Question

我遇到以下SQL问题。为什么在地球上减去2个完全相同的查询会返回非空结果？我尝试过“UNION ALL”而不是UNION，我尝试了很多其他的东西，但都没有。请指教。

    SELECT y.segment1 po_num, fad.seq_num seq, fdst.short_text st
              FROM applsys.fnd_attached_documents fad,
                   applsys.fnd_documents fd,
                   applsys.fnd_documents_short_text fdst,
                   po_headers_all y
             WHERE     1 = 1
                   AND fad.pk1_value(+) = y.po_header_id
                   AND fad.entity_name = 'PO_HEADERS'
                   AND fad.document_id = fd.document_id
                   AND fd.datatype_id = 1
                   and fad.seq_num>=100
                   AND fdst.media_id = fd.media_id
                   and y.type_lookup_code='STANDARD'
                   AND NVL(y.CANCEL_FLAG,'N')='N'
                 --  and y.segment1 in (100,1000,100,650,26268)
                --   and y.segment1=1000
            UNION 
            SELECT poh.segment1, 1, '1' --null, null
              FROM    po.po_headers_all poh
                   LEFT JOIN
                      (SELECT fad1.pk1_value
                         FROM applsys.fnd_attached_documents fad1,
                              applsys.fnd_documents fd1
                        WHERE     1 = 1
                              AND fad1.entity_name = 'PO_HEADERS'
                              AND fad1.document_id = fd1.document_id
                              and fad1.seq_num>=100
                              AND fd1.datatype_id = 1) sub1
                   ON poh.po_header_id = sub1.pk1_value
             WHERE sub1.pk1_value IS NULL 
                        and poh.type_lookup_code='STANDARD'
                        AND NVL(poh.CANCEL_FLAG,'N')='N'
                      --  and poh.segment1 in (100,1000,100,650,26268) 
                      --  and poh.segment1=1000                       
          --   and poh.segment1=650)              
          minus
   SELECT y.segment1 po_num, fad.seq_num seq, fdst.short_text st
              FROM applsys.fnd_attached_documents fad,
                   applsys.fnd_documents fd,
                   applsys.fnd_documents_short_text fdst,
                   po_headers_all y
             WHERE     1 = 1
                   AND fad.pk1_value(+) = y.po_header_id
                   AND fad.entity_name = 'PO_HEADERS'
                   AND fad.document_id = fd.document_id
                   AND fd.datatype_id = 1
                   and fad.seq_num>=100
                   AND fdst.media_id = fd.media_id
                   and y.type_lookup_code='STANDARD'
                   AND NVL(y.CANCEL_FLAG,'N')='N'
                   --and y.segment1 in (100,1000,100,650,26268)
                   --and y.segment1=1000
            UNION 
            SELECT poh.segment1, 1, '1'--null,null
              FROM    po.po_headers_all poh
                   LEFT JOIN
                      (SELECT fad1.pk1_value
                         FROM applsys.fnd_attached_documents fad1,
                              applsys.fnd_documents fd1
                        WHERE     1 = 1
                              AND fad1.entity_name = 'PO_HEADERS'
                              AND fad1.document_id = fd1.document_id
                              and fad1.seq_num>=100
                              AND fd1.datatype_id = 1) sub1
                   ON poh.po_header_id = sub1.pk1_value
             WHERE sub1.pk1_value IS NULL 
                        and poh.type_lookup_code='STANDARD'
                        AND NVL(poh.CANCEL_FLAG,'N')='N'
                      --  and poh.segment1 in (100,1000,100,650,26268)
                      --  and poh.segment1=1000
                      --   and poh.segment1=650)

Answer 1

使用括号。现在，当您打算((set1 UNION set2) MINUS set1) UNION set2时，您正在进行(set1 UNION set2) MINUS (set1 UNION set2)。

换句话说，你将set1和set2联合起来，从中删除set1并将set2与之结合，而你可能想要取一个set1和set2的并集，并从中删除set1和set2的并集。 UNION和MINUS具有相同的优先级，并按照遇到的顺序进行处理。

Answer 2

问题在于，首先，你执行前两个查询的UNION，然后，你执行第三个的MINUS，然后使用第四个查询对结果进行UNION。

Answer 3

请注意，UNION [ALL]和MINUS具有相同的优先级。基本上，如果我们忽略细节，你就是这样做：

SELECT query1
UNION
SELECT query2
MINUS
SELECT query1
UNION
SELECT query2

由于所有这些集合运算符具有相同的优先级，因此它们逐个进行评估。 SELECT query1 MINUS SELECT query2 UNION SELECT query1应该返回query2。然后，应用了最后一个UNION，结果为SELECT query2 UNION SELECT query2，当然是query2。

要解决这个问题，你必须做这样的事情：

SELECT * FROM (SELECT query1
               UNION
               SELECT query2)
MINUS
SELECT * FROM (SELECT query1
               UNION
               SELECT query2)

2个相同查询的减号（除外）返回非空结果集

3 个答案: