Java PreparedStatement錯誤

Question

我有下面的代碼。它試圖基本處理從jsf表單提交的數據，然後將數據插入到數據庫中的表中。現在我們安裝的數據庫已經很老了......它是sql server 2000，所以我用jtds作爲驅動程序。的代碼：

public void dbConnect(String db_connect_string, 
      String db_userid, 
      String db_password) 
    { 
     try { 
     Class.forName("net.sourceforge.jtds.jdbc.Driver"); 
     Connection conn = DriverManager.getConnection(db_connect_string, 
        db_userid, db_password); 
     System.out.println("connected"); 
     dateSubmitted = "dasdasd"; 
     /*HERE THE STATEMENTS TO INSERT*/ 
     java.sql.PreparedStatement insertData = null; 
     String insertStatement = "INSERT INTO SLMDATA VALUES(?,?,?,?,?,?,?,?,?,?," 
       + "?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?)"; 
     insertData = conn.prepareStatement(insertStatement); 
     insertData.setString(1, serviceLogger); 
     insertData.setString(2, dateSubmitted); 
     insertData.setString(3, serviceName); 
     insertData.setString(4, department); 
     insertData.setString(5, division); 
     insertData.setString(6, businessServiceDescription); 
     insertData.setString(7, technicalServiceDescription); 
     insertData.setString(8, serviceCustomer); 
     insertData.setString(9, serviceCriticality); 
     insertData.setString(10, serviceRequestSystem); 
     insertData.setString(11, serviceCategory); 
     insertData.setString(12, serviceHours); 
     insertData.setString(13, availabilityTarget); 
     insertData.setString(14, lowIncidentResponseScale); 
     insertData.setString(15, lowIncidentResponseHrsType); 
     insertData.setString(16, lowResponseTimeText); 
     insertData.setString(17, mediumIncidentResponseScale); 
     insertData.setString(18, mediumIncidentResponseHrsType); 
     insertData.setString(19, mediumResponseTimeText); 
     insertData.setString(20, highIncidentResponseScale); 
     insertData.setString(21, highIncidentResponseHrsType); 
     insertData.setString(22, highResponseTimeText); 
     insertData.setString(23, lowIncidentResolutionScale); 
     insertData.setString(24, lowIncidentResolutionHrsType); 
     insertData.setString(25, lowResolutionTimeText); 
     insertData.setString(26, mediumIncidentResolutionScale); 
     insertData.setString(27, mediumIncidentResolutionHrsType); 
     insertData.setString(28, mediumResolutionTimeText); 
     insertData.setString(29, highIncidentResolutionScale); 
     insertData.setString(30, highIncidentResolutionHrsType); 
     insertData.setString(31, highResolutionTimeText); 
     insertData.setString(32, blockTime); 
     insertData.setString(33, impactedServices); 
     insertData.setString(34, operationalDependencies); 
     insertData.setString(35, disasterAvailable); 
     insertData.setString(36, configurationItems); 
     insertData.executeUpdate(); 

     conn.close(); 
     FacesContext.getCurrentInstance().getExternalContext().dispatch("thankyou.xhtml"); 

     } catch (Exception e) { 
     e.printStackTrace(); 
     } 
    } 


    public void callConnection(){ 

     dbConnect("jdbc:jtds:sqlserver://host;", "username", 
        "password"); 
    } 

和GlassFish服務器控制檯輸出顯示：

java.sql.SQLException中：無效的對象名稱SLMDATA'。

SLMDATA是表的名稱，並且確定它是正確的。當我通過斷點功能遍歷代碼時，我會到達executeQuery，然後事情就會失去控制。我研究了所有變量的內容，並且都是有效的。任何想法是怎麼回事？一直卡在這一段時間

Answer 1

您應該嘗試將模式名稱放在表schemaName.table中，或者在數據庫中爲模式創建同義詞。