从 Excel 获取数据时出错

Question

我正在尝试使用以下方法从 Excel 文件中获取数据：

OpenFileDialog OpenCSVDialog = new OpenFileDialog();
            OpenCSVDialog.Filter = "Excel |*.xlsx";
            OpenCSVDialog.ShowDialog();
            ExcelFileName = System.IO.Path.GetFileName(OpenCSVDialog.FileName);
            string path = System.IO.Path.GetDirectoryName(OpenCSVDialog.FileName);
            fullpath = Path.Combine(path, ExcelFileName);

            dt_data = new DataSet("CSV File");
            var connString = string.Format(
                @"Provider=Microsoft.Jet.OleDb.4.0; Data Source={0};Extended Properties=""Text;HDR=YES;FMT=Delimited""",
                Path.GetDirectoryName(fullpath)
            );
            using (var conn = new OleDbConnection(connString))
            {
                conn.Open();
                var query = "SELECT * FROM [" + Path.GetFileName(fullpath) + "]";
                using (var adapter = new OleDbDataAdapter(query, conn))
                {
                    adapter.Fill(dt_data);
                }

            } 

但在浏览时看到此行出现错误：adapter.Fill(dt_data);:

我在相关路径中有 Excel 文件。

Answer 1

FROM [...] 子句指定您要读取的工作表（实际上是范围）。它应该包含工作表的名称，而不是工作簿的文件名。

如果您的工作表名为Sheet1，则应使用

var query = "SELECT * FROM [Sheet1$]";

SO 中有很多答案显示如何提取第一张纸的名称（如果不知道等）。

更新

我还注意到您在连接字符串中使用文件的目录路径而不是完整路径。您应该使用完整的文件路径：

 var connString = string.Format(
            @"Provider=Microsoft.Jet.OleDb.4.0; Data Source={0};Extended Properties=""Text;HDR=YES;FMT=Delimited""",
            fullpath
        );

Answer 2

我已更改代码。现在它的工作。

OpenFileDialog OpenCSVDialog = new OpenFileDialog();
            OpenCSVDialog.Filter = "Excel |*.xls";
            OpenCSVDialog.ShowDialog();
            ExcelFileName = System.IO.Path.GetFileName(OpenCSVDialog.FileName);
            string path = System.IO.Path.GetDirectoryName(OpenCSVDialog.FileName);
            fullpath = Path.Combine(path, ExcelFileName);

            dt_data = new DataSet("CSV File");

            string connString = "Provider=Microsoft.Jet.OLEDB.4.0;" +
          "Data Source=" + fullpath +
          ";Extended Properties='Excel 8.0;'";

            using (var conn = new OleDbConnection(connString))
            {
                conn.Open();
                string query = string.Format("select * from [{0}$]", "Sheet1");

                using (var adapter = new OleDbDataAdapter(query, conn))
                {
                    adapter.Fill(dt_data);
                }
            }

Answer 3

请参阅下面的课程。查看将完整文件名拆分为路径和文件名。

public class CSVReader
    {

        public DataSet ReadCSVFile(string fullPath, bool headerRow)
        {

            string path = fullPath.Substring(0, fullPath.LastIndexOf("\\") + 1);
            string filename = fullPath.Substring(fullPath.LastIndexOf("\\") + 1);
            DataSet ds = new DataSet();

            try
            {
                if (File.Exists(fullPath))
                {
                    string ConStr = string.Format("Provider=Microsoft.Jet.OLEDB.4.0;Data Source={0}" + ";Extended Properties=\"Text;HDR={1};FMT=Delimited\\\"", path, headerRow ? "Yes" : "No");
                    string SQL = string.Format("SELECT * FROM {0}", filename);
                    OleDbDataAdapter adapter = new OleDbDataAdapter(SQL, ConStr);
                    adapter.Fill(ds, "TextFile");
                    ds.Tables[0].TableName = "Table1";
                }
                foreach (DataColumn col in ds.Tables["Table1"].Columns)
                {
                    col.ColumnName = col.ColumnName.Replace(" ", "_");
                }
            }

            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
            return ds;
        }
    }​