如何通過查詢Java中的數據庫填充對象構造函數

Question

我想通過查詢數據庫來創建UserInfo對象構造函數，並且我一直收到UserInfo thisUserInfo = new UserInfo()部分的cannot find symbol錯誤。我試圖採取ResultSet rs並填寫構造函數爲登錄的用戶創建「會話」。

我在這裏做了什麼錯？這裏是我的代碼：

private UserInfo getUserInfo(HttpServletRequest request, HttpServletResponse response) throws SQLException { 
    String userName = request.getParameter("userName"); 
    String nullString = null; 
    char nullChar = ' '; 

    ResultSet rs = null; 

    stmt = conn.createStatement(); 
    String getInfoSQL = "SELECT * FROM " + studentsTable + " WHERE USERNAME = '" + userName + "'"; 
    rs = stmt.executeQuery(getInfoSQL); 
    if(rs.next()) { 
     UserInfo thisUserInfo = new UserInfo(rs.getString(userName), rs.getString(passWord), rs.getString(lastName), rs.getString(firstName), rs.getString(age), rs.getString(sex), rs.getString(email)); 
     rs.close(); 
     stmt.close(); 
    } else { 
     UserInfo thisUserInfo = new UserInfo(nullString, nullString, nullString, nullString, nullString, nullChar, nullString); 
     rs.close(); 
     stmt.close(); 
    } 
    return thisUserInfo; 
} 

這裏是我的UserInfo類：

public class UserInfo { 
private final String userName; 
private final String passWord; 
private final String lastName; 
private final String firstName; 
private final String age; 
private final char sex; 
private final String email; 

public UserInfo(String userName, String passWord, String lastName, String firstName, String age, char sex, String email) { 
    this.userName = userName; 
    this.passWord = passWord; 
    this.lastName = lastName; 
    this.firstName = firstName; 
    this.age = age; 
    this.sex = sex; 
    this.email = email; 
} 

public String getUserName() { 
    return this.userName; 
} 
public String getPassWord() { 
    return this.passWord;     
} 
public String getLastName() { 
    return this.lastName; 
} 
public String getFirstName() { 
    return this.firstName; 
} 
public String getAge() { 
    return this.age;  
} 
public char getSex() { 
    return this.sex; 
} 
public String getEmail() { 
    return this.email; 
} 

}

Answer 1

我不能左右一切100％，但是這看起來我錯了......

if(rs.next()) { 
    // UserInfo is declared within a local context 
    UserInfo thisUserInfo = new UserInfo(rs.getString(userName), rs.getString(passWord), rs.getString(lastName), rs.getString(firstName), rs.getString(age), rs.getString(sex), rs.getString(email)); 
    rs.close(); 
    stmt.close(); 
} else { 
    // UserInfo is declared within a local context 
    UserInfo thisUserInfo = new UserInfo(nullString, nullString, nullString, nullString, nullString, nullChar, nullString); 
    rs.close(); 
    stmt.close(); 
} 
// thisUserInfo does not exist 
return thisUserInfo; 

它應該更像......

UserInfo thisUserInfo = null; 
if(rs.next()) { 
    thisUserInfo = new UserInfo(rs.getString(userName), rs.getString(passWord), rs.getString(lastName), rs.getString(firstName), rs.getString(age), rs.getString(sex), rs.getString(email)); 
    rs.close(); 
    stmt.close(); 
} else { 
    thisUserInfo = new UserInfo(nullString, nullString, nullString, nullString, nullString, nullChar, nullString); 
    rs.close(); 
    stmt.close(); 
} 
return thisUserInfo; 

你要知道，我會親自做到這一點更像....

try { 
    stmt = conn.createStatement(); 
    String getInfoSQL = "SELECT * FROM " + studentsTable + " WHERE USERNAME = '" + userName + "'"; 
    rs = stmt.executeQuery(getInfoSQL); 
    UserInfo thisUserInfo = null; 
    if(rs.next()) { 
     thisUserInfo = new UserInfo(rs.getString(userName), rs.getString(passWord), rs.getString(lastName), rs.getString(firstName), rs.getString(age), rs.getString(sex), rs.getString(email)); 
    } else { 
     thisUserInfo = new UserInfo(nullString, nullString, nullString, nullString, nullString, nullChar, nullString); 
    } 
} finally { 
    try { 
     rs.close(); 
    } catch (Exception exp) { 
    } 
    try { 
     rs.close(); 
    } catch (Exception exp) { 
     stmt.close(); 
    } 
} 
return thisUserInfo; 

正如（現在）已經指出... UserInfo需要一個char爲sex參數，但你傳遞一個String

你應該嘗試提取使用類似String sexValue = rs.getString(sex)數據庫中的值，如果它不爲空，則使用sexValue.getCharAt(0)。

我也有不知道在哪裏userName，passWord，lastName，firstName，age，sex和email定義。這讓我有點擔心......

Answer 2

您需要發佈的代碼爲您UserInfo類起動器。我敢打賭，你沒有定義一個構造函數來接受這些參數。

編輯你試圖暗中從String轉換爲char我懷疑你不能這樣做。而不是rs.getString(sex)嘗試rs.getChar(sex)。或者，你可以添加一個額外的構造函數，接受性別參數String。

編輯哦，我現在看到了這個問題。你從來沒有定義任何其他變量，如passWord等。這些需要是字符串列索引的字符串或整數。

此外，這是不相關的，但從來沒有通過類似的查詢通過unsanitized輸入。你打開自己的SQL Injection。您應該使用prepared statements instead 或確保至少避開輸入。