我有一个XMLTYPE,其中包含以下内容:
<?xml version="1.0"?>
<users>
<user>
<name>user1</name>
</user>
<user>
<name>user2</name>
</user>
<user>
<name>user3</name>
</user>
</users>
如何通过所有元素“user”循环PL / SQL?感谢
答案 0 :(得分:20)
您可以使用EXTRACT
和XMLSequence
(将XML拆分为不同的块 - 此处为用户)循环遍历元素,如下所示:
SQL> SELECT extractvalue(column_value, '/user/name') "user"
2 FROM TABLE(XMLSequence(XMLTYPE(
3 '<?xml version="1.0"?>
4 <users>
5 <user>
6 <name>user1</name>
7 </user>
8 <user>
9 <name>user2</name>
10 </user>
11 <user>
12 <name>user3</name>
13 </user>
14 </users>').extract('/users/user'))) t;
user
--------
user1
user2
user3
答案 1 :(得分:16)
您可以使用XQuery。查看下面的选择声明。 v_xml_doc是包含XML数据的XMLTYPE变量。
select name
from XMLTable('for $i in /users/user
return $i'
passing v_xml_doc
columns name varchar2(200) path 'name'
)
答案 2 :(得分:12)
这个怎么样:
PROCEDURE xmltest IS
v_userlist XMLType;
v_count NUMBER(38) := 1;
BEGIN
/* define XML variable */
v_userlist := XMLType('<?xml version="1.0"?>
<users>
<user>
<name>user1</name>
</user>
<user>
<name>user2</name>
</user>
<user>
<name>user3</name>
</user>
</users>');
/* for each user, print out their name (each element can be extracted using xpath '//user[1]' '//user[2]' etc) */
WHILE v_userlist.existsNode('//user[' || v_count || ']') = 1 LOOP
dbms_output.put_line(v_userlist.extract('//user[' || v_count || ']/name/text()').getStringVal());
v_count := v_count + 1;
END LOOP;
END;
答案 3 :(得分:-2)
ITS VERY GOOD!!
CADENA CLOB;
BEGIN
SELECT CASE
WHEN EXISTSNODE (:NEW.MENSAJE, '/Body') <> 0 THEN 'ERROR'
ELSE NULL
END
INTO :NEW.DESCRIPCION_ERROR
FROM DUAL;
CADENA := :NEW.MENSAJE.EXTRACT ('/Body/xmlOriginal/text()').getStringVal ();
CADENA := REPLACE (CADENA, '<', '<');
CADENA := REPLACE (CADENA, '>', '>');