檢索值

Question

可能重複：
retrieving values from database in java

我正在AGAIN另一個程序，檢索來自我創建的數據庫字段的輸入的數據/值。但是這一次，我輸入的值將來自我創建的JtextField。我不知道這裏有什麼問題，因爲當我運行它時，輸出始終爲空。

在這個程序中，我會將我的JTextField的輸入值轉換爲int。這就是：

public class ButtonHandler implements ActionListener 
{ 
public void actionPerformed(ActionEvent e) 
{ 
if(e.getSource() == extendB) 
{ 
    ExtensionForm extend = new ExtensionForm(); 
    extend.setVisible(true); 
    } 
else if(e.getSource()== searchB)    
{ 
//get text from the textField 
String guest = guestIDTF.getText(); 
//parse the string to integer for retrieving of data 
int id = Integer.parseInt(guest); 
GuestsInfo guestInfo = new GuestsInfo(id); 
Room roomInfo = new Room(id); 
String labels[] = {guestInfo.getFirstName()+" "+guestInfo.getLastName(),""+roomInfo.getRoomNo(),roomInfo.getRoomType(),guestInfo.getTime(),"11:00"}; for(int z = 0; z<labels.length; z++) { labelDisplay[z].setText(labels[z]); } 

在我的第二類

它就從我創建 這裏的數據庫字段的值輸入是程序代碼： 進口的java.sql。*;

公共教室 {

private String roomType; 
private int guestID, roomNo; 

private Connection con; 
private PreparedStatement statement; 

public Room(){ 
    try { 
     Class.forName("com.mysql.jdbc.Driver"); 
     con = DriverManager.getConnection(
         "jdbc:mysql://localhost:3306/3moronsdb","root","");   
    } 
    catch (Exception e) { 
     e.printStackTrace(); 
    } 
} 
public Room(int guestsID) 
{ 
    this(); 
    try{ 
     statement = con.prepareStatement("SELECT * FROM guest WHERE guestID=?"); 
     statement.setInt(1, guestID); 
     ResultSet rs = statement.executeQuery(); 
     while(rs.next()){ 
      this.guestID = rs.getInt(1); 
      this.roomType = rs.getString(2); 
      this.roomNo = rs.getInt(3); 
     } 
    }catch(Exception e){ 
     System.out.print(e); 
    } 
} 
//Constructor for setting rate 
public Room(String roomType, int roomNo) 
{ 
    this(); 
    try 
    { 
     statement = con.prepareStatement("Insert into room(roomType, roomNo) values(?,?)"); 
      statement.setString(1, roomType); 
      statement.setInt(2, roomNo); 
      statement.executeUpdate(); 
    } 
    catch(Exception e) 
    { 
     e.printStackTrace();   
     return; 
    } 
} 
//getting roomType 
public String getRoomType(){ 
    return roomType; 
} 
//getting roomNo 
public int getRoomNo(){ 
    return roomNo; 
} 
//getting guestID 
public int getGuestId(){ 
    return guestID; 
} 

}

我已經在我的3moronsdb它們是（1，經典，103）插入一些值。這裏是我的測試主類：

public class TestMain { 
    public static void main(String [] a){ 

     GuestsInfo guest = new GuestsInfo(1); //note that this instantiation is the other class which i just 
ask the other day.. (https://stackoverflow.com/questions/12762835/retrieving-values-from-database-in-java) 
     Room rum = new Room(1); 
    System.out.print(rum.getRoomType()+" "+ guest.getFirstName()); 
} 
} 

當我運行它，它只是給我的房類空輸出，但我得到的GuestsInfo類，這是「埃裏卡」的輸出。你能幫助我嗎？我知道我昨天問這種問題，但我現在真的不知道這裏有什麼問題..

Answer 1

此方法中的select語句看起來不正確。它不應該從Room表中選擇信息？

public Room(int guestsID) 
{ 
    this(); 
    try{ 
     // This line looks wrong, shouldn't this be selecting from the 
     // room table?? 
     // statement = con.prepareStatement("SELECT * FROM room WHERE guestID=?"); 
     // instead??? 
     statement = con.prepareStatement("SELECT * FROM guest WHERE guestID=?"); 
     statement.setInt(1, guestID); 
     ResultSet rs = statement.executeQuery(); 
     while(rs.next()){ 
      this.guestID = rs.getInt(1); 
      this.roomType = rs.getString(2); 
      this.roomNo = rs.getInt(3); 
     } 
    }catch(Exception e){ 
     System.out.print(e); 
    } 
} 

而不是使用列索引來檢索值，您應該嘗試使用列名稱。雖然可能非常罕見，但數據庫可能會以不同的順序返回列，而您希望在代碼中使用這些列。

（這可能發生在任何數量的原因，更新數據庫，該數據庫是重新創建，數據庫引擎是介意的變化：P）

while(rs.next()){ 
    this.guestID = rs.getInt("guestID"); 
    this.roomType = rs.getString("roomType"); 
    this.roomNo = rs.getInt("roomNumber"); 
} 

現在，很明顯，我對你的數據庫結構一無所知，所以你需要適當地更新列名。這會突出顯示當你犯了一個錯誤，並從錯誤的表中選擇...

另外。你應該釋放你的數據庫資源，當他們不再需要

PreparedStatement statement = null; 
try{ 
    statement = con.prepareStatement("SELECT * FROM guest WHERE guestID=?"); 
    statement.setInt(1, guestID); 
    ResultSet rs = statement.executeQuery(); 
    while(rs.next()){ 
     this.guestID = rs.getInt(1); 
     this.roomType = rs.getString(2); 
     this.roomNo = rs.getInt(3); 
    } 
}catch(Exception e){ 
    System.out.print(e); 
} finally { 
    try { 
     statement.close(); 
    }catch(Exception e){ 
    } 
} 

否則你運行耗盡數據庫資源的風險：P

我也不會創建Connection每類。我要麼爲應用程序創建一個連接，要麼使用某種連接池