1

我正在做一个项目..我需要将 datagridview 记录导出到 excel 文件..我尝试了以下解决方案

using System;
using System.Windows.Forms;
using System.Data;
using System.Data.SqlClient;
using Excel = Microsoft.Office.Interop.Excel; 

namespace WindowsApplication1
{
public partial class Form1 : Form
{
    public Form1()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {
        SqlConnection cnn ;
        string connectionString = null;
        string sql = null;

        connectionString = "data source=servername;initial catalog=databasename;user                                   id=username;password=password;";
        cnn = new SqlConnection(connectionString);
        cnn.Open();
        sql = "SELECT * FROM Product";
        SqlDataAdapter dscmd = new SqlDataAdapter(sql, cnn);
        DataSet ds = new DataSet();
        dscmd.Fill(ds);

        dataGridView1.DataSource = ds.Tables[0];
    }

    private void button2_Click(object sender, EventArgs e)
    {
        Excel.Application xlApp ;
        Excel.Workbook xlWorkBook ;
        Excel.Worksheet xlWorkSheet ;
        object misValue = System.Reflection.Missing.Value;

        xlApp = new Excel.ApplicationClass();
        xlWorkBook = xlApp.Workbooks.Add(misValue);
        xlWorkSheet = (Excel.Worksheet)xlWorkBook.Worksheets.get_Item(1);
        int i = 0;
        int j = 0; 

        for (i = 0; i <= dataGridView1.RowCount  - 1; i++)
        {
            for (j = 0; j <= dataGridView1.ColumnCount  - 1; j++)
            {
                DataGridViewCell cell = dataGridView1[j, i];
                xlWorkSheet.Cells[i + 1, j + 1] = cell.Value;
            }
        }

        xlWorkBook.SaveAs("csharp.net-informations.xls",      Excel.XlFileFormat.xlWorkbookNormal, misValue, misValue, misValue, misValue, Excel.XlSaveAsAccessMode.xlExclusive, misValue, misValue, misValue, misValue, misValue);
        xlWorkBook.Close(true, misValue, misValue);
        xlApp.Quit();

        releaseObject(xlWorkSheet);
        releaseObject(xlWorkBook);
        releaseObject(xlApp);

        MessageBox.Show("Excel file created , you can find the file c:\\csharp.net-informations.xls");
    }

    private void releaseObject(object obj)
    {
        try
        {
            System.Runtime.InteropServices.Marshal.ReleaseComObject(obj);
            obj = null;
        }
        catch (Exception ex)
        {
            obj = null;
            MessageBox.Show("Exception Occured while releasing object " + ex.ToString());
        }
        finally
        {
            GC.Collect();
        }
    }

}

}

但上述解决方案总是覆盖相同的 Report.xls 文件..我每次都需要将记录复制到一个新的 excel 文件中..我需要对此的专家建议..

4

1 回答 1

0

在这一行中，您始终使用相同的文件名：

xlWorkBook.SaveAs("csharp.net-informations.xls",      Excel.XlFileFormat.xlWorkbookNormal, misValue, misValue, misValue, misValue, Excel.XlSaveAsAccessMode.xlExclusive, misValue, misValue, misValue, misValue, misValue);

您可以尝试将时间戳记到文件名中，例如：

string fileName = "csharp.net-informations" + DateTime.Now.ToString("MM-dd-yyyy-hh-mm-ss") + ".xls";
xlWorkBook.SaveAs(filename, Excel.XlFileFormat.xlWorkbookNormal, misValue, misValue, misValue, misValue, Excel.XlSaveAsAccessMode.xlExclusive, misValue, misValue, misValue, misValue, misValue);
于 2012-10-30T10:16:32.987 回答