Question

我想从按日期分组的表中获取数据，但由于我的日期列数据类型是datetime。

"select to_char(datePerformed, 'yyyy-mm-dd'), sum(sendAmount) from AppTransaction group by to_char(datePerformed, 'yyyy-mm-dd')"

因为datePerformed带有时间戳，其组包含时间戳，所以我用上面的查询

group by to_char(datePerformed, 'yyyy-mm-dd')"

我得到这样的例外，

java.lang.IllegalArgumentException: Cannot format given Object as a Date

我希望这个查询是用没有时间戳的数据对数据集进行分组，我该怎么做？

Answer 1

尝试：

group by datePerformed::date

或更好

group by cast(datePerformed as date)

Answer 2

Here是我使用date_trunc和Hibernate sqlGroupProjection的任何可能的postgres时间范围的完整解决方案。

public enum PostgresDateTrunc {

    second,
    minute,
    hour,
    day,
    week,
    month,
    quarter,
    year;

    public String getPostgresGroupExpression(String dtField, String alias) {
        return String.format("date_trunc('%s', %s) as %s", name(), dtField, alias);
    }

}

public class GenericRepository {

    public List calculateTimeGroupingStats(PostgresDateTrunc trunc, String timeGroupField, String...calculatedFields) {
        Criteria criteria = createCriteria();
        // adjust criteria object here

        ProjectionList projection = Projections.projectionList().
            add(Projections.alias(Projections.sqlGroupProjection(
                trunc.getPostgresGroupExpression(timeGroupField, "timeGroupField"),
                "timeGroupField", new String[]{"timeGroupField"}, new Type[]{StandardBasicTypes.DATE}),
                "timeGroupField")).
            add(Projections.alias(Projections.rowCount(), "count"));

        for (String calculatedField : calculatedFields) {
            projection.
                add(Projections.alias(Projections.sum(calculatedField), calculatedField + "Sum")).
                add(Projections.alias(Projections.avg(calculatedField), calculatedField + "Avg"));
        }

        criteria.addOrder(Order.asc("timeGroupField"));

        criteria.setProjection(projection);

        return criteria.list();
    }

}

Answer 3

我遇到了同样的问题，这种方式对我有用：

group by l.launchTime::date

我也尝试过这种方式，但不起作用：

group by cast(l.launchTime as date)

或

{{1}}

如何在hibernate查询语言中按日期分组而没有时间戳？

3 个答案: