Oracle XMLTYPE 提取节点深度级别数

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*

这是对另一个问题的扩展要求。

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 FARE="10" Start="2020-08-07" End="2020-10-30" Mon="true" Tue="true" Weds="true" Thur="true" Fri="true" Sat="true" Sun="true">
            <STUDENT>
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="1" />
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="2" />
            </STUDENT>
        </S_DAYS>
        <S_DAYS FARE="20" Start="2020-08-07" End="2020-10-30" Mon="true" Tue="true" Weds="true" Thur="true" Fri="true" Sat="true" Sun="true">
            <STUDENT>
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="1" />
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="2" />
            </STUDENT>
        </S_DAYS>
    </PlanCode>
</SSO_XML>') 
 from dual
 )

select h.PlanCodeCode
,b.Original
,b.code
,b.s_code
,node_level
 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
    cross join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns 
                    ORIGINAL number path  '@ORIGINAL',
                    Code            varchar2(100) path '@Code',
                    S_CODE  number path '@S_CODE',
                    node_level      for ordinality
            ) b;

我正在尝试获取节点深度/级别数。 在这里，我有 2 个 S_DAYS 节点。 我想按顺序打印级别编号。

实际结果：

预期结果：

任何帮助都感激不尽。

Answer 1

FOR ORDINALITY仅返回生成的行号。 所以它必须在你想要计算的级别上生成。

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 FARE="10" Start="2020-08-07" End="2020-10-30" Mon="true" Tue="true" Weds="true" Thur="true" Fri="true" Sat="true" Sun="true">
            <STUDENT>
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="1" />
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="2" />
            </STUDENT>
        </S_DAYS>
        <S_DAYS FARE="20" Start="2020-08-07" End="2020-10-30" Mon="true" Tue="true" Weds="true" Thur="true" Fri="true" Sat="true" Sun="true">
            <STUDENT>
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="1" />
                <DIVISION ORIGINAL="150.05" Code="Flat" S_CODE="2" />
            </STUDENT>
        </S_DAYS>
    </PlanCode>
</SSO_XML>') 
 from dual
 )

select h.PlanCodeCode
,b.Original
,b.code
,b.s_code
,x.node_level
 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
    cross join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS'
             passing h.attributes
             columns 
                    node_level      for ordinality,
                    attributes xmltype path '/S_DAYS'
            ) x
    cross join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'S_DAYS/STUDENT/DIVISION'
             passing x.attributes
             columns 
                    ORIGINAL number path  '@ORIGINAL',
                    Code            varchar2(100) path '@Code',
                    S_CODE  number path '@S_CODE'
            ) b;