1

我想要实现的是为这个特定的 xml 创建一个包含所有多项目子节点的单独行的表：

<ABCD>
<EMPLOYEE id="11" date="25-Apr-1983"> 
<NameDetails> 
<Name NameType="a"> 
<NameValue> 
<FirstName>ABCD</FirstName> 
<Surname>PQR</Surname> 
<OriginalName>TEST1</OriginalName> 
<OriginalName>TEST2</OriginalName> 
</NameValue> 
</Name> 
 <Name NameType="b"> 
 <NameValue> 
<FirstName>TEST3</FirstName> 
<Surname>TEST3</Surname> 
</NameValue> 
 <NameValue> 
<FirstName>TEST5</FirstName> 
<MiddleName>TEST6</MiddleName> 
<Surname>TEST7</Surname> 
<OriginalName>JAB1</OriginalName> 
</NameValue> 
 <NameValue> 
<FirstName>HER</FirstName> 
<MiddleName>HIS</MiddleName> 
<Surname>LOO</Surname> 
</NameValue> 
</Name>  <Name NameType="c"> 
<NameValue> 
<FirstName>CDS</FirstName> 
<MiddleName>DRE</MiddleName> 
<Surname>QWE</Surname> 
</NameValue> 
 <NameValue> 
<FirstName>CCD</FirstName> 
<MiddleName>YTD</MiddleName> 
<Surname>QQA</Surname> 
</NameValue> 
 <NameValue> 
<FirstName>DS</FirstName> 
<Surname>AzDFz</Surname> 
</NameValue> 
</Name> 
</NameDetails> 

</EMPLOYEE >
</ABCD>

我尝试使用查询：

SELECT t.personid,n.nametypeid,t.firstname,t.middlename,t.surname,t.maidenname,t.originalName
FROM xml_files p,master_nametypes n,
     XMLTable(
      'for $i in ADCD/Employee/NameDetails/Name/NameValue
       return <row>
       {
          $i/../../../@id,
          $i/../@NameType,
          $i/FirstName,
          $i/MiddleName,
          $i/OriginalName
          $i/Surname,
          $i/MaidenName,
          $i/Suffix,
          $i/SingleStringName,
          $i/EntityName

       } 
       </row>' 
      PASSING p.filecontent
      COLUMNS 
              personid  number PATH '@id',
              nametypeid    VARCHAR2(255)  PATH '@NameType',
              firstname    VARCHAR2(4000)  PATH 'FirstName',
              middlename    VARCHAR2(4000)  PATH 'MiddleName',
              surname    VARCHAR2(4000)  PATH 'Surname',
              maidenname    VARCHAR2(4000)  PATH 'MaidenName',
              originalName    VARCHAR2(4000)  PATH '.'

              ) t  where t.nametypeid = n.nametype and n.recordtype = 'Employee'
;

但是当“NAMEVALUE”节点下有多个像“ORIGINALNAME”这样的子节点时，这将引发错误。如何根据父节点在单独的行中检索这些值。有人可以帮我纠正这个查询。任何帮助，将不胜感激。

4

2 回答 2

1

您也可以尝试这样的事情（它会使表格反规范化）：

这是一个 sqlfiddle 演示

SELECT t.personid,t.nametypeid,t.firstname,t.middlename,t.surname,t.maidenname,t.originalName
FROM xml_files p,
     XMLTable(
      '
       for $m in $ADCD//NameValue/OriginalName
       |  $ADCD//NameValue[not(exists(OriginalName))]
       return
       <row>
       {
          ($m/../../../../@id , $m/../../../@id)[1],
          ($m/../../@NameType,$m/../@NameType)[1],
          ($m/../FirstName,$m/FirstName)[1],
          ($m/../MiddleName,$m/MiddleName)[1],
          ($m,"")[1],
          ($m/../Surname,$m/Surname)[1],
          ($m/../MaidenName,$m/MaidenName)[1],
          ($m/../Suffix,$m/Suffix)[1],
          ($m/../SingleStringName,$m/SingleStringName)[1],
          ($m/../EntityName,$m/EntityName)[1]
       } 
       </row>' 

      PASSING p.filecontent as "ADCD"
      COLUMNS 

              personid  number PATH '@id',
              nametypeid    VARCHAR2(255)  PATH '@NameType',
              firstname    VARCHAR2(4000)  PATH 'FirstName',
              middlename    VARCHAR2(4000)  PATH 'MiddleName',
              surname    VARCHAR2(4000)  PATH 'Surname',
              maidenname    VARCHAR2(4000)  PATH 'MaidenName',
              originalName    VARCHAR2(4000)  PATH 'OriginalName'


              ) t  
;
于 2013-05-30T14:57:02.500 回答
0

尝试这个：

SQL> SELECT t.personid, t.firstname, t.middlename,
  2         t.surname,t.maidenname,
  3         replace(replace(t.originalName, '<OriginalName>'),
  4                 '</OriginalName>', ' ') originalName
  5  FROM xml_files p,
  6       XMLTABLE (
  7        --'ABCD/EMPLOYEE/NameDetails/Name/NameValue'
  8        'for $i in ABCD/EMPLOYEE/NameDetails/Name/NameValue
  9         return <row>
 10                    {$i/../../../@id}
 11                    {$i/../@NameType}
 12                    {$i/FirstName}{$i/MiddleName}{$i/OriginalName}
 13                    {$i/Surname}{$i/MaidenName}
 14                </row>'
 15        PASSING p.filecontent
 16        COLUMNS
 17                personid     NUMBER         PATH '@id',
 18                nametypeid   VARCHAR2(255)  PATH '@NameType',
 19                firstname    VARCHAR2(4000) PATH 'FirstName',
 20                middlename   VARCHAR2(4000) PATH 'MiddleName',
 21                surname      VARCHAR2(4000) PATH 'Surname',
 22                maidenname   VARCHAR2(4000) PATH 'MaidenName',
 23                originalName XMLTYPE        PATH 'OriginalName'
 24                ) t;

  PERSONID FIRSTNAME  MIDDLENAME  SURNAME  MAIDENNAME  ORIGINALNAME
---------- ---------- ----------- -------- ----------- ------------
        11 ABCD                   PQR                  TEST1 TEST2
        11 TEST3                  TEST3                
        11 TEST5      TEST6       TEST7                JAB1
        11 HER        HIS         LOO                  
        11 CDS        DRE         QWE                  
        11 CCD        YTD         QQA
于 2013-05-30T13:19:55.007 回答