2011-06-23 47 views
0

我遇到了一些簡單的代碼問題,但無法弄清楚我究竟做了什麼錯誤。 以下是我的代碼:Java SQL錯誤1064

String userName = "username"; 
     String password = "password"; 
     String url = "jdbc:mysql://computername/proejct"; 
     Connection conn=null; 
     String value="some_value"; 

     try 
     { 
      Class.forName("com.mysql.jdbc.Driver").newInstance(); 
      conn = DriverManager.getConnection(url, userName, password); 

      String query1= "SELECT * FROM table WHERE column_Name=?"; 
      PreparedStatement psmt1 = conn.prepareStatement(query1); 
      psmt1.setString(1, value); 

      Boolean count=psmt1.execute(query1); 
      if(count==false) 
      { 
       //do something 
      } 
      psmt1.close(); 

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

這是我收到的錯誤:

[email protected]: SELECT * FROM source2 WHERE File_Name='Messages.java' 
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1 
    at sun.reflect.GeneratedConstructorAccessor3.newInstance(Unknown Source) 
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(Unknown Source) 
    at java.lang.reflect.Constructor.newInstance(Unknown Source) 
    at com.mysql.jdbc.Util.handleNewInstance(Util.java:409) 
    at com.mysql.jdbc.Util.getInstance(Util.java:384) 
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054) 
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3566) 
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3498) 
    at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:1959) 
    at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2113) 
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2562) 
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2512) 
    at com.mysql.jdbc.StatementImpl.execute(StatementImpl.java:781) 
    at com.mysql.jdbc.StatementImpl.execute(StatementImpl.java:624) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:98) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveUpdate(SearchAndUpdate.java:115) 
    at SearchAndUpdate.recursiveTraversal(SearchAndUpdate.java:55) 
    at SearchAndUpdate.main(SearchAndUpdate.java:40) 

我做得不對的preparedstements?

我打印出來的準備語句,這是它的樣子: SELECT * FROM源2 WHERE列名= 'Messages.java'

+1

這是'字符串QUERY1 = 「SELECT * FROM表WHERE列名=?」;'你的實際類的聲明?您的錯誤消息'SELECT * FROM source2 WHERE File_Name ='Messages.java'顯示不同的語句。 –

回答

3

基於這一行:

check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1 

它看起來像你的陳述沒有被準備。 ?正在發送到MySQL。

我想這:

Boolean count=psmt1.execute(query1); 

應該是(編輯)

ResultSet count=psmt1.executeQuery(); 
+0

也只是想添加你的變量不應該是一個布爾值它應該是一個ResultSet,因爲你正在做一個選擇語句。我不知道這是否會導致錯誤,但應該考慮到 – RMT

+1

@RMT您是對的。 – NullRef

+0

謝謝。有效。 – sap