Obtener la primera hoja de un documento de Excel independientemente del nombre de la hoja con OleDb

Question

Tengo usuarios que nombran a sus hojas todo tipo de locuras, pero quiero poder obtener la primera hoja del documento de Excel independientemente de cómo se llame.

Actualmente uso:

OleDbDataAdapter adapter = new OleDbDataAdapter(
"SELECT * FROM [sheetName$]", connString);

¿Cómo haría para obtener la primera hoja sin importar cómo se llame?

Gracias.

Answer 1

terminó usando esto:

using (OleDbConnection conn = new OleDbConnection(connString))
{
    conn.Open();
    dtSchema = conn.GetOleDbSchemaTable(OleDbSchemaGuid.Tables, new object[] { null, null, null, "TABLE" });
    Sheet1= dtSchema.Rows[0].Field<string>("TABLE_NAME");
}

Answer 2

OleDbConnection oconn = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source="Path"; Extended Properties=Excel 12.0;Persist Security Info=False;");

oconn.Open();
myCommand.Connection = oconn;
DataTable dbSchema = oconn.GetOleDbSchemaTable(OleDbSchemaGuid.Tables,  null);
if (dbSchema == null || dbSchema.Rows.Count < 1)
{
    throw new Exception("Error: Could not determine the name of the first worksheet.");
}
string firstSheetName = dbSchema.Rows[0]["TABLE_NAME"].ToString();

Answer 3

Este código ha funcionado bien donde he usado la cuadrícula de datos " DataGridView1 " cargar todo el contenido de la hoja

Dim MyConnection As System.Data.OleDb.OleDbConnection
Dim DtSet As System.Data.DataSet : Dim filteext As String = ""

 ''check for the file type
 If IO.Path.GetExtension(fileName) = "xls" Then
                filteext = "Excel 8.0"
 ElseIf IO.Path.GetExtension(fileName) = ".xlsx" Then
                filteext = "Excel 12.0"
 End If

''open connection

 MyConnection = New System.Data.OleDb.OleDbConnection _
               ("provider=Microsoft.Jet.OLEDB.4.0;Data Source='" & fileName & "';Extended Properties=" & filteext & ";")
            MyConnection.Open()

  Dim myTableName = MyConnection.GetSchema("Tables").Rows(0)("TABLE_NAME")

  Dim MyCommand As OleDbDataAdapter = New OleDbDataAdapter(String.Format("SELECT * FROM [{0}]", myTableName), MyConnection)


   MyCommand.TableMappings.Add("Table", "TestTable")
            DtSet = New System.Data.DataSet

    MyCommand.Fill(DtSet)

    DataGridView1.DataSource = DtSet.Tables(0)
            'DtSet.DataSetName. 

    MyConnection.Close()

Answer 4

Puede usar GetOleDbSchemaTable (VB) o GetOleDbSchemaTable (C #).

Usando la tabla Enum, devolverá una lista de todos los nombres de las hojas de trabajo, que luego puede usar para generar dinámicamente el SQL requerido.

Puedes usar:

 MySchemaTable = MyConnection.GetOleDbSchemaTable (OleDbSchemaGuid.Tables, New Object () {Nothing, Nothing, Nothing, " TABLE "})

Todos los nombres de las Hojas de trabajo se devolverán como parte de una DataTable a través de la cual puede iterar.

El uso de la información OleDbSchemaGuid se puede recuperar en

  

  
• Columnas
  
• Claves foráneas
  
• Índices
  
• Claves primarias
  
• Tablas
  
• Vistas
  

Documentación completa de MSDN disponible aquí

Answer 5

Básicamente, una copia de la respuesta de Anirudh Gaur. Realicé un reformateo del código y lo puse en una función. Agregué un StringBuilder para poder hacer más con la instrucción SELECT.

Los votos a favor van a Anirudh Gaur

Private Function Load_XLS(FileName As String) As DataTable
    Dim DataTable As New DataTable
    Dim Format As String = ""
    If IO.Path.GetExtension(FileName) = ".xls" Then
        Format = "Excel 8.0"
    ElseIf IO.Path.GetExtension(FileName) = ".xlsx" Then
        Format = "Excel 12.0"
    End If

    Using Connection As New OleDb.OleDbConnection("provider=Microsoft.Jet.OLEDB.4.0;Data Source='" & FileName & "';Extended Properties=" & Format & ";")
        Connection.Open()

        Dim TableName As String = Connection.GetSchema("Tables").Rows(0)("TABLE_NAME")

        Dim SQLCommand As New Text.StringBuilder
        SQLCommand.AppendLine("SELECT *")
        SQLCommand.AppendLine("FROM [{0}]")

        Dim Command As OleDb.OleDbDataAdapter = New OleDb.OleDbDataAdapter(String.Format(SQLCommand.ToString, TableName), Connection)

        Command.Fill(DataTable)

        Connection.Close()
    End Using
    Return DataTable
End Function

Answer 6

Es mi solución & # 9660; (Fácil, rápido, ejecutable, comprensible)

internal static DataTable GetExcelSheet(string excelFile,string sheetName = "")
{
    string fullPathToExcel = Path.GetFullPath(excelFile);
    string connString = string.Format("Provider=Microsoft.ACE.OLEDB.12.0;Data Source={0};Extended Properties='Excel " + (excelFile.ToLower().EndsWith("x") ? "12.0" : "8.0") + ";HDR=yes'", fullPathToExcel);
    return GetDataTable(connString, "SELECT * FROM [" + (string.IsNullOrEmpty(sheetName) ? GetTableName(connString, 0) : sheetName + "<*>quot;) + "]");
}

private static DataTable GetDataTable(string connectionString, string sql)
{
    DataTable dt = new DataTable();

    using (OleDbConnection conn = new OleDbConnection(connectionString))
    {
        conn.Open();
        using (OleDbCommand cmd = new OleDbCommand(sql, conn))
        {
            using (OleDbDataReader rdr = cmd.ExecuteReader())
            {
                dt.Load(rdr);
                return dt;
            }
        }
    }
}
private static string GetTableName(string connectionString, int row = 0)
{
    OleDbConnection conn = new OleDbConnection(connectionString);
    try
    {
        conn.Open();
        return conn.GetSchema("Tables").Rows[row]["TABLE_NAME"] + "";
    }
    catch { }
    finally { conn.Close();}
    return "sheet1";
}

Answer 7

Puede usar este enfoque también para obtener el nombre de la hoja. Ver comentarios para una mayor comprensión

myExcelConn.Open()

//GET DATA FROM EXCEL SHEET.
Dim str As String = String.Empty
Dim Sheets As DataTable = myExcelConn.GetOleDbSchemaTable(OleDbSchemaGuid.Tables, Nothing)
For i As Integer = 0 To Sheets.Rows.Count - 1
    str += Sheets.Rows(i)("TABLE_NAME").ToString() + "," //It will return sheet1,sheet2,sheet3 according to my excel file
Next

Dim objOleDB As New OleDbCommand("SELECT *FROM [" + str.Split(",")(0) + "]", myExcelConn) //It will select sheet1
Me.Label1.Text = str.Split(",")(0).Replace("<*>quot;, "")
// READ THE DATA EXTRACTED FROM THE EXCEL FILE.
Dim objBulkReader As OleDbDataReader
objBulkReader = objOleDB.ExecuteReader

Dim dt As DataTable = New DataTable
dt.Load(objBulkReader)

//FINALLY, BIND THE EXTRACTED DATA TO THE GRIDVIEW.
GridView1.DataSource = dt
GridView1.DataBind()
//If you want sheet2 data
 Dim objOleDB1 As New OleDbCommand("SELECT *FROM [" + str.Split(",")(1).Split(",")(0) + "]", myExcelConn)
//If you want sheet3 data
 Dim objOleDB2 As New OleDbCommand("SELECT *FROM [" + str.Split(",")(2).Split(",")(0) + "]", myExcelConn)

Answer 8

Puede obtener el nombre de la hoja 1 como este y utilizarlo de esta manera. Si desea obtener otros nombres de hoja, puede aumentar el valor de 0,1,2 ..

 Dim myExcelConn As OleDbConnection = _
                New OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" & _
                    Server.MapPath(".") & "\" & FileUpload1.FileName() & _
                    ";Extended Properties=Excel 12.0;")
    Dim Sheets As DataTable = myExcelConn.GetOleDbSchemaTable(OleDbSchemaGuid.Tables, Nothing)
                Sheet1 = Sheets.Rows(0)("TABLE_NAME").ToString()
  Dim objOleDB As New OleDbCommand("SELECT *FROM [" + Sheet1 + "]", myExcelConn)