基於值和條件的 Oracle XMLTYPE 提取

Question

SELECT * FROM v$version;
Oracle Database 12c Enterprise Edition Release 12.1.0.2.0 - 64bit Production
PL/SQL Release 12.1.0.2.0 - Production
"CORE   12.1.0.2.0  Production"
TNS for Linux: Version 12.1.0.2.0 - Production
NLSRTL Version 12.1.0.2.0 - Production

我有 XML 的示例查詢，如下所示：

with t(xml) as 
(
select xmltype(
'<SSO_XML
    xmlns:xsd="http://www.w3.org/2001/XMLSchema"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
TimeStamp="2020-08-05T21:57:23Z" 
Target="Production" 
Version="1.0" 
TransactionIdentifier="PLAN_A" 
SequenceNmbr="123456"
    xmlns="http://www.w3.org/2001/XMLSchema">
    <PlanCode PlanCodeCode="CHOICE">
        <S_DAYS PCODE="P123">
            <STUDENT>
                <DIVISION Amount="150.05" Code="Flat" S_CODE="1" />
                <DIVISION Amount="250.05" Code="Flat" S_CODE="2" />
            </STUDENT>
        </S_DAYS>
        <S_DAYS PCODE="P1234">
            <STUDENT>
                <DIVISION Amount="150.05" Code="Flat" S_CODE="1" />
                <DIVISION Amount="250.05" Code="Flat" S_CODE="2" />
            </STUDENT>
        </S_DAYS>
          <S_DAYS PCODE="Child1">
            <AdditonalFare>
              <AdditonalFareAmount Amount="100"/>
            </AdditonalFare>
          </S_DAYS> 
          <S_DAYS PCODE="Child2">
            <AdditonalFare>
              <AdditonalFareAmount Amount="130"/>
            </AdditonalFare>
          </S_DAYS> 
    </PlanCode>
</SSO_XML>') 
 from dual
 )

select h.PlanCodeCode
,b.*
 from   t
    cross join
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode'
            ) h
    left join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
                    , child1_amount number path './../../@Amount[1]' --->Child1
                    , child2_amount number path './../../@Amount[2]' --->Child2
            ) b on 1=1;

XML 預計會有 S_DAYS 節點和 STUDENT -> DIVISION，我們從 XML 獲取金額值。

有可選節點 S_DAYS 與 S_DAYS 與 PCODE="Child1" 或 PCODE="Child2"

當 Child1 或 Child2 的節點 PCODE 存在時，我們必須自己申請現有的行。

實際結果：

預期結果：

任何幫助都感激不盡。 謝謝。

Answer 1

您可以回到學生s_days節點的兄弟節點：

select h.PlanCodeCode, b.amount, b.pcode, b.child1_amount, b.child2_amount
 from   t
    cross join
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode'
            ) h
    left join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
                    , child1_amount number path './../../../S_DAYS[@PCODE="Child1"]/AdditonalFare/AdditonalFareAmount/@Amount'
                    , child2_amount number path './../../../S_DAYS[@PCODE="Child2"]/AdditonalFare/AdditonalFareAmount/@Amount'
            ) b on 1=1;

或者，您可以從第一個 XMLTable 中獲取孩子，如果即使沒有學生節點也總是想查看它們：

select h.PlanCodeCode, b.amount, b.pcode, h.child1_amount, h.child2_amount
 from   t
    cross join
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode',
                     child1_amount number path './PlanCode/S_DAYS[@PCODE="Child1"]/AdditonalFare/AdditonalFareAmount/@Amount',
                     child2_amount number path './PlanCode/S_DAYS[@PCODE="Child2"]/AdditonalFare/AdditonalFareAmount/@Amount'
            ) h
    left join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
            ) b on 1=1;

順便說一句，當您在 12c 上時，您可以使用cross apply和outer apply - 后者而不是on 1=1條件下使用 dummy 進行外部連接。

select h.PlanCodeCode, b.amount, b.pcode, h.child1_amount, h.child2_amount
 from   t
    cross apply
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode',
                     child1_amount number path './PlanCode/S_DAYS[@PCODE="Child1"]/AdditonalFare/AdditonalFareAmount/@Amount',
                     child2_amount number path './PlanCode/S_DAYS[@PCODE="Child2"]/AdditonalFare/AdditonalFareAmount/@Amount'
            ) h
    outer apply xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
            ) b;

其中任何一個都與您的樣本數據得到相同的結果：

PLANCODECODE | AMOUNT | PCODE | CHILD1_AMOUNT | CHILD2_AMOUNT
:----------- | -----: | :---- | ------------: | ------------:
CHOICE       | 150.05 | P123  |           100 |           130
CHOICE       | 250.05 | P123  |           100 |           130
CHOICE       | 150.05 | P1234 |           100 |           130
CHOICE       | 250.05 | P1234 |           100 |           130

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