oracle10g第二版中支持xmlquery查询,产生相应的xml文件
如下创建xml文件目录
BEGIN
IF(DBMS_XDB.CREATEFOLDER('/public/employees')) THEN
DBMS_OUTPUT.PUT_LINE('Folder is created');
ELSE
DBMS_OUTPUT.PUT_LINE('Cannot create folder');
END IF;
COMMIT;
END;创建xml文件
DECLARE
XMLdoc XMLType;
BEGIN
SELECT XMLQuery(
'for $j in 1
return (
<EMPLOYEES> {
for $i in ora:view("YANGYUN", "ADDRESS")/ROW
where $i/ADDRESS_SEQ <= 2
return (<EMPLOYEE>
<EMPNO>{xs:integer($i/ADDRESS_SEQ)}</EMPNO>
<ENAME>{xs:string($i/FIRST_NAME)}</ENAME>
<SAL>{xs:string($i/LAST_NAME)}</SAL>
</EMPLOYEE>)} </EMPLOYEES>)'
RETURNING CONTENT) INTO XMLdoc FROM DUAL;
IF(DBMS_XDB.CREATERESOURCE('/public/employees/employees.xml', XMLdoc)) THEN
DBMS_OUTPUT.PUT_LINE('Resource is created');
ELSE
DBMS_OUTPUT.PUT_LINE('Cannot create resource');
END IF;
COMMIT;
END;
现在我想把oracle服务器上的/public/employees/employees.xml取出来,怎么取???
如下创建xml文件目录
BEGIN
IF(DBMS_XDB.CREATEFOLDER('/public/employees')) THEN
DBMS_OUTPUT.PUT_LINE('Folder is created');
ELSE
DBMS_OUTPUT.PUT_LINE('Cannot create folder');
END IF;
COMMIT;
END;创建xml文件
DECLARE
XMLdoc XMLType;
BEGIN
SELECT XMLQuery(
'for $j in 1
return (
<EMPLOYEES> {
for $i in ora:view("YANGYUN", "ADDRESS")/ROW
where $i/ADDRESS_SEQ <= 2
return (<EMPLOYEE>
<EMPNO>{xs:integer($i/ADDRESS_SEQ)}</EMPNO>
<ENAME>{xs:string($i/FIRST_NAME)}</ENAME>
<SAL>{xs:string($i/LAST_NAME)}</SAL>
</EMPLOYEE>)} </EMPLOYEES>)'
RETURNING CONTENT) INTO XMLdoc FROM DUAL;
IF(DBMS_XDB.CREATERESOURCE('/public/employees/employees.xml', XMLdoc)) THEN
DBMS_OUTPUT.PUT_LINE('Resource is created');
ELSE
DBMS_OUTPUT.PUT_LINE('Cannot create resource');
END IF;
COMMIT;
END;
现在我想把oracle服务器上的/public/employees/employees.xml取出来,怎么取???
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你想怎么取?在哪里取?取到哪?