2012-03-26 66 views
0

我有下面的代码。它试图基本处理从jsf表单提交的数据,然后将数据插入到数据库中的表中。现在我们安装的数据库已经很老了......它是sql server 2000,所以我用jtds作为驱动程序。的代码:Java PreparedStatement错误

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服务器控制台输出显示:

java.sql.SQLException中:无效的对象名称SLMDATA'。

SLMDATA是表的名称,并且确定它是正确的。当我通过断点功能遍历代码时,我会到达executeQuery,然后事情就会失去控制。我研究了所有变量的内容,并且都是有效的。任何想法是怎么回事?一直卡在这一段时间

+1

可能的重复项:[here](http://stackoverflow.com/questions/909599/java-mssql-java-sql-sqlexception-invalid-object-name-tablename)和[there](http:// stackoverflow .com/questions/6647846/invalid-object-name) - 你可以尝试一个简单的'SELECT * FROM SLMDATA'来查看你是否得到相同的错误。 – assylias 2012-03-26 09:48:51

+0

谢谢你做了@assylias:D真的很感激它,很抱歉忽略那个:) – 2012-03-26 09:53:13

回答

1

您应该尝试将模式名称放在表schemaName.table中,或者在数据库中为模式创建同义词。