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enter code here我被缺少参数“@comment”的错误所困扰。但是，我确信它不是空值，并且我已将所有必要的值传递到相应的参数中。

扫描代码几个小时后，我确定我的 SQL 命令没有丢失错误所述的“@comment”。

主程序将调用下面的方法并将相应的值传递给该方法。根据这些属性值，它将调用要使用的 8 条 sql 命令中的 1 条。

在这种情况下该怎么做有什么建议吗？谢谢。

在将属性值传递给方法之前，我使用以下代码验证了我的属性值。

//-------Checklist 5-------
        itemID = "dnaIN"; type = "in";
        if (string.IsNullOrEmpty(lblD5.Text) == false)
        { objR.CheckBy = lblD5.Text; }
        else if (string.IsNullOrEmpty(lblD5.Text) == true)
        { objR.CheckBy = "0"; }

        if (string.IsNullOrEmpty(lblV5.Text) == false)
        { objR.VerifiedBy = lblV5.Text; }
        else if (string.IsNullOrEmpty(lblV5.Text) == true)
        { objR.VerifiedBy = "0"; }

        if (string.IsNullOrEmpty(lblsbC.Text) == false) // check if user type comments
        { objR.Comment = lblsbC.Text; }
        else
        { objR.Comment = "0"; }
        update = objR.update(formID, itemID, type);

//------------传递给下面的代码----------

string strConn = Convert.ToString(ConfigurationManager.ConnectionStrings["InternshipConnectionString"]);
SqlConnection conn = new SqlConnection(strConn);

SqlCommand cmd1 = new SqlCommand();

//program pass in 0 as null for checkby & verifiedby
if (_CheckBy != "0" && _VerifiedBy != "0" && type == "in") // user enter everything, system in 
{
cmd1 = new SqlCommand("UPDATE SystemIn SET CheckBy =@checkBy, VerifiedBy =@verifiedBy WHERE FormID =@formID AND ItemID =@itemID", conn);
 }

 else if (_CheckBy != "0" && _VerifiedBy == "0" && type == "in") // user enter checkBy only, 
 {
 cmd1 = new SqlCommand("UPDATE SystemIn SET CheckBy =@checkBy, VerifiedBy = NULL WHERE FormID =@formID AND ItemID =@itemID", conn);
 }
 else if (type == "in")
 {   
 cmd1 = new SqlCommand("UPDATE SystemIn SET CheckBy = NULL, VerifiedBy =NULL WHERE FormID =@formID AND ItemID =@itemID", conn);
  }

  else if (_CheckBy != "0" && _VerifiedBy != "0" && _Comment != "0" && type == "out") 
  {
  cmd1 = new SqlCommand("UPDATE SystemOut SET CheckBy =@checkBy, VerifiedBy =@verifiedBy, Comment=@comment WHERE FormID =@formID AND ItemID =@itemID", conn);
  }
  else if (_CheckBy != "0" && _VerifiedBy == "0" && _Comment == "0" && type == "out") 
  {
  cmd1 = new SqlCommand("UPDATE SystemOut SET CheckBy =@checkBy, VerifiedBy = NULL, Comment=NULL WHERE FormID =@formID AND ItemID =@itemID", conn);
  }
  else if (_CheckBy != "0" && _Comment != "0" && _VerifiedBy == "0" && type == "out")
  {
  cmd1 = new SqlCommand("UPDATE SystemOut SET CheckBy =@checkBy, VerifiedBy=NULL, Comment=@comment WHERE FormID =@formID AND ItemID =@itemID", conn);
  }
  else if (_CheckBy != "0" && _Comment == "0" && _VerifiedBy != "0" && type == "out")
  {
  cmd1 = new SqlCommand("UPDATE SystemOut SET CheckBy =@checkBy, VerifiedBy =@verifiedBy, Comment=NULL WHERE FormID =@formID AND ItemID =@itemID", conn);
   }
   else if (type == "out")
   {   
   cmd1 = new SqlCommand("UPDATE SystemOUT SET CheckBy = NULL, VerifiedBy = NULL, Comment=NULL WHERE FormID =@formID AND ItemID =@itemID", conn);
    }

    cmd1.Parameters.AddWithValue("@formID", formID);
    cmd1.Parameters.AddWithValue("@itemID", itemID);

    if (_CheckBy != "0")
    { cmd1.Parameters.AddWithValue("@checkBy", _CheckBy); }
    if (_VerifiedBy != "0")
    { cmd1.Parameters.AddWithValue("@verifiedBy", _VerifiedBy); }
    if (_Comment != "0")
    { cmd1.Parameters.AddWithValue("@comment", _Comment); }
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1 回答 1

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显然，代码中的最后两行对您的错误消息很重要。仅当is
@comment时才会丢失。 如果您总是必须通过，则将其更改为： _Comment"0"
@comment

if (_Comment != "0")
    { cmd1.Parameters.AddWithValue("@comment", _Comment); }
else
    { cmd1.Parameters.AddWithValue("@comment", null); }
于 2013-11-07T09:26:21.047 回答