43

让 T1 和 T2 是DataTable具有以下字段的 s

T1(CustID, ColX, ColY)

T2(CustID, ColZ)

我需要联表

TJ (CustID, ColX, ColY, ColZ)

如何以简单的方式在 C# 代码中完成此操作?谢谢。

4

5 回答 5

57

如果允许您使用 LINQ,请查看以下示例。它创建两个具有整数列的 DataTable,用一些记录填充它们,使用 LINQ 查询连接它们并将它们输出到控制台。

    DataTable dt1 = new DataTable();
    dt1.Columns.Add("CustID", typeof(int));
    dt1.Columns.Add("ColX", typeof(int));
    dt1.Columns.Add("ColY", typeof(int));

    DataTable dt2 = new DataTable();
    dt2.Columns.Add("CustID", typeof(int));
    dt2.Columns.Add("ColZ", typeof(int));

    for (int i = 1; i <= 5; i++)
    {
        DataRow row = dt1.NewRow();
        row["CustID"] = i;
        row["ColX"] = 10 + i;
        row["ColY"] = 20 + i;
        dt1.Rows.Add(row);

        row = dt2.NewRow();
        row["CustID"] = i;
        row["ColZ"] = 30 + i;
        dt2.Rows.Add(row);
    }

    var results = from table1 in dt1.AsEnumerable()
                 join table2 in dt2.AsEnumerable() on (int)table1["CustID"] equals (int)table2["CustID"]
                 select new
                 {
                     CustID = (int)table1["CustID"],
                     ColX = (int)table1["ColX"],
                     ColY = (int)table1["ColY"],
                     ColZ = (int)table2["ColZ"]
                 };
    foreach (var item in results)
    {
        Console.WriteLine(String.Format("ID = {0}, ColX = {1}, ColY = {2}, ColZ = {3}", item.CustID, item.ColX, item.ColY, item.ColZ));
    }
    Console.ReadLine();

// Output:
// ID = 1, ColX = 11, ColY = 21, ColZ = 31
// ID = 2, ColX = 12, ColY = 22, ColZ = 32
// ID = 3, ColX = 13, ColY = 23, ColZ = 33
// ID = 4, ColX = 14, ColY = 24, ColZ = 34
// ID = 5, ColX = 15, ColY = 25, ColZ = 35
于 2009-03-20T12:01:24.767 回答
34

我想要一个无需您使用匿名类型选择器定义列即可连接表的函数,但很难找到任何列。我最终不得不自己做。希望这将有助于将来搜索此内容的任何人:

private DataTable JoinDataTables(DataTable t1, DataTable t2, params Func<DataRow, DataRow, bool>[] joinOn)
{
    DataTable result = new DataTable();
    foreach (DataColumn col in t1.Columns)
    {
        if (result.Columns[col.ColumnName] == null)
            result.Columns.Add(col.ColumnName, col.DataType);
    }
    foreach (DataColumn col in t2.Columns)
    {
        if (result.Columns[col.ColumnName] == null)
            result.Columns.Add(col.ColumnName, col.DataType);
    }
    foreach (DataRow row1 in t1.Rows)
    {
        var joinRows = t2.AsEnumerable().Where(row2 =>
            {
                foreach (var parameter in joinOn)
                {
                    if (!parameter(row1, row2)) return false;
                }
                return true;
            });
        foreach (DataRow fromRow in joinRows)
        {
            DataRow insertRow = result.NewRow();
            foreach (DataColumn col1 in t1.Columns)
            {
                insertRow[col1.ColumnName] = row1[col1.ColumnName];
            }
            foreach (DataColumn col2 in t2.Columns)
            {
                insertRow[col2.ColumnName] = fromRow[col2.ColumnName];
            }
            result.Rows.Add(insertRow);
        }
    }
    return result;
}

您可以如何使用它的一个示例:

var test = JoinDataTables(transactionInfo, transactionItems,
               (row1, row2) =>
               row1.Field<int>("TransactionID") == row2.Field<int>("TransactionID"));

一个警告:这肯定不是优化的,所以当行数超过 20k 时要注意。如果您知道一张桌子会比另一张大,请尝试将较小的一张放在第一位,然后将较大的一张放在第二位。

于 2012-07-16T13:56:51.277 回答
5

这是我的代码。不完美,但工作良好。我希望它可以帮助某人:

    static System.Data.DataTable DtTbl (System.Data.DataTable[] dtToJoin)
    {
        System.Data.DataTable dtJoined = new System.Data.DataTable();

        foreach (System.Data.DataColumn dc in dtToJoin[0].Columns)
            dtJoined.Columns.Add(dc.ColumnName);

        foreach (System.Data.DataTable dt in dtToJoin)
            foreach (System.Data.DataRow dr1 in dt.Rows)
            {
                System.Data.DataRow dr = dtJoined.NewRow();
                foreach (System.Data.DataColumn dc in dtToJoin[0].Columns)
                    dr[dc.ColumnName] = dr1[dc.ColumnName];

                dtJoined.Rows.Add(dr);
            }

        return dtJoined;
    }
于 2013-07-19T09:13:41.863 回答
1

此函数将使用已知的连接字段连接 2 个表,但这不能允许两个表上的 2 个具有相同名称的字段,除了连接字段,一个简单的修改是保存带有计数器的字典,只需将数字添加到同名领域。

public static DataTable JoinDataTable(DataTable dataTable1, DataTable dataTable2, string joinField)
{
    var dt = new DataTable();
    var joinTable = from t1 in dataTable1.AsEnumerable()
                            join t2 in dataTable2.AsEnumerable()
                                on t1[joinField] equals t2[joinField]
                            select new { t1, t2 };

    foreach (DataColumn col in dataTable1.Columns)
        dt.Columns.Add(col.ColumnName, typeof(string));

    dt.Columns.Remove(joinField);

    foreach (DataColumn col in dataTable2.Columns)
        dt.Columns.Add(col.ColumnName, typeof(string));

    foreach (var row in joinTable)
    {
        var newRow = dt.NewRow();
        newRow.ItemArray = row.t1.ItemArray.Union(row.t2.ItemArray).ToArray();
        dt.Rows.Add(newRow);
    }
    return dt;
}
于 2017-01-18T12:16:06.420 回答
0

我试着用下一种方式做到这一点

public static DataTable JoinTwoTables(DataTable innerTable, DataTable outerTable)
        {
            DataTable resultTable = new DataTable();
            var innerTableColumns = new List<string>();
            foreach (DataColumn column in innerTable.Columns)
            {
                innerTableColumns.Add(column.ColumnName);
                resultTable.Columns.Add(column.ColumnName);
            }

            var outerTableColumns = new List<string>();
            foreach (DataColumn column in outerTable.Columns)
            {
                if (!innerTableColumns.Contains(column.ColumnName))
                {
                    outerTableColumns.Add(column.ColumnName);
                    resultTable.Columns.Add(column.ColumnName);
                }                    
            }

            for (int i = 0; i < innerTable.Rows.Count; i++)
            {
                var row = resultTable.NewRow();
                innerTableColumns.ForEach(x =>
                {
                    row[x] = innerTable.Rows[i][x];
                });
                outerTableColumns.ForEach(x => 
                {
                    row[x] = outerTable.Rows[i][x];
                });
                resultTable.Rows.Add(row);
            }
            return resultTable;
        }
于 2019-03-22T16:07:57.383 回答