Oracle XMLTYPE 基于值和条件提取

Oracle XMLTYPE extract based on value and condition

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 的示例查询,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 预期有 S_DAYS 节点与 STUDENT -> DIVISION ,我们从 XML.

中获取 Amount 值

有可选节点 S_DAYS S_DAYS PCODE="Child1" 或 PCODE="Child2"

当存在 Child1 或 Child2 的节点 PCODE 时,我们必须自己申请现有行。

实际结果:

预期结果:

任何帮助将不胜感激。 谢谢

你可以走回学生s_days节点的兄弟姐妹:

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;

或者您可以从第一个 XMLTable 中获取 children,如果您即使没有学生节点也总是希望看到它们:

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;

顺便说一句,因为你在 12c 上,你可以使用 cross applyouter apply - 后者而不是带有虚拟 on 1=1 条件的外部连接。

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;

其中任何一个都得到与您的示例数据相同的结果:

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

db<>fiddle