animal
id name
1 dog
1 cat
2 lion
怎么得到相同id的name list呢?也就是想得到id 为1的list为dog, cat格式。
1. 首先Oracle有一个方法,wm_concat()
select a.id , wm_concat(a.name) as namelist from animal a group by a.id 但很遗憾,这个方法现在已经不support了. 2. 第二个方法XMLAGG() select a.id , (rtrim(xmlagg(xmlelement("xml", a.name||', ')order by a.id).extract ('//text()'),', ')) as namelistfrom animal agroup by a.id; 3. LISTAGG()这个方法有些版本不支持,比如我的电脑和学校的。。下面是官方用法:SELECT department_id "Dept.",
LISTAGG(last_name, '; ') WITHIN GROUP (ORDER BY hire_date) "Employees"
FROM employees
GROUP BY department_id
ORDER BY department_id;