SQL準備好的語句不採用'？'在where子句

我有這樣的事情在我的sql查詢SQL準備好的語句不採用'？'在where子句

private static final String QUERY_PHYSICIAN_INFO= "SELECT * FROM PHYSICIAN_INFO WHERE ? = ?";

，但以下情況不工作雙方..

Connection conn = null; 
    PreparedStatement stmt = null; 
    String logintype; 
     if(isInteger(id)) 
     { 
      logintype="BADGEID"; 

     }else{ 
      logintype="ID"; 
     } 
     stmt=conn.prepareStatement(QUERY_PHYSICIAN_INFO); 
     stmt.setString(1, logintype); 
     stmt.setString(2, id); 
     ResultSet rs = stmt.executeQuery(); 
     Physician phs = null;

對此有什麼特殊原因嗎？在此先感謝。

來源

2012-09-03 user1065490

？用於傳遞參數，而不是字段名稱。

如果你必須這樣做，構建SQL作爲

"SELECT * FROM PHYSICIAN_INFO WHERE " + "BADGEID" + " = ?"

來源

2012-09-03 10:50:56 podiluska

非常感謝您的關注，我試過這種方式，但它是靜態SQL查詢更新爲空。字符串logintype; String query =「SELECT * FROM PHYSICIAN_INFO WHERE」+ logintype +「=？」現在在某些方法中，我更新logintype爲其他值，最後生成的查詢是SELECT * FROM PHYSICIAN_INFO WHERE null =？ – user1065490

您需要更改SQL字符串，一旦您確定了什麼logintype是 – podiluska

你可以嘗試這樣的事情，因爲？只能用於傳遞參數，或者可以使用數據庫函數，然後將值傳遞給函數。

private static final String QUERY_PHYSICIAN_INFO_BADGE= "SELECT * FROM PHYSICIAN_INFO WHERE BADGEID = ?"; 

private static final String QUERY_PHYSICIAN_INFO_ID= "SELECT * FROM PHYSICIAN_INFO WHERE ID = ?"; 

Connection conn = null; 
    PreparedStatement stmt = null; 
    String logintype; 
     if(isInteger(id)) 
     {    
      stmt=conn.prepareStatement(QUERY_PHYSICIAN_INFO_BADGE); 

     }else{    
      stmt=conn.prepareStatement(QUERY_PHYSICIAN_INFO_ID); 
     } 
     stmt.setString(1, id); 
     ResultSet rs = stmt.executeQuery(); 
     Physician phs = null;

來源

2012-09-03 10:57:34 user75ponic

SQL準備好的語句不採用'？'在where子句

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