我如何检查DataGridView的重复值并将其计算为一个？

2 回答

• 0

  它有点复杂，但有效：

  private void button2_Click(object sender, EventArgs e)
      {
          //Contains the text of column0 e.g 001, 002 etc..
          List<string> myListCol0Text = new List<string>();
          //Contains the row index e.g 0, 1 ,2 ,3 etc...
          List<int> myListCol0Index = new List<int>();
          //Contains the strings of the resulting column0 datagrid
          List <string> col0 = new List<string>();
          //Contains the doubles of the resulting column2 datagrid
          List <double> col2 = new List<double>();
          int i = 0, k = 0;
  
          foreach (DataGridViewRow row in InputDataGrid.Rows)
          {
              if (row.Cells[0].Value == null)
              {
                  break;
              }
  
              //Fill two lists with data
              myListCol0Text.Add(row.Cells[0].Value.ToString ());
              myListCol0Index.Add(i);
  
              i++;
          }
  
          double dbl = 0D;
          int count = 0;
          bool bl = false ;
  
          while (myListCol0Text.Count != 0)
          {
              count = myListCol0Text.Count;
              k = 0;
              bl = false;
              dbl = 0D;
              //Compares the first element of myListCol0Text with ALL the next ones.
              //If it finds a dublicate it removes it and in the end removes the first
              //element also
              for (i = 0; i < count - 1; i++)
              {
                  if (myListCol0Text[0] == myListCol0Text[k + 1]) //Dublicate
                  {
                      if(!bl) //Only the first dublicate is added to col0 list
                      {
                          col0.Add(myListCol0Text[0].ToString());
                          dbl = Convert.ToDouble(dataGridView1.Rows[myListCol0Index[0]].Cells[2].Value);
                          dbl += Convert.ToDouble(dataGridView1.Rows[myListCol0Index[k + 1]].Cells[2].Value);
                          col2.Add(dbl);
                          bl = true;
                      }
                      else{ //dont add to list but make the addition of the first and third column
                          dbl += Convert.ToDouble(dataGridView1.Rows[myListCol0Index[k + 1]].Cells[2].Value);
                          col2[col2.Count -1] = dbl;
                      }
  
                      //Removes dublicate
                      myListCol0Text.RemoveAt(k + 1);
                      myListCol0Index.RemoveAt(k + 1);
                  }
                  else
                  {
                      k++;
                  }
              }
  
              //Removes first element
              myListCol0Text.RemoveAt(0);
              myListCol0Index.RemoveAt(0);
          }
  
          //Add col0 and col2 to the resulting datagrid
          DataGridViewRow row_;
  
          for (i = 0; i < col0.Count; i++)
          {
              row_ = new DataGridViewRow();
  
              row_.CreateCells(ResultDatagrid);
              row_.Cells[0].Value = col0[i];
              row_.Cells[1].Value = col2[i].ToString ();
  
              ResultDatagrid.Rows.Add(row_);
          }
      }
  

  回复于 2024-04-28T01:26:22+08:00
• 0

  我认为最简单的原因是使用linq只用几行代码对 DataGridView 的底层数据源执行求和操作，然后将结果用作另一个网格的数据源：

  public class Program
  {
      public static void Main()
      {
          var items = new List<DataItem>
          {
              new DataItem { Position = "001", Description = "test", Value = 2.5m },
              new DataItem { Position = "002", Description = "", Value = 1m },
              new DataItem { Position = "001", Description = "hello", Value = 1.5m },
              new DataItem { Position = "002", Description = "test2", Value = 2m }
          };
  
          var results = 
              items
              .GroupBy(x => x.Position, (key, values) => new { Position = key, Values = values })
              .Select(g => new DataItem() 
              { 
                  Position = g.Position, 
                  Value = g.Values.Sum(x => x.Value) 
              });
      }
  }
  
  public class DataItem
  {
      public string Position { get; set; }
      public string Description { get; set; }
      public decimal Value { get; set; }
  }
  

  回复于 2024-04-28T01:26:22+08:00