Question

无论如何在Java中指定日期格式？我不是在询问字符串格式。当我在我的REST服务中检查日期值（在调试模式下）时，它看起来如下：

Mon Sep 01 00:00:00 IST 2014

当我将此日期传递给hibernate查询时，它不会返回任何结果，尽管有符合条件的有效结果。以下是我的hibernate查询：

select this_.ID as ID1_59_0_, this_.SEVERITY as SEVERITY5_59_0_, this_.EVENT as EVENT2_59_0_, this_.PROCESS_NAME   as PROCESS3_59_0_, this_.HOST_NAME as HOST6_59_0_, this_.TIME as TIME4_59_0_ 
from SYSTEM_LOG this_ 
where this_.TIME>? and this_.TIME<?
order by this_.TIME asc

我在mySQL中运行此查询并传递日期，如下所示，它不会返回任何数据，我相信这是应用程序中发生的事情：

    select this_.ID as ID1_59_0_, this_.SEVERITY as SEVERITY5_59_0_, this_.EVENT as EVENT2_59_0_, this_.PROCESS_NAME   as PROCESS3_59_0_, this_.HOST_NAME as HOST6_59_0_, this_.TIME as TIME4_59_0_ 
from SYSTEM_LOG this_ 
where this_.TIME> 'Mon Sep 01 00:00:00 IST 2014' and this_.TIME<'Mon Sep 11 00:00:00 IST 2014'
order by this_.TIME asc;

但如果我按如下方式更改日期，则查询将返回数据：

     select this_.ID as ID1_59_0_, this_.SEVERITY as SEVERITY5_59_0_, this_.EVENT as EVENT2_59_0_, this_.PROCESS_NAME   as PROCESS3_59_0_, this_.HOST_NAME as HOST6_59_0_, this_.TIME as TIME4_59_0_ 
from SYSTEM_LOG this_ 
where this_.TIME> '2014/9/1' and this_.TIME<'2014/9/11'
order by this_.TIME asc;

如何使用正确的日期格式？我尝试了以下操作，但日期仍然以相同的格式显示：

 DateFormat formatter = new SimpleDateFormat("MM-dd-yyyy", Locale.ENGLISH);
        Date sDate = formatter.parse(startDate);
        Date eDate = formatter.parse(endDate);

更新： FOllowing是我用来检索数据的hibernate方法：

 public PagedDataResult getSystemLogsByDate(Date startDate, Date endDate, String sortColumn, int searchBySeverity,
                                           String searchByProcessName, String searchByHostName, String searchByDescription,
                                           int page, int pageSize) {
    String sortField = "time";
    List<Criterion> criteria = new ArrayList<Criterion>();
    criteria.add(Restrictions.gt(sortField, startDate));
    criteria.add(Restrictions.lt(sortField, endDate));
    if(searchBySeverity >= 0)
        criteria.add(Restrictions.eq("alarmSeverities.id", searchBySeverity));
    if(!searchByProcessName.isEmpty())
        criteria.add(Restrictions.like("processName", searchByProcessName));
    if(!searchByHostName.isEmpty())
        criteria.add(Restrictions.like("servers.hostName", searchByHostName));
    if(!searchByDescription.isEmpty())
        criteria.add(Restrictions.like("event", searchByDescription));
    return (PagedDataResult)getAllBySort(sortColumn, page, pageSize, criteria);
}

Answer 1

我所理解的是你有一个日期为String，因为你已经指定了，并且你想将它用于一个选择查询，那么你需要以下面的方式将它改为sql格式日期

首先从StringDate转换为java.util.Date

public static java.util.Date transformFromStringToDate(String dateInString){
SimpleDateFormat dateFormat = new SimpleDateFormat("dd-MM-yyyy", Locale.ENGLISH);
java.util.Date utilDate = null;
try {
   if(dateInString != null && !(dateInString.equals(""))) {
   utilDate = dateFormat.parse(dateInString);
 }
 } catch (ParseException e) {
     e.printStackTrace();
 }
    return utilDate;
}

然后将java.util.Date转换为java.sql.Date

  public static java.sql.Date convertUtilDateToSqlDate(java.util.Date date){
    if(date != null && !(date.equals(""))) {
       Date sqlDate = new Date(date.getTime());
       return sqlDate;
     }
        return null;
   }

最后在查询PreparedStatement

时使用java.sql.Date

如果直接使用java.util.Date，则跳过第一步

修改

如果列是timestamp，则需要将java.util.Date转换为java.sql.Timestamp

public java.sql.Timestamp convertUtilDateToSqlTimestamp(java.util.Date date){
    if(date != null && !(date.equals(""))) {
       Timestamp sqlTimestamp = new Timestamp(date.getTime());
       return sqlTimestamp ;
     }
        return null;
   }
public PagedDataResult getSystemLogsByDate(Date startDate, Date endDate,..) {
  java.sql.Timestamp startTimestamp = new YourClassName()
                                      .convertUtilDateToSqlTimestamp(startDate);
  java.sql.Timestamp endTimestamp =  new YourClassName()
                                      .convertUtilDateToSqlTimestamp(endDate);
...
criteria.add(Restrictions.gt(sortField, startTimestamp));
criteria.add(Restrictions.lt(sortField, endTimestamp));
...
}

让此方法将Date转换为sql时间戳并使用转换时间戳进行查询

Answer 2

以下是我将使用TimeStamp

检索的一些代码

java.util.Date date = getItSomehow();
Timestamp timestamp = new Timestamp(date.getTime());
preparedStatement = connection.prepareStatement("SELECT * FROM tbl WHERE ts > ?");
preparedStatement.setTimestamp(1, timestamp);

方法getItSomehow将使用SimpleDateFormat将您的String转换为日期。

基于您从REST系统获取的String，掩码将是

EEE MMM HH:mm:ss yyyy Z -> Mon Sep 01 00:00:00 IST 2014

根据您更新的代码，尝试

criteria.add(Restrictions.gt(sortField, new Timestamp(startDate.getTime ())));
criteria.add(Restrictions.lt(sortField,  new Timestamp(endDate.getTime ())));

Answer 3

这可能是因为您使用的是java.util.Date而不是java.sql.Date。尝试将导入更改为java.sql.Date并查看其是否有效。

Answer 4

默认情况下，mySql使用瑞典语语言环境，所有日期必须以“2014-08-26”提交。将日期格式化程序更改为SimpleDateFormat（“yyyy-MM-dd”，Locale.ENGLISH）;

将日期对象传递给hibernate查询

4 个答案: