如何在datatable中查找具有相同id和sum值的行 [英] how to find rows having same id and sum values in datatable

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本文介绍了如何在datatable中查找具有相同id和sum值的行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

您好b $ b

我的数据表有值



日期值

1 / 7/11 100

2/7/11 10

2/7/11 700

3/7/11 40



我希望数据表为



日期值

1/7/11 100

2/7/11 710

3/7/11 40



i想要将价值列总和havind重复的行



任何帮助都将不胜感激



问候

chinnu

解决方案





尝试这个可能有帮助...

 DataTable x = new DataTable(); 
IEnumerable< DataRow> dtrow = x.AsEnumerable();
var dt = dtrow.GroupBy(dtl => dtl.Field< string>(TransDate))
.Select(dtg => new dtg
{
dtl .Field< string>(TransDate)= dtg.Key,
dtrow.Sum(d => d.Field< double>(Amount))
})。ToList();







问候,



Algem


hi Al Moje



i used Dictionary< key,value>


< blockquote>

 var result = dt_UnSort.AsEnumerable()。GroupBy(dr => dr.Field< string>(Datevalue))。选择(group => new {Datevalue = group.Key, Count = group.Sum(col => col.Field< int>(Count))}); 

DataTable dt_Sorted = dt_UnSort.Clone();
foreach(结果中的var grp)
dt_Sorted.Rows.Add(grp.Datevalue,grp.Count);


Hi
I have a datatable having values

Date value
1/7/11 100
2/7/11 10
2/7/11 700
3/7/11 40

I want the datatable as

Date value
1/7/11 100
2/7/11 710
3/7/11 40

i want to sum the value column havind duplicate rows

Any help wil b appreciated

regards
chinnu

解决方案

Hi,

Try this if could help...

DataTable x = new DataTable();
         IEnumerable<DataRow> dtrow = x.AsEnumerable();
         var dt = dtrow.GroupBy(dtl => dtl.Field<string>("TransDate"))
             .Select(dtg => new dtg
             {
                 dtl.Field<string>("TransDate") = dtg.Key,
                  dtrow.Sum(d => d.Field<double>("Amount"))
             }).ToList();




Regards,

Algem


hi Al Moje

i used Dictionary<key,value>


var result = dt_UnSort.AsEnumerable().GroupBy(dr => dr.Field<string>("Datevalue")).Select(group => new { Datevalue= group.Key, Count = group.Sum(col => col.Field<int>("Count")) });

                DataTable dt_Sorted = dt_UnSort.Clone();
                foreach (var grp in result)
                    dt_Sorted.Rows.Add(grp.Datevalue, grp.Count);


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