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Insert value into SQL Server

开发者 https://www.devze.com 2022-12-10 22:41 出处:网络
I want to insert value into SQL Server but there is problem is I pass both value in parameter then it\'s not insert otherewise if i selecting one value then it\'s insert my database name is sample and

I want to insert value into SQL Server but there is problem is I pass both value in parameter then it's not insert otherewise if i selecting one value then it's insert my database name is sample and table is item

this is perfect insert statement or not ?

try
{
    int val = stmt.executeUpdate("INSERT item (patientid,itemid) VALUES(nPatientID," + LrBn.TestID+ ")");
    out.println("1 row affected");
}
catch (SQLException s)
{
   System.out.println("SQL statement is not executed!");
}

<%
if(testname!=null)
    {
            LrBn.beginInsert();
        for(int i=0; i<testname.length; i++)
          {
                     nCount++ ;
               LrBn.ResultID=0;
               try
                     { 
                 LrBn.TestID = Integer.parseInt(testname[i]) ;
             } catch( NumberFormatException ex)
             { 
                 LrBn.TestID = 0 ;
             }

                         LrBn.GroupID = nGroupID ;
                         LaBn.locateRecord(LrBn.TestID) ;

                        short nemergencyType =  com.hims.emergencyType.normal ;
                        try
                        {
                     nemergencyType = Short.parseShort(request.getParameter("emergencyType"));
                  }
                        catch( NumberFormatException ex)
                        { 
                         nemergencyType =  com.hims.emergencyType.normal ;
                        }    
             LrBn.Emergency = nemergencyType;
                     LrBn.ResultType =  LaBn.TestResultType ;
                         LrBn.PatientID = nPatientID ;
                         LrBn.DoctorID = LogUsr.DoctorID;
                     LrBn.UnitID = LogUsr.UnitID ;
                         LrBn.RequestTime = com.webapp.utils.DateHelper.requestDateTime(request, "RequestTime"); 
               LrBn.CollectionTime = null;
             LrBn.ResultTime = null;
                         LrBn.CollectedBy = 0 ;
             LrBn.TestDoneBy = 0 ;
             LrBn.PathologyUnitID = 0 ;
                         LrBn开发者_运维问答.BoolValue = 0;
                         LrBn.ScalarValue = null ;
                         LrBn.DescValue = null ;
                         LrBn.TestStatus = com.hims.TestStatusType.REQUESTED ;
                         LrBn.TestCharges = LaBn.TestCharge ;
             LrBn.PaymentStatus = com.hims.PaymentStatus.PENDING ;                       
                         LrBn.continueInsert();

                        Class.forName("net.sourceforge.jtds.jdbc.Driver");
                    Connection conn = DriverManager.getConnection("jdbc:jtds:sqlserver://localhost:1433/sample", "sa", "sa1234");
                        java.sql.Statement stmt = conn.createStatement();
                        try
                        {
                         int val = stmt.executeUpdate("INSERT item (patientid,itemid) VALUES(nPatientID," + LrBn.TestID+ ")");
                         out.println("1 row affected");
                }
                catch (SQLException s)
                        {
                            System.out.println("SQL statement is not executed!");
                }
                        stmt.close();
                        conn.close();

        } // end for
                LrBn.endInsert();
    }
%>


Looking at your first line of code:

int val = stmt.executeUpdate("INSERT item (patientid,itemid) VALUES(nPatientID,"
        + LrBn.TestID+ ")");    

If you are getting SQL Exceptions AND patientid is some sort of IDENTITY field you can solve that by doing stmt.executeUpdate("INSERT item (itemid) VALUES(" + LrBn.TestID + ")"); instead.

I apologize if this wasn't the issue, you may have to clarify your question.

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