Check the error from this code

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4.50/5 (2 votes)

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see the below code:
I want to insert three values, such as Id,Name,address but using this code it insert only address column value in the database.

XML

public static int Dynamicarray(string connectionString, Dictionary<string, object> parameters)
    {
        using (SqlConnection con = Database.GetConnection())
        {

            string column_name="";
            string value="";
            foreach (KeyValuePair<string, object> kvp in parameters)
            {
                column_name =RemoveSpecialChars(kvp.Key + ",".TrimEnd(','));
                value = kvp.Key+",".TrimEnd(',');
            }
            SqlCommand cmdExecute = new SqlCommand("insert into tbl2 ("+column_name+") values("+value+")",con);
            foreach (KeyValuePair<string, object> kvp in parameters)
            {
                cmdExecute.Parameters.AddWithValue(kvp.Key, kvp.Value);
            }
               return cmdExecute.ExecuteNonQuery();
        }
    }

Please help me in this regard.

Thanks in Advance.

Posted 17-May-11 2:26am

rahul dev123

Updated 17-May-11 2:31am

Ramalinga Koushik

v2

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OriginalGriff · Answer 1 · 2011-05-17T02:50:00

If we ignore that you are concatentating strings (please, look at using a StringBuilder - it's a lot more efficient for that kind of thing), then your problem is here:

C#

column_name =RemoveSpecialChars(kvp.Key + ",".TrimEnd(','));

The "column_name" string will always be the last key, followed by a comma...

nit_singh · Answer 2 · 2011-05-17T02:54:00

Solution 2

I think the problem is here

foreach (KeyValuePair<string,> kvp in parameters)
            {
                column_name += RemoveSpecialChars(kvp.Key + ",".TrimEnd(','));
                value += kvp.Key+",".TrimEnd(',');
            }

Posted 17-May-11 2:54am

nit_singh

Comments

Manfred Rudolf Bihy 17-May-11 9:03am

Correct! 5+
Well spotted!

rahul dev123 17-May-11 9:10am

I use this but it not separated the string. It should be like that:
IdNameAddresss or values(@Id@Name@Address).
Can you help me..