Question

我在postgresql中使用嵌套的左连接有一个奇怪的问题...这很难解释，但很容易显示=）这里我们是：

SELECT * FROM
(
    SELECT 1 as key1
) sub1
LEFT JOIN 
(
    SELECT sub3.key3, sub4.value2 FROM
    (
        SELECT 1 as key3
    ) sub3
    LEFT JOIN 
    (
        SELECT sub5.key5, COALESCE(sub6.value1, 1) as value2
        FROM
        (
            SELECT 1 as key5
        ) sub5
        LEFT JOIN
        (
            SELECT 1 as key6, value1
            FROM
            (
                SELECT NULL::integer as value1
            ) sub7
            WHERE false
        ) sub6 ON false

    )
    sub4 ON sub4.key5=sub3.key3
)
sub2 ON sub1.key1 = sub2.key3

此查询的结果：

key1;key3;value2
1;1;NULL

这就是问题 - 因为sub4中的COALESCE，value2不能为NULL（至少我认为它不能是=））无论如何，如果我们要改变

SELECT sub3.key3, sub4.value2 FROM

与

SELECT sub3.key3, value2 FROM

我们会得到正确的结果：

key1;key3;value2
1;1;1

我的思绪和手上有什么问题吗？还是一个bug？

提前致谢！

Answer 1

我已将您的查询翻译为Oracle语法，我得到了预期的结果

+------+------+--------+
| KEY1 | KEY3 | VALUE2 |
+------+------+--------+
|    1 |    1 |      1 |
+------+------+--------+

产生的

SELECT * FROM (
    SELECT 1 as key1 from dual
) sub1
LEFT JOIN (
    SELECT sub3.key3, sub4.value2 FROM (
        SELECT 1 as key3 from dual
    ) sub3
    LEFT JOIN (
        SELECT sub5.key5, COALESCE(sub6.value1, 1) as value2
        FROM (
            SELECT 1 as key5 from dual
        ) sub5
        LEFT JOIN (
            SELECT 1 as key6, value1
            FROM (
                SELECT cast(NULL as NUMBER(7)) as value1 from dual
            ) sub7
            WHERE 1=0
        ) sub6 ON 1=0
    )
    sub4 ON sub4.key5 = sub3.key3
)
sub2 ON sub1.key1 = sub2.key3

另外，通过阅读这个相当奇怪的查询，我仍然认为你从Postgres得到的结果是错误的。我认为您可以安全地将此文件作为错误提交给Postgres

Answer 2

我从postgresql团队得到了答案。判决：

This is on HEAD from today.  Clearly there's a problem.

所以，它是（而且是）一个错误。谢谢大家，谁参与了这个问题！ =）

嵌套左连接和合并的问题

2 个答案: