如何从 XMLType 节点提取元素路径
1个回答
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您可以实现的XMLTable函数从Oracle XML DB XQuery 函数设置帮助:
select * from
XMLTable(
'
declare function local:path-to-node( $nodes as node()* ) as xs:string* {
$nodes/string-join(ancestor-or-self::*/name(.), ''/'')
};
for $i in $rdoc//name
return <ret><name_path>{local:path-to-node($i)}</name_path>{$i}</ret>
'
passing
XMLParse(content '
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>'
)
as "rdoc"
columns
name_path varchar2(4000) path '//ret/name_path',
name_value varchar2(4000) path '//ret/name'
)
select * from
XMLTable(
'
declare function local:path-to-node( $nodes as node()* ) as xs:string* {
$nodes/string-join(ancestor-or-self::*/name(.), ''/'')
};
for $i in $rdoc//name
return <ret><name_path>{local:path-to-node($i)}</name_path>{$i}</ret>
'
passing
XMLParse(content '
<users><user><name>user1</name></user>
<user><name>user2</name></user>
<group>
<user><name>user3</name></user>
</group>
<user><name>user4</name></user>
</users>'
)
as "rdoc"
columns
name_path varchar2(4000) path '//ret/name_path',
name_value varchar2(4000) path '//ret/name'
)
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