List<RoomTicketEntity> list = roomTicketService.list(new LambdaQueryWrapper<RoomTicketEntity>().eq(RoomTicketEntity::getStatus, StatusEnum.DID.type) .eq(RoomTicketEntity::getArea, ticketDemand.getArea()) .orderByAsc(BaseEntity::getCreateDate) 帮改一下，先按照status等于1排序，然后按照createDate从近到远排序

可以将代码改为：

List<RoomTicketEntity> list = roomTicketService.list(new LambdaQueryWrapper<RoomTicketEntity>()
        .eq(RoomTicketEntity::getArea, ticketDemand.getArea())
        .eq(RoomTicketEntity::getStatus, StatusEnum.DID.type)
        .orderByDesc(BaseEntity::getCreateDate)
        .orderByAsc(RoomTicketEntity::getStatus));

这样就可以先按照状态从小到大排序（1在前面），再按照创建时间从近到远排序。注意，orderBy方法可以连续调用多次，表示多个排序条件，调用顺序即为排序的优先级。

相关问题

public ApiResult<List<ViewUser>> FindUser(int id) { List<ViewUser> viewuser = DbConfig.Db.Queryable<ViewUser>().Where(it => it.id == id).ToList(); return new ApiResult<ViewUser> { Status = ResultStatus.Success, Data = viewuser }; }帮我查看一下这段代码的问题

这段代码的问题在于返回类型不匹配，应该返回 ApiResult<List<ViewUser>> 而不是 ApiResult<ViewUser>。正确的代码应该是：

public ApiResult<List<ViewUser>> FindUser(int id)
{
List<ViewUser> viewuser = DbConfig.Db.Queryable<ViewUser>().Where(it => it.id == id).ToList();
return new ApiResult<List<ViewUser>> { Status = ResultStatus.Success, Data = viewuser };
}

List<Orders> orderList = orderService.selectOrderByUserId(userId);用LambdaQueryWrapper<Employee> queryWrapper = new LambdaQueryWrapper();实现

LambdaQueryWrapper<Orders> queryWrapper = new LambdaQueryWrapper<>();
queryWrapper.eq(Orders::getUserId, userId);
List<Orders> orderList = orderService.list(queryWrapper);