Snowflake XMLGET() 获取多个内部标签值

Question

下面是 xmldata 列中的 XML 数据。

<Inquiry>
  <MemberCode>56</MemberCode>
  <MemberName>COMP</MemberName>
  <Date>16</Date>
</Inquiry>

如何使用 XMLGET 从内部标记中仅提取值

SELECT TOP 100
    KEY,
    GET(b.VALUE,'@')::STRING as name, --returns Inquiry
    b.VALUE:"$"[1]."$"[1],--returns null
    GET(XMLGET(b.VALUE, GET(b.VALUE,'@')::STRING), '$') as value1,--returns null
    -- XMLGET(VALUE,GET(b.VALUE,'@')::STRING):"$" as value,--returns null
    b.value as children,-- returns full xmldata
    b.this
    
FROM
    (SELECT  
        KEY,
        VALUE
        ,this
    from
         FULL_XML,
        LATERAL FLATTEN(XMLGET(XMLGET(XML_DATA,'Report'),'Inquiries'):"$") )b ;

值列的预期输出是：

56COMP16

Answer 1

如果您的对象是单级事物，那么要访问该事物的成员，您只需命名它们，转换为字符串，然后连接它们：

with data_in_cte as (
    select parse_xml('<Inquiry>
  <MemberCode>56</MemberCode>
  <MemberName>COMP</MemberName>
  <Date>16</Date>
</Inquiry>') as XML_DATA
)
select
    get(xmlget(xml_data,'MemberCode'),'$')::text as a,
    get(xmlget(xml_data,'MemberName'),'$')::text as b,
    get(xmlget(xml_data,'Date'),'$')::text as  c,
    a || b || c as result
from data_in_cte