PostgreSQL LEFT OUTER JOIN查询语法

Question

可以说我有一个table1：

  id      name
-------------
  1       "one"
  2       "two"
  3       "three"

并table2带有第一个外键的：

id    tbl1_fk    option   value
-------------------------------
 1      1         1        1
 2      2         1        1
 3      1         2        1
 4      3         2        1

现在，我想将其作为查询结果：

table1.id | table1.name | option | value
-------------------------------------
      1       "one"        1       1
      2       "two"        1       1
      3       "three"    
      1       "one"        2       1
      2       "two"    
      3       "three"      2       1

我该如何实现？

我已经尝试过：

SELECT
  table1.id,
  table1.name,
  table2.option,
  table2.value
FROM table1 AS table1
LEFT outer JOIN table2 AS table2 ON table1.id = table2.tbl1fk

但结果似乎忽略了空值：

1    "one"    1   1
2    "two"    1   1
1    "one"    2   1
3    "three"  2   1

求助：感谢Mahmoud Gamal：（加上GROUP BY）用此查询解决了

SELECT 
  t1.id,
  t1.name,
  t2.option,
  t2.value
FROM
(
  SELECT t1.id, t1.name, t2.option
  FROM table1 AS t1
  CROSS JOIN table2 AS t2
) AS t1
LEFT JOIN table2 AS t2  ON t1.id = t2.tbl1fk
                       AND t1.option = t2.option
group by t1.id, t1.name, t2.option, t2.value
ORDER BY t1.id, t1.name

Answer 1

您必须使用CROSS JOIN以获得name第一个表与option第二个表的所有可能组合。然后将LEFT JOIN这些与第二张表组合。就像是：

SELECT 
  t1.id,
  t1.name,
  t2.option,
  t2.value
FROM
(
  SELECT t1.id, t1.name, t2.option
  FROM table1 AS t1
  CROSS JOIN table2 AS t2
) AS t1
LEFT JOIN table2 AS t2  ON t1.id = t2.tbl1_fk
                       AND t1.option = t2.option