在我的项目中,我以周的方式存储和检索时间表。我有一张这样的桌子
Id projectId activityId Date Spenttime
1 1 1 2014-11-10 8
2 1 1 2014-11-11 8
3 1 1 2014-11-12 8
4 1 1 2014-11-13 8
5 1 1 2014-11-14 8
6 1 1 2014-11-15 8
7 1 1 2014-11-16 8
8 1 2 2014-11-10 8
9 1 2 2014-11-11 8
10 1 2 2014-11-12 8
11 1 2 2014-11-13 8
12 1 2 2014-11-14 8
13 1 2 2014-11-15 8
14 1 2 2014-11-16 8
15 2 1 2014-11-15 8
16 2 1 2014-11-16 8 我想要上面表格的结果,如下所示
projectId activityId 2014-11-10 2014-11-11 2014-11-12 2014-11-13 2014-11-14 2014-11-15 2014-11-16
1 1 8 8 8 8 8 8 8
1 2 8 8 8 8 8 8 8
2 1 0 0 0 0 0 8 8上表的hibernate代码
List<Timesheet> timesheetList=sessionfactory.getCurrentSession.createCriteria(Timesheet.class)
.add(Restrctions.between("date",formatter.parse("2014-11-09"),formatter.parse("2014-11-16"))
.list(); 检索逻辑:
List<DisplayTable> dispaly=new ArrayList<DisplayTable>();
for(int i=0;i<timesheetList.size();i+=7)
{
DisplayTable disp=new DisplayTable();
disp.setProjectId(timesheetList.get(i).getProjectId());
disp.setActivityId(timesheetList.get(i).getActivityId());
disp.setSpentTimeDate1(timesheetList.get(i).getSpentTime());
disp.setSpentTimeDate2(timesheetList.get(i+1).getSpentTime());
disp.setSpentTimeDate3(timesheetList.get(i+2).getSpentTime());
disp.setSpentTimeDate4(timesheetList.get(i+3).getSpentTime());
disp.setSpentTimeDate5(timesheetList.get(i+4).getSpentTime());
disp.setSpentTimeDate6(timesheetList.get(i+5).getSpentTime());
disp.setSpentTimeDate7(timesheetList.get(i+6).getSpentTime());
}上面的逻辑在抛出索引出绑定异常的前两个iteration.after中工作得很好。我知道异常是抛出的,因为项目2只包含2行。
他们有什么方法可以在hibernate3中达到预期的结果吗?任何帮助都将不胜感激!
发布于 2014-11-18 01:27:59
将条件更改为
i < timesheetList.size() - 6因为你不希望go +7比大小
https://stackoverflow.com/questions/26978332
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