Question

我已经尝试了大约三个小时来解决这个问题，但找不到解决方案。

如何在表格中显示每个名字的第一次出现（最短时间）的整行（全部20列）？

例如，我想做这样的事情，但这不起作用：

       public partial class p_GetTaresForReturnResult : INotifyPropertyChanged
    {

              ...

        [global::System.Data.Linq.Mapping.ColumnAttribute(Storage="_IsReturned", DbType="Bit")]
        public System.Nullable<bool> IsReturned
        {
            get
            {
                return this._IsReturned;
            }
            set
            {
                if ((this._IsReturned != value))
                {
                    this._IsReturned = value;
                    NotifyPropertyChanged("IsReturned");
                }
            }
        }

        public event PropertyChangedEventHandler PropertyChanged;
        private void NotifyPropertyChanged(String propertyName = "")
        {
            if (PropertyChanged != null)
            {
                PropertyChanged(this, new PropertyChangedEventArgs(propertyName));
            }
        }
    }

Answer 1

您必须首先获得每个名称的最短时间，然后加入到原始名称/时间匹配的表格中。

获得最短时间：

SELECT name, MIN(time) AS minTime
FROM myTable
GROUP BY name;

然后，获取所有列：

SELECT m.*
FROM myTable m
JOIN(
   SELECT name, MIN(time) AS minTime
   FROM myTable
   GROUP BY name) tmp ON tmp.name = m.name AND tmp.minTime = m.time;

Answer 2

大多数数据库都支持ANSI标准窗口函数。有了这些，您可以这样做：

select t.*
from (select t.*, row_number() over (partition by name order by time) as seqnum
      from table t
     ) t
where seqnum = 1;

从表中显示整个记录，组中的时间戳最小

2 个答案: