我有这样的疑问:
SELECT c.name, i.name FROM inv_invoice_items c
LEFT OUTER JOIN inv_invoice_items i ON i.name = c.name
WHERE c.id_invoice = 2108 AND i.id_invoice = (SELECT id_invoice FROM inv_invoices WHERE id = 2108)
对于此查询,我有这样的结果:
name | name
-------------
pen | pen
但是对于没有加入的选择:
SELECT c.name FROM inv_invoice_items c
WHERE c.id_invoice = 2108
结果是:
name
------
pen
pencil
和第二个查询:
SELECT i.name FROM inv_invoice_items i
WHERE i.id_invoice = (SELECT id_invoice FROM inv_invoices WHERE id = 2108)
给出结果:
name
------
pen
我希望第一次加入查询结果:
name | name
---------------
pen | pen
pencil | NULL
如何实现这样的结果?我认为这种方式应该工作LEFT OUTER JOIN。感谢您提前提出任何建议。
聚苯乙烯。我需要记录纠正和更正(相关)发票的发票项目的差异。
一些示例数据:
create table inv_invoices (id bigint, id_invoice bigint, primary key(id));
create table inv_invoice_items (id bigint, id_invoice bigint NOT NULL, name character varying(100) NOT NULL, primary key (id));
insert into inv_invoices values (2105, NULL), (2106, NULL), (2107, NULL), (2108, 2106);
insert into inv_invoice_items values (1000, 2105, 'pen'), (1001, 2105, 'pencil'), (1002,2106, 'pen'), (1003, 2107, 'rubber'),
(1004, 2107, 'pencil'), (1005, 2108, 'pen'), (1006, 2108, 'pencil');
答案 0 :(得分:3)
改为尝试:
SELECT DISTINCT c.name, i.name
FROM inv_invoice_items AS c
LEFT OUTER JOIN inv_invoice_items AS i ON i.name = c.name
AND c.id_invoice = 2108
LEFT OUTER JOIN inv_invoices AS i2 ON i.id_invoice = i2.id_invoice
AND i2.id = 2108;
击> <击> 撞击>
试试这个:
SELECT
t.name AS name1,
v.name AS name2
FROM
(
SELECT c.name
FROM inv_invoice_items c
WHERE c.id_invoice = 2108
) AS t
LEFT JOIN
(
SELECT i.name
FROM inv_invoice_items i
WHERE i.id_invoice = (SELECT id_invoice
FROM inv_invoices
WHERE id = 2108)
) AS v ON t.name = v.name;
这会给你:
| NAME1 | NAME2 |
-------------------
| pen | pen |
| pencil | (null) |