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        XML Oracle:多子节点提取

        时间:2023-11-02
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                  本文介绍了XML Oracle:多子节点提取的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  我有一个 xml 代码:

                  I have an xml code :

                  <begin>
                      <entry>
                          <lastname>gordon</lastname>
                          <NumberList>
                              <number>100</number>
                              <codelist>
                                   <code>213</code>
                                   <code>214</code>
                              <codelist>
                              <login>
                                   <user>user1</user>
                                   <user>user2</user>
                              </login>
                          <NumberList>
                          <address>
                              <addresslist>Jl. jalan pelan-pelan ke Bekasi, Indonesia</addresslist>
                          </address>
                      </entry>
                      <entry>
                          <lastname>mark</lastname>
                          <address>
                              <addresslist>Jl. jalan cepet-cepet ke Jakarta, Indonesia</addresslist>
                          </address>
                      </entry>
                  </begin>
                  

                  我的代码:

                  FOR r IN (SELECT VALUE(p) col_val,
                                   EXTRACT(VALUE(P), '/entry/codelist') AS code,
                                   EXTRACT(VALUE(P), '/entry/login') AS login
                             FROM TABLE(XMLSequence(Extract(x,'/begin/entry'))) p)
                  LOOP
                     IF r.col_val.existsnode('/entry/lastname/text()') > 0 
                     THEN
                        vc_lastname := r.col_val.extract('/sdnEntry/lastname/text()').getstringval();
                     END IF;
                  
                     IF r.col_val.existsnode('/entry/address/addresslist/text()') > 0 
                     THEN
                      vc_address := r.col_val.extract('/sdnEntry/address/addresslist/text()').getstringval();
                     END IF;
                  
                     IF r.col_val.existsnode('/entry/codelist/id/code/text()') > 0 AND r.col_val.existsnode('/entry/login/user/text()') > 0 
                     THEN
                        FOR R1 IN (SELECT EXTRACTVALUE(VALUE(T1), '/codelist/code/text()') AS code
                                     FROM TABLE(XMLSEQUENCE(EXTRACT(R.code, '/codelist'))) T1)
                        LOOP
                           DBMS_OUTPUT.PUT_LINE(vc_uid||' - '||vc_firstName||' - '||R1.code||' - '||R2.address);
                        END LOOP;
                  
                        FOR R2 IN (SELECT
                                          EXTRACTVALUE(VALUE(T1), '/login/user/text()') AS user
                                     FROM TABLE(XMLSEQUENCE(EXTRACT(R.address, 'login/'))) T1)
                        LOOP
                           DBMS_OUTPUT.PUT_LINE(vc_uid||' - '||vc_firstName||' - '||R2.user||' - '||R2.address);
                        END LOOP;
                    ELSE
                          DBMS_OUTPUT.PUT_LINE(vc_uid||' - '||vc_firstName);
                    END IF;
                  

                  我的问题:如何循环子节点使数据变成这样:

                  My problem : How to loop child nodes so the data will become like this :

                  LastName | Number | code    | user  |   address
                  gordon   | 100    | 213     | user1 |Jl. jalan pelan-pelan ke Bekasi, Indonesia
                  gordon   | 100    | 213     | user2 |Jl. jalan pelan-pelan ke Bekasi, Indonesia
                  gordon   | 100    | 214     | user1 |Jl. jalan pelan-pelan ke Bekasi, Indonesia
                  gordon   | 100    | 214     | user2 |Jl. jalan pelan-pelan ke Bekasi, Indonesia
                  mark     | Null   | null    | null  |Jl. jalan cepet-cepet ke Jakarta, Indonesia
                  

                  任何帮助将不胜感激.

                  推荐答案

                  你可以使用 XMLTable() 函数来达到预期的效果:

                  You can achieve desired result by using XMLTable() function:

                  select q.Lastname
                       , q.Numberid
                       , s.codeid
                       , w.LoginId
                       , q.address
                    from t1 t
                    left join xmltable('/begin/entry'
                                        passing t.xml_col 
                                        columns LastName   varchar2(21)  path 'lastname',
                                                NumberId   number        path 'NumberList/number',
                                                Address    varchar2(201) path 'address/addresslist',
                                                CodeList   XmlType       Path 'NumberList/codelist/code',
                                                Logins     XmlType       Path 'NumberList/login/user'
                                        ) q
                      on (1=1) 
                    left join xmltable('/code'
                                        passing q.CodeList
                                        columns CodeId number path '.') s
                      on (1=1)
                    left join   xmltable('/user'
                                          passing q.Logins
                                          columns LoginId varchar2(11) path '.') w
                      on (1=1)
                  

                  结果:SQLFiddle 演示

                  Lastname Numberid Codeid Loginid Address 
                  ---------------------------------------------------------------------------
                  gordon   100      213    user1   Jl. jalan pelan-pelan ke Bekasi, Indonesia 
                  gordon   100      213    user2   Jl. jalan pelan-pelan ke Bekasi, Indonesia 
                  gordon   100      214    user1   Jl. jalan pelan-pelan ke Bekasi, Indonesia 
                  gordon   100      214    user2   Jl. jalan pelan-pelan ke Bekasi, Indonesia 
                  mark     null     null   null    Jl. jalan cepet-cepet ke Jakarta, Indonesia 
                  

                  了解更多关于XMLTable() 函数.

                  注意:使用 11.2.0.2 之前的 Oracle 版本,当 cursor_sharing 参数设置为 FORCESIMILAR(从 11.2 开始弃用).将cursor_sharing参数设置为EXACT(默认值),即可解决问题.

                  Note: Working with Oracle releases prior to 11.2.0.2, you can encounter ORA-1780 error(bug 8545377) on certain types of XML queries when cursor_sharing parameter is set to FORCE or SIMILAR(deprecated starting from 11.2). Setting cursor_sharing parameter to EXACT(default value), will solve the problem.

                  这篇关于XML Oracle:多子节点提取的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

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