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Home/ Questions/Q 6138601
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Editorial Team
  • 0
Editorial Team
Asked: May 23, 20262026-05-23T17:53:15+00:00 2026-05-23T17:53:15+00:00

If you please help me out I have an error and I am trying

  • 0

If you please help me out I have an error and I am trying to understand what the error is all about so I can fix it since I am a beginner.

My error is:

Must declare the scalar variable "@Details"

And my function goes like this:

public static void CreateReview(string paperId, string grate, string criteriaId, string Details)
{
    var sqlCon = new SqlConnection(System.Configuration.ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString);

             // GET CONFERENCE ROLE ID
             SqlCommand cmd = new SqlCommand();
             cmd.Connection = sqlCon;
             cmd.CommandText = "select Conference_Role_ID from AuthorPaper where Paper_ID = @PaperId";
             cmd.Parameters.AddWithValue("@PaperId", paperId);
             cmd.Connection.Open();
             string ConferenceRoleId = cmd.ExecuteScalar().ToString();
             cmd.Connection.Close();
             cmd.Dispose();

             string query2 = @"insert into ReviewPaper(Overall_Rating,Paper_id,Conference_role_id,Details)
             values(0,@paperId,@ConferenceRoleId,@Details); select
 SCOPE_IDENTITY() as RPID";

             cmd = new SqlCommand(query2, sqlCon);
                 cmd.Parameters.AddWithValue("@paperId",
 paperId);
                 cmd.Parameters.AddWithValue("@ConferenceRoleId",
 ConferenceRoleId);

             string ReviewPaperId;

                 try
                 {
                     cmd.Connection.Open();
                     ReviewPaperId = cmd.ExecuteScalar().ToString();
                     cmd.Connection.Close();
                 }
                 catch (Exception ee) { throw ee; }
                 finally { cmd.Dispose(); } }

If i delete details my function works perfectly. Thanking you in advance.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T17:53:16+00:00Added an answer on May 23, 2026 at 5:53 pm

    You need a line:

    cmd.Parameters.AddWithValue("@Details", Details);
    

    You are executing statement that takes 3 parameters but you are supplying only two.

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