ADO.NET :: Clone Object In Linq To Sql?

Oct 21, 2010

I have two tables in the dbml, Table1 and Table2. Table1 has a Table1ID as primary key and Table2 has Table1ID as foreign key.I have an instance of Table1 called table1,

[Code]....

then I create another instance of Table1 called anotherTable1,

[Code]....

At this point, my intention is to have table1 containing a list of Table2 and anotherTable1 containing 0 Table2. However, both table1.Table2 and anotherTable1.Table2 become 0 count.

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txtLastName.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Last_Name.ToString();

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txtZip.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Zip_Code.ToString();
txtSSNo.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Ssn.ToString();
txtMRNo.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Mr_No.ToString();
txtDOB.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Birth_Date.ToString();
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RBMaritalStatus.SelectedItem.Value = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Marital_Status.ToString();
txtHomePhone.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Home_Phone.ToString();
txtWorkPhone.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).WorkPhone.ToString();
txtCellPhone.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).CellPhone.ToString();
ddlSuffix.SelectedItem.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Suffix.ToString();
txtReligion.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Religion.ToString();
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txtSSOName.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Ssn.ToString();
txtPatientKin.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Patient_Kin.ToString();
txtRelationWithPatient.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Kin_Relation.ToString();
txtPhoneNo.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Kin_Phone.ToString();
ddlPreferredPharmacy.SelectedItem.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Id_Pharmacy.ToString();
txtEmergencyContactName.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Emergency_Contact_Name.ToString();
txtEmergencyContactNumber.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Emergency_Contact_Phone.ToString();
txtRelation.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Kin_Relation.ToString();
txtRace.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Race.ToString();
ddlHowDidPatientFindUs.SelectedItem.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).How_Found.ToString();
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txtEmployerOccupation.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Employer_Occupation.ToString();
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txtEmployerCity.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Employer_City.ToString();
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txtEmployerZip.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Employer_Zip.ToString();
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txtPGFirstName.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Ins_Grn_Name.ToString();
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txtPGAddress.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Ins_Grn_Address.ToString();
txtPGCity.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Ins_Grn_City.ToString();
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txtPGZip.Text = db.usp_ATI_FetchPatientDetails(iPatientID).ElementAt(0).Ins_Grn_Zip.ToString();
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[code]....

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AS
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[Code]....

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[Code]....

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[code]...

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EDIT:

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[Code]....

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ADO.NET :: Unable To Create Object Variable For Linq Data Class?

Sep 23, 2010

For some unknown reason I suddenly am unable to create an object variable for a Linq Data Class. If I create a new project and try the process again with the same database everything works as advertized.The code I am using is Dim db As New RoggDataContext which gives me an error on the object variable "db": Overload resolution failed because no accessibility 'NEW' accepts this number of arguments.I have looked inside every file in the Web Project like the web config to check the connection string, gobal, and designer.vb looking for any errors whit no JOY. I also tried to create a new form and code behind with a button and received the same error.This is mind blowing because as I already stated, I can create a new Web Site and the problem goes away.

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