Insertion Multiple records into database using asp.net

Sourav Kumar Das

Insertion of Multiple Records into database using Asp.Net?

When i select multiple check boxes in grid view, that it is shown on another grid view of the selected rows, when I submit all those records only the last record value gets inserted but not other records into database.

First GridView: From Database-1

Second GridView: To Database-2

Where I am filling out the details of text boxes and Pay it out.

Then it shows Paid successfully.

While in database it shows only the last RequestNo is inserted, other two records is not inserted..

As I need to insert all three records of RequestNo to be inserted in database.

So, if there's a solution or any answer please send me it......

9 Replies

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Piyush Pansuriya8y ago
Accepted answer
Hello,
To generate comma saperated string
StringBuilder sb = new StringBuilder();

    for ( int i = 0; i < MyList.Items.Count; ++i )
    {
        ListItem item = MyList.Items[ i ];
        sb.Append( item.Value );
        if ( i < MyList.Items.Count - 1 )
        {
            sb.Append( ", " );
        }
    }
Piyush Pansuriya8y ago
Bro.. That was the sample.
MyList means you have to put your collection which is given to the DataGridView.
I shared you a sample code. Change it as per your need.
Sourav Kumar Das8y ago
Will you please send me the code of it by using for loop.....

@Piyush Pansuriya
Piyush Pansuriya8y ago
Hello,
For that you have to make comma saperated string and stroed into session, beacause you are getting it from session

objHelper.cmdCollection.Parameters.AddWithValue("@RequestNo", Session["RequestNo"].ToString());

To generate comma saperated string, you have to use for loop on DataGridView records and add the RequestNo with comma in one variable.
Sourav Kumar Das8y ago
Yes...or else like this....

ReferenceId RequestNo
106 3,4,11

@Piyush
Piyush Pansuriya8y ago
Okay,
so you want to insert 3 records with the referenceId 106 and ReqestNo are 3,4,11 ?

ReferenceId Request No
106 3
106 4
106 11

Like this?
Sourav Kumar Das8y ago
This is the cs code...

protected void btnPay_Click(object sender, EventArgs e)
{
try
{
objHelper.CreateConnection(hfDistrict.Value);
objHelper.cmdCollection.CommandType = CommandType.StoredProcedure;
objHelper.cmdCollection.Connection = objHelper.conCollection;
objHelper.cmdCollection.CommandText = "spDMS_PaymentDetails";
objHelper.cmdCollection.Parameters.AddWithValue("@QueryType", "Insert");
objHelper.cmdCollection.Parameters.AddWithValue("@ReferenceId", lblRefId.Text.Trim());
objHelper.cmdCollection.Parameters.AddWithValue("@RequestNo", Session["RequestNo"].ToString());
objHelper.cmdCollection.Parameters.AddWithValue("@ChequeDate", txtChequeDate.Text.Trim());
objHelper.cmdCollection.Parameters.AddWithValue("@ChequeNo", txtChequeNumber.Text.Trim());
objHelper.cmdCollection.Parameters.AddWithValue("@RecipientName", txtRecipientName.Text.Trim());
objHelper.cmdCollection.Parameters.AddWithValue("@BankName", txtBankName.Text.Trim());
objHelper.cmdCollection.Parameters.AddWithValue("@AmountPaid", txtAmountPaid.Text.Trim());
objHelper.cmdCollection.Parameters.AddWithValue("@District", hfDistrict.Value);
objHelper.cmdCollection.Parameters.AddWithValue("@Panchayat", hfPanchayat.Value);
objHelper.cmdCollection.Parameters.AddWithValue("@rStatus", "Y");
objHelper.cmdCollection.Parameters.AddWithValue("@UserId", hfUserId.Value);
objHelper.cmdCollection.Parameters.AddWithValue("@CreatedBy", hfUserName.Value);
SqlParameter SQLReturn = new SqlParameter("@SQLReturn", SqlDbType.NVarChar, 50);
SQLReturn.Direction = ParameterDirection.Output;
objHelper.cmdCollection.Parameters.Add(SQLReturn);
objHelper.conCollection.Open();
int iReturn = objHelper.cmdCollection.ExecuteNonQuery();
string password = SQLReturn.Value.ToString();
if (password == "Success")
{
ScriptManager.RegisterStartupScript(this, GetType(), "showalert", "alert('Paid Successfully.');", true);
}
else if (password == "Already")
{
ScriptManager.RegisterStartupScript(this, GetType(), "showalert", "alert('Already Paid.');", true);
}
bindUpdatedCompletedRequest();
bindCompletedRequest();
Reset();
GetRefId();
}
catch (Exception ex)
{
Page.Controls.Add(new LiteralControl(""));
return;
}
}

@ Piyush
Sourav Kumar Das8y ago
Hmmm K done.....
@Piyush Pansuriya
Piyush Pansuriya8y ago
Hello,
Can you please share the code of insertion?

ReferenceId	Request No
106	3
106	4
106	11