Sql 无法使用extractvalue函数获取数据

我有一个表，在Oracle DB中有一个XML类型的列。我试图在该表上使用extractvalue和updatexml函数

XML

有人能告诉我我做错了什么吗

---

整理完摘录后，我想将描述从计划付款替换为计划前付款。

您正在做的是工作-它将为您提供具有该属性值的记录节点的文本值。但该节点是空的：

<record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>

但是，它早就被弃用了，所以无论如何，您都不应该真正使用它

可以使用查看整个匹配记录节点；由于节点没有内容，因此获取其内容没有多大意义，正如当前查询所做的那样：

select XMLQuery('$x//record[@DESCRIPTION="Plan Payment"]'
    passing xmlcontent as "x"
    returning content
  ) as result
from table_xml;

RESULT                                                                          
--------------------------------------------------------------------------------
<record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>

对示例XML文档使用CTE，并将其包装在XMLSerialize调用中以保留格式以保持可读性：

with table_xml (xmlcontent) as (
  select xmltype(q'[<?xml version = '1.0' encoding = 'UTF-8'?>
<custominfo>
   <singlerecord  ZIP="51100"/>
   <multiplerecord type="ONE_TIME_CHARGES_LIST">
      <record DESCRIPTION="Device Payment" RATE="1000000" TAX="0"/>
      <record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>
      <record DESCRIPTION="Deposit" RATE="1000000" TAX="0"/>
   </multiplerecord>
</custominfo>]'
  ) from dual
)
select XMLSerialize(
    document
    XMLQuery('copy $i := $x modify
      (for $j in $i//record[@DESCRIPTION="Plan Payment"]/@DESCRIPTION
        return replace value of node $j with $r)
      return $i'
      passing xmlcontent as "x", 'Pre Plan Payment' as "r"
      returning content
    )
    indent
  ) as result
from table_xml;

RESULT                                                                          
--------------------------------------------------------------------------------
<?xml version="1.0" encoding="UTF-8"?>
<custominfo>
  <singlerecord ZIP="51100"/>
  <multiplerecord type="ONE_TIME_CHARGES_LIST">
    <record DESCRIPTION="Device Payment" RATE="1000000" TAX="0"/>
    <record DESCRIPTION="Pre Plan Payment" RATE="480000" TAX="0"/>
    <record DESCRIPTION="Deposit" RATE="1000000" TAX="0"/>
  </multiplerecord>
</custominfo>

---

如果要更新表中的值，可以将同一查询用作update语句的一部分，并对匹配的行进行筛选

您正在做的是工作-它将使用该属性值为您提供记录节点的文本值。但该节点是空的：

<record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>

但是，它早就被弃用了，所以无论如何，您都不应该真正使用它

可以使用查看整个匹配记录节点；由于节点没有内容，因此获取其内容没有多大意义，正如当前查询所做的那样：

select XMLQuery('$x//record[@DESCRIPTION="Plan Payment"]'
    passing xmlcontent as "x"
    returning content
  ) as result
from table_xml;

RESULT                                                                          
--------------------------------------------------------------------------------
<record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>

对示例XML文档使用CTE，并将其包装在XMLSerialize调用中以保留格式以保持可读性：

with table_xml (xmlcontent) as (
  select xmltype(q'[<?xml version = '1.0' encoding = 'UTF-8'?>
<custominfo>
   <singlerecord  ZIP="51100"/>
   <multiplerecord type="ONE_TIME_CHARGES_LIST">
      <record DESCRIPTION="Device Payment" RATE="1000000" TAX="0"/>
      <record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>
      <record DESCRIPTION="Deposit" RATE="1000000" TAX="0"/>
   </multiplerecord>
</custominfo>]'
  ) from dual
)
select XMLSerialize(
    document
    XMLQuery('copy $i := $x modify
      (for $j in $i//record[@DESCRIPTION="Plan Payment"]/@DESCRIPTION
        return replace value of node $j with $r)
      return $i'
      passing xmlcontent as "x", 'Pre Plan Payment' as "r"
      returning content
    )
    indent
  ) as result
from table_xml;

RESULT                                                                          
--------------------------------------------------------------------------------
<?xml version="1.0" encoding="UTF-8"?>
<custominfo>
  <singlerecord ZIP="51100"/>
  <multiplerecord type="ONE_TIME_CHARGES_LIST">
    <record DESCRIPTION="Device Payment" RATE="1000000" TAX="0"/>
    <record DESCRIPTION="Pre Plan Payment" RATE="480000" TAX="0"/>
    <record DESCRIPTION="Deposit" RATE="1000000" TAX="0"/>
  </multiplerecord>
</custominfo>

---

如果要更新表中的值，可以将同一查询用作update语句的一部分，并对匹配的行进行筛选

记录节点没有内容，只有属性；那么您希望看到什么输出呢？您的记录节点没有内容，只有属性；那么您希望看到什么样的输出呢？谢谢@Alex。我能够根据您给定的语法使用updatexml进行更新。虽然我不能使用XMLQuery，但它仍然返回nullThank@Alex。我能够根据您给定的语法使用updatexml进行更新。虽然我不能使用XMLQuery，但它仍然返回null

with table_xml (xmlcontent) as (
  select xmltype(q'[<?xml version = '1.0' encoding = 'UTF-8'?>
<custominfo>
   <singlerecord  ZIP="51100"/>
   <multiplerecord type="ONE_TIME_CHARGES_LIST">
      <record DESCRIPTION="Device Payment" RATE="1000000" TAX="0"/>
      <record DESCRIPTION="Plan Payment" RATE="480000" TAX="0"/>
      <record DESCRIPTION="Deposit" RATE="1000000" TAX="0"/>
   </multiplerecord>
</custominfo>]'
  ) from dual
)
select XMLSerialize(
    document
    XMLQuery('copy $i := $x modify
      (for $j in $i//record[@DESCRIPTION="Plan Payment"]/@DESCRIPTION
        return replace value of node $j with $r)
      return $i'
      passing xmlcontent as "x", 'Pre Plan Payment' as "r"
      returning content
    )
    indent
  ) as result
from table_xml;

RESULT                                                                          
--------------------------------------------------------------------------------
<?xml version="1.0" encoding="UTF-8"?>
<custominfo>
  <singlerecord ZIP="51100"/>
  <multiplerecord type="ONE_TIME_CHARGES_LIST">
    <record DESCRIPTION="Device Payment" RATE="1000000" TAX="0"/>
    <record DESCRIPTION="Pre Plan Payment" RATE="480000" TAX="0"/>
    <record DESCRIPTION="Deposit" RATE="1000000" TAX="0"/>
  </multiplerecord>
</custominfo>