Postgresql 是否可以使用基于列'；的子查询获取列值；钥匙是什么？

我有下表：

id   |     fruit |  parent |
----------------------------
id_1 |     apple |         |
id_2 |           |    id_3 |
id_3 | pineapple |         |
id_4 |      plum |    id_5 |
id_5 |      plum |         |

使用子查询/横向联接是否可以获得以下输出：

id_1 apple
id_2 pineapple
id_4 plum

因此，如果结果为空，则获取父对象的结果。尝试通过子查询获取此结果，收集链接的父对象以从中获取水果值，但在这种情况下，这些值与它们的ID成对出现，而不是与“孩子”ID成对出现。比如说：

id_1 apple
id_3 pineapple
id_4 plum

select id, 
       coalesce(fruit, (select t2.fruit
                        from the_table t2
                        where t2.id = t1.parent
                        limit 1)) as fruit
from the_table t1

---

如果这只是一个级别，您可以这样做：

id_1 apple
id_3 pineapple
id_4 plum

select id, 
       coalesce(fruit, (select t2.fruit
                        from the_table t2
                        where t2.id = t1.parent
                        limit 1)) as fruit
from the_table t1

---

如果这只是一个级别，您可以这样做：

id_1 apple
id_3 pineapple
id_4 plum

select id, 
       coalesce(fruit, (select t2.fruit
                        from the_table t2
                        where t2.id = t1.parent
                        limit 1)) as fruit
from the_table t1

---

您可以这样做：

id_1 apple
id_3 pineapple
id_4 plum

select id, 
       coalesce(fruit, (select t2.fruit
                        from the_table t2
                        where t2.id = t1.parent
                        limit 1)) as fruit
from the_table t1

编辑，抱歉，错过了postgres。这很有效

SELECT t1.id, (CASE WHEN t1.parent is NULL THEN t1.fruit ELSE t2.fruit END) as fruit
FROM fruits as t1
LEFT JOIN fruits as t2 ON t1.parent = t2.id;

---

您可以这样做：

id_1 apple
id_3 pineapple
id_4 plum

select id, 
       coalesce(fruit, (select t2.fruit
                        from the_table t2
                        where t2.id = t1.parent
                        limit 1)) as fruit
from the_table t1

编辑，抱歉，错过了postgres。这很有效

SELECT t1.id, (CASE WHEN t1.parent is NULL THEN t1.fruit ELSE t2.fruit END) as fruit
FROM fruits as t1
LEFT JOIN fruits as t2 ON t1.parent = t2.id;

---

@一匹没有名字的马，是的，谢谢你指出它是Postgres而不是MySQL。更新我的answer@a_horse_with_no_name，是的，谢谢你指出这是Postgres而不是MySQL。更新了我的答案