Object reference not set to an instance of an object.

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i am inserting record in sql while inserting into database following error is coming
"

Object reference not set to an instance of an object.

"
below is my c# code

C#

if (radioButton1.Checked)
            {
                sqlconnection sqlconn;
                SqlCommand sqlcmd = sqlconn.CreateCommand();
                //dd.putquertdata("insert into GI_sub0 values('001','" + textBox1.Text + "','" + textBox2.Text + "','admin','2011-03-04 00:00:00.000','10')");
                sqlcmd.CommandText = ("insert into GI_sub0 values('001','" + textBox1.Text + "','" + textBox2.Text + "','admin','2011-03-04 00:00:00.000','10')");
                try
                {
                    sqlcmd.ExecuteNonQuery();
                }
                catch (SqlException err)
                {
                    MessageBox.Show(err.Message);
                }
                MessageBox.Show("Record Entered Successfull");
            }
else messagebox.show("please select radioobtn");

Posted 21-Aug-14 6:46am

Member 10690757

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Comments

Ashi0891 21-Aug-14 13:59pm

have you provided the connection string? I don't think so!
and later add "sqlconn.open();" before "sqlcmd.ExecuteNonQuery();" or before try{}, and also add
finally
{
sqlconn.close();
}
after catch{}

or simply do as Solution 2.

2 solutions

Solution 1

You are not opening the Sql Connection.

Add sqlconn.open() before sqlcmd.ExecuteNonQuery()

Posted 21-Aug-14 6:49am

SteveyJDay

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Richard Deeming · Accepted Answer · 2014-08-21T07:11:00

Ignoring the typos in your question, you've declared a SqlConnection variable, but you haven't actually set it to anything before you try to call a method on it.

Also, your code is susceptible to SQL Injection[^]. You should be using parameterized queries instead of string concatenation.

C#

if (radioButton1.Checked)
{
    using (SqlConnection sqlconn = new SqlConnection("YOUR CONNECTION STRING HERE"))
    using (SqlCommand sqlcmd = sqlconn.CreateCommand())
    {
        // TODO: Give the parameters meaningful names.
        sqlcmd.CommandText = "insert into GI_sub0 values (@p0, @p1, @p2, @p3, @p4, @p5)";

        sqlcmd.Parameters.AddWithValue("@p0", "001");
        sqlcmd.Parameters.AddWithValue("@p1", textBox1.Text);
        sqlcmd.Parameters.AddWithValue("@p2", textBox2.Text);
        sqlcmd.Parameters.AddWithValue("@p3", "admin");
        sqlcmd.Parameters.AddWithValue("@p4", new DateTime(2011, 3, 4));
        sqlcmd.Parameters.AddWithValue("@p5", "10");

        try
        {
            sqlconn.Open();
            sqlcmd.ExecuteNonQuery();
            MessageBox.Show("Record Entered Successfull");
        }
        catch (SqlException err)
        {
            MessageBox.Show(err.Message);
        }
    }
}