SQL Oracle XML空标记和属性

Question

我有SQL / XML和Oracle的问题。 我使用Oracle的示例和dept和emp。 一开始的情况：我想要一个部门的xml和员工：

SELECT Xmlelement("employees", dept.deptno, Xmlagg(
       Xmlelement("employee", ename))).
       getStringVal()
FROM   emp
       INNER JOIN dept
               ON dept.deptno = emp.deptno
GROUP  BY dept.deptno; 

现在我想要所有部门，即使没有任何员工（右外连接）。 此外，我想要员工的一些属性：

SELECT XMLELEMENT("employees", dept.deptno, 
       Xmlagg(
        XMLELEMENT("employee",
             xmlattributes(empno AS "empno", sal AS "salary"), ename)
        )
        ).getStringVal()
FROM   emp
       right outer join dept
                     ON dept.deptno = emp.deptno
GROUP  BY dept.deptno; 

但现在没有员工的部门有一个空标记“员工”。 我有一个解决这个问题的方法：

SELECT           xmlelement("employees", 
                    xmlattributes(dept.deptno as "Abt"), 
                    xmlagg( xmlforest(ename AS "employee") )
                 ).getstringval()
FROM             emp
RIGHT OUTER JOIN dept
ON               dept.deptno = emp.deptno
GROUP BY         dept.deptno;

现在我没有空标签，但我无法放置我想要的属性。 这个问题有什么解决方案吗？

最好的问候ses

Answer 1

使用此答案，您可以将DELETXML添加到第二个查询中

SELECT DELETEXML(
     XMLELEMENT("employees", dept.deptno, 
       Xmlagg(
        XMLELEMENT("employee",
             xmlattributes(empno AS "empno", sal AS "salary"), ename)
        )
        ),
         '//*[not(text())][not(*)]'
       ).getStringVal()
FROM   emp
       right outer join dept
                     ON dept.deptno = emp.deptno
GROUP  BY dept.deptno;

样本输出

<employees>1<employee empno="1" salary="100">name1</employee></employees>
<employees>2<employee empno="2" salary="100">name2</employee></employees>
<employees>3<employee empno="3" salary="100">name3</employee></employees>
<employees>4</employees>
<employees>5</employees>

Answer 2

您可以使用案例表达式来首先不生成流氓<employee/>节点：

SELECT XMLELEMENT("employees", dept.deptno, 
         Xmlagg(
           CASE WHEN emp.empno IS NOT NULL THEN
             XMLELEMENT("employee",
               xmlattributes(empno AS "empno", sal AS "salary"), ename)
           END
         )
       ).getStringVal()
FROM   emp
       right outer join dept
                     ON dept.deptno = emp.deptno
GROUP  BY dept.deptno; 

XMLELEMENT("EMPLOYEES",DEPT.DEPTNO,XMLAGG(CASEWHENEMP.EMPNOISNOTNULLTHENXMLELEMENT("EMPLOYEE",XMLATTRIBUTES(EMPNOAS"EMPNO",SALAS"SALARY"),ENAME)END)).GETSTRINGVAL()
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
<employees>10<employee empno="7839" salary="5000">KING</employee><employee empno="7934" salary="1300">MILLER</employee><employee empno="7782" salary="2450">CLARK</employee></employees>
<employees>20<employee empno="7566" salary="2975">JONES</employee><employee empno="7876" salary="1100">ADAMS</employee><employee empno="7369" salary="800">SMITH</employee><employee empno="7902" salary="3000">FORD</employee><employee empno="7788" salary="3000">SCOTT</employee></employees>
<employees>30<employee empno="7698" salary="2850">BLAKE</employee><employee empno="7900" salary="950">JAMES</employee><employee empno="7844" salary="1500">TURNER</employee><employee empno="7654" salary="1250">MARTIN</employee><employee empno="7521" salary="1250">WARD</employee><employee empno="7499" salary="1600">ALLEN</employee></employees>
<employees>40</employees>