系统管理-二级管理员模块代码中SQL语法不兼容ORACLE
1、具体界面如下图：2、代码如：
//超级用户可查看全部
if(userName.equals("admin")) {
sql = "select * from t_s_depart_auth_group fg LIMIT 0,50";//这是Mysql写法
departAuthGroupList = this.systemService.findForJdbc(sql);
String chkSql = "select dag.* from t_s_depart_auth_group as dag join t_s_depart_authg_manager as dam on dam.group_id=dag.id where user_id = ?"; //这是Mysql写法
chkDepartAuthGroupList = this.systemService.findForJdbc(chkSql,userName);
recursiveGroup(dataList, departAuthGroupList,chkDepartAuthGroupList,"0");
} else {
sql = "select dag.* from t_s_depart_auth_group as dag join t_s_depart_authg_manager as dam on dam.group_id=dag.id where user_id = ? LIMIT 0,50";
departAuthGroupList = this.systemService.findForJdbc(sql,userName);
String chkSql = "select dag.* from t_s_depart_auth_group as dag join t_s_depart_authg_manager as dam on dam.group_id=dag.id where user_id = ?";
chkDepartAuthGroupList = this.systemService.findForJdbc(chkSql,userName);
recursiveGroup(dataList, departAuthGroupList,chkDepartAuthGroupList,"0");
}
3、T_S_DEPART_AUTH_GROUP 表中的LEVEL（级别）字段，在ORACLE是关键字。在代码departAuthGroup = systemService.getEntity(TSDepartAuthGroupEntity.class, id); 会报ORA-01747: user.table.column, table.column 或列说明无效
希望JEECG团队有望改进，谢谢！
好的，我们确认下，感谢
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