Oracle XMLTYPE екстракт въз основа на стойност и състояние

Можете да се върнете пеша до брата или сестрата на ученика s_days възел:

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;

Или можете да получите дъщерните елементи от първата XMLTable, ако искате винаги да ги виждате, дори ако няма студентски възли:

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;

Между другото, тъй като сте на 12c, можете да използвате cross apply и outer apply - последното вместо външното съединение с фиктивно on 1=1 състояние.

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;

Всеки от тях получава същия резултат с вашите примерни данни:

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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