2013-04-24 77 views
-2

我需要此代碼的幫助才能建立與數據庫的連接。到數據庫的Java連接不能正常工作

try 
{ 
String str="SELECT * FROM Author WHERE city LIKE 'P%'"; 
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver"); 
Connection con = 
DriverManager.getConnection  
("jdbc:odbc:myds;databaseName=Library;user=sa;password=password123"); 
Statement =con.createStatement(); 
ResultSet rs= executeQuery(); 
System.out.println("Author ID\tAuthor Name\tCity"); 
while (rs.next()) 
{ 
String id=rs.getString("au_id"); 
String name=rs.getString("au_name"); 
String city=rs.getString("city"); 
System.out.print(id+"\t"); 
if (name.length() <=7) 
System.out.print(name+"\t\t"); 
else 
System.out.print("\t"+name+"\t"); 
System.out.println(city); 
} 
} 
catch(Exception ex) 
{ 
System.out.println("Error occurred"); 
System.out.println("Error:"+ex); 

上述代碼有什麼問題?

+0

看看請避免大聲喊叫。另外,閱讀正確使用大寫和小寫的文本要容易得多。非常感謝您的幫助。 – 2013-04-24 23:43:08

+1

什麼是錯誤信息/例外/等等。? – TheEwook 2013-04-24 23:43:49

+0

嗯,所以有錯誤信息?它不應該做什麼? – 2013-04-24 23:44:05

回答

1

我認爲這是前人的精力類似的東西

String str="SELECT * FROM Author WHERE city LIKE 'P%'"; 
// Some stuff 
Statement stmt = con.createStatement(); //stmt added 
// Some more stuff 
ResultSet rs = stmt.executeQuery(str); // stmt and str added 

你應該Execute SQL Queries with Java Application.