先做这样的假设,学生的学号和姓名可以唯一确定一个学生。Oracle存储过程create or replace function get_stu_birth(vid varchar,vname varchar) return date is
vbirth date;
n number;begin
select count(*),birth into n,pbirth from student where id = vid and name = vname;
if n>0 then
vbirth:=pbirth;
else
null;
end if;
return vbirth;
end;
StudentSqlMapper.xml<sqlMap namespace="studentMapper">
<parameterMap id="stuparam" class="java.util.Map">
<parameter property="vbirth" jdbcType="DATE" javaType="java.lang.String" mode="OUT" />
<parameter property="vid" jdbcType="VARCHAR" javaType="java.lang.String" mode="IN" />
<parameter property="vname" jdbcType="VARCHAR" javaType="java.lang.String" mode="IN" />
</parameterMap>
<procedure id="stuproc" parameterMap="stuparam">
<![CDATA[
{? = call get_stu_birth(?,?) }
]]>
</procedure>
</sqlMap>
Student.javaMap<String, String> mapIn = new HashMap<String, String>();
mapIn.put("vid", "100");
mapIn.put("vname", "xy");
mapIn.put("vbirth", "");
baseDao.selectObject("studentMapper.stuproc", mapIn);
Date birth = DateUtil.toDate(mapIn.get("vbirth").substring(0, mapIn.get("vbirth").length() - 2), "yyyy-mm-dd hh24:mi:ss");在执行完这句话之后baseDao.selectObject("studentMapper.stuproc", mapIn),mapIn.get("vbirth")的值已经被装进去了。格式如2012-1-1 12:12:12.0,所以要处理一下。Oracle中两表之间值互相插入和表自身插入值Oracle中的程序包相关资讯 Oracle基础知识
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