGHT JOIN employees mgr
ON emps.manager_id = mgr.employee_id
结果
Employee LAST_NAME Manager LAST_NAME
---------- ------------------------- ---------- -------------------------
101 Kochhar 100 King
102 De Haan 100 King
103 Hunold 102 De Haan
104 Ernst 103 Hunold
105 Austin 103 Hunold
106 Pataballa 103 Hunold
107 Lorentz 103 Hunold
108 Greenberg 101 Kochhar
109 Faviet 108 Greenberg
110 Chen 108 Greenberg
111 Sciarra 108 Greenberg
112 Urman 108 Greenberg
113 Popp 108 Greenberg
114 Raphaely 100 King
115 Khoo 114 Raphaely
116 Baida 114 Raphaely
117 Tobias 114 Raphaely
118 Himuro 114 Raphaely
119 Colmenares 114 Raphaely
120 Weiss 100 King
121 Fripp 100 King
122 Kaufling 100 King
123 Vollman 100 King
124 Mourgos 100 King
125 Nayer 120 Weiss
126 Mikkilineni 120 Weiss
127 Landry 120 Weiss
128 Markle 120 Weiss
129 Bissot 121 Fripp
130 Atkinson 121 Fripp
131 Marlow 121 Fripp
132 Olson 121 Fripp
133 Mallin 122 Kaufling
134 Rogers 122 Kaufling
135 Gee 122 Kaufling
136 Philtanker 122 Kaufling
137 Ladwig 123 Vollman
138 Stiles 123 Vollman
139 Seo 123 Vollman
140 Patel 123 Vollman
141 Rajs 124 Mourgos
142 Davies 124 Mourgos
143 Matos 124 Mourgos
144 Vargas 124 Mourgos
145 Russell 100 King
146 Partners 100 King
147 Errazuriz 100 King
148 Cambrault 100 King
149 Zlotkey 100 King
150 Tucker 145 Russell
151 Bernstein 145 Russell
152 Hall 145 Russell
153 Olsen 145 Russell
154 Cambrault 145 Russell
155 Tuvault 145 Russell
156 King 146 Partners
157 Sully 146 Partners
158 McEwen 146 Partners
159 Smith 146 Partners
160 Doran 146 Partners
161 Sewall 146 Partners
162 Vishney 147 Errazuriz
163 Greene 147 Errazuriz
164 Marvins 147 Errazuriz
165 Lee 147 Errazuriz
166 Ande 147 Errazuriz
167 Banda 147 Errazuriz
168 Ozer 148 Cambrault
169 Bloom 148 Cambrault
170 Fox 148 Cambrault
171 Smith 148 Cambrault
172 Bates 148 Cambrault
173 Kumar 148 Cambrault
174 Abel 149 Zlotkey
175 Hutton 149 Zlotkey
176 Taylor 149 Zlotkey
177 Livingston 149 Zlotkey
178 Grant 149 Zlotkey
179 Johnson 149 Zlotkey
180 Taylor 120 Weiss
181 Fleaur 120 Weiss
182 Sullivan 120 Weiss
183 Geoni 120 Weiss
184 Sarchand 121 Fripp
185 Bull 121 Fripp
186 Dellinger 121 Fripp
187 Cabrio 121 Fripp
188 Chung 122 Kaufling
189 Dilly 122 Kaufling
190 Gates 122 Kaufling
191 Perkins 122 Kaufling
192 Bell 123 Vollman
193 Everett 123 Vollman
194 McCain 123 Vollman
195 Jones 123 Vollman
196 Walsh 124 Mourgos
197 Feeney 124 Mourgos
198 OConnell 124 Mourgos
199 Grant 124 Mourgos
200 Whalen 101 Kochhar
201 Hartstein 100 King
202 Fay 201 Hartstein
203 Mavris 101 Kochhar
204 Baer 101 Kochhar
205 Higgins 101 Kochhar
206 Gietz 205 Higgins
162 Vishney
133 Mallin
136 Philtanker
154 Cambrault
196 Walsh
104 Ernst
184 Sarchand
172 Bates
197 Feeney
150 Tucker
142 Davies
143 Matos
191 Perkins
119 Colmenares
200 Whalen
183 Geoni
180 Taylor
152 Hall
137 Ladwig
139 Seo
126 Mikkilineni
125 Nayer
170 Fox
175 Hutton
129 Bissot
163 Greene
105 Austin
176 Taylor
188 Chung
116 Baida
115 Khoo
144 Vargas
195 Jones
174 Abel
157 Sully
182 Sullivan
156 King
194 McCain
193 Everett
187 Cabrio
117 Tobias
179 Johnson
135 Gee
159 Smith
131 Marlow
190 Gates
169 Bloom
166 Ande
151 Bernstein
204 Baer
203 Mavris
160 Doran
155 Tuvault
107 Lorentz
185 Bull
128 Markle
134 Rogers
140 Patel
168 Ozer
178 Grant
141 Rajs
181 Fleaur
165 Lee
138 Stiles
173 Kumar
206 Gietz
164 Marvins
202 Fay
112 Urman
189 Dilly
110 Chen
153 Olsen
161 Sewall
186 Dellinger
109 Faviet
177 Livingston
198 OConnell
106 Pataballa
111 Sciarra
118 Himuro
132 Olson
192 Bell
113 Popp
171 Smith
127 Landry
167 Banda
130 Atkinson
158 McEwen
199 Grant
195 rows selected
答案 0 :(得分:2)
在我看来,right join
的查询会让经理和员工感到困惑。这就是right join
似乎无法回答明确答案的原因。使用left join
,您需要"所有员工,如果有相关经理,也需要经理"。使用right join
,无论他们是否是经理,您仍然可以获得"所有员工和#34;所以"右边的意思"是错的。
当然,表格的目的是包含两种类型,但可能通过更好的分离可以获得更清晰的图像。
说,经理是没有manager_id
的所有人(如果你的桌子不是深树,那就是这种情况)。然后,看看这个修改:
SELECT emps.employee_id as "Employee", emps.last_name, mgr.employee_id as "Manager", mgr.last_name
FROM employees emps
RIGHT JOIN (SELECT * FROM employees WHERE manager_id IS NULL) mgr
ON emps.manager_id = mgr.employee_id
像这样,right join
的数据基础将是所有经理的正确选择。即使他们没有员工,他们也会向员工展示。然而,这最后一次"甚至"你所选择的那种关系不会发生。
然后,我完全同意@Ameya Desphande有一个身份证号为156的第二位国王。这更令人费解; - )