在SQLite中存储试题

Question

我正在创建我的第一个android应用程序。这将是一个小测验。

有40个不同的问题类别，每个类别都有自己的数据库表。

我把所有的问题都写进了这样的sqlite数据库：

private void addQuestions() {
    // Category 1 (ACE) Questions
    // First Question
    String aceQuestion1 = MainActivity.getContext().getResources().getString(R.string.ace_Question1);
    String aceQuestion1a = MainActivity.getContext().getResources().getString(R.string.ace_Question1a);
    String aceQuestion1b = MainActivity.getContext().getResources().getString(R.string.ace_Question1b);
    String aceQuestion1c = MainActivity.getContext().getResources().getString(R.string.ace_Question1c);
    String aceQuestion1d = MainActivity.getContext().getResources().getString(R.string.ace_Question1d);
    Question ace1 = new Question(aceQuestion1, aceQuestion1a, aceQuestion1b, aceQuestion1c, aceQuestion1d, aceQuestion1b);
    this.addACEQuestion(ace1);

    // Second Question
    String aceQuestion2 = MainActivity.getContext().getResources().getString(R.string.ace_Question2);
    String aceQuestion2a = MainActivity.getContext().getResources().getString(R.string.ace_Question2a);
    String aceQuestion2b = MainActivity.getContext().getResources().getString(R.string.ace_Question2b);
    String aceQuestion2c = MainActivity.getContext().getResources().getString(R.string.ace_Question2c);
    String aceQuestion2d = MainActivity.getContext().getResources().getString(R.string.ace_Question2d);
    Question ace2 = new Question(aceQuestion2, aceQuestion2a, aceQuestion2b, aceQuestion2c, aceQuestion2d, aceQuestion2d);
    this.addACEQuestion(ace2);

    // Category 2 (Androgen) Questions
    // First Question
    String androgensQuestion1 = MainActivity.getContext().getResources().getString(R.string.androgens_Question1);
    String androgensQuestion1a = MainActivity.getContext().getResources().getString(R.string.androgens_Question1a);
    String androgensQuestion1b = MainActivity.getContext().getResources().getString(R.string.androgens_Question1b);
    String androgensQuestion1c = MainActivity.getContext().getResources().getString(R.string.androgens_Question1c);
    String androgensQuestion1d = MainActivity.getContext().getResources().getString(R.string.androgens_Question1d);
    Question androgens1 = new Question(androgensQuestion1, androgensQuestion1a, androgensQuestion1b, androgensQuestion1c, androgensQuestion1d, androgensQuestion1c);
    this.addAndrogensQuestion(androgens1);

    // Second Question
    String androgensQuestion2 = MainActivity.getContext().getResources().getString(R.string.androgens_Question2);
    String androgensQuestion2a = MainActivity.getContext().getResources().getString(R.string.androgens_Question2a);
    String androgensQuestion2b = MainActivity.getContext().getResources().getString(R.string.androgens_Question2b);
    String androgensQuestion2c = MainActivity.getContext().getResources().getString(R.string.androgens_Question2c);
    String androgensQuestion2d = MainActivity.getContext().getResources().getString(R.string.androgens_Question2d);
    Question androgens2 = new Question(androgensQuestion2, androgensQuestion2a, androgensQuestion2b, androgensQuestion2c, androgensQuestion2d, androgensQuestion2b);
    this.addAndrogensQuestion(androgens2);
}

// Add Category 1 (ACE) Questions
public void addACEQuestion(Question quest) {
    addQuestion(quest, TABLE_ACE);
}

// Add Category 2 (Androgens) Questions
public void addAndrogensQuestion(Question quest) {
    addQuestion(quest, TABLE_ANDROGENS);
}

public void addQuestion(Question quest, String table) {
    //SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(QUES, quest.getQUESTION());
    values.put(OPTA, quest.getOPTA());
    values.put(OPTB, quest.getOPTB());
    values.put(OPTC, quest.getOPTC());
    values.put(OPTD, quest.getOPTD());
    values.put(ANSWER, quest.getANSWER());
    // Inserting Rows
    database.insert(table, null, values);
}

这很好，但是我会为每个类别添加大约30个问题，所以如果我这样实现它，将会有超过1000个问题，并且代码将变得无法管理。

我试图为每个类别实现一个for循环，但不幸的是，没有进行管理。

我如何简化这段代码，使其更少重复？

问题对象

public class Question {
private int ID;
private String QUESTION;
private String OPTA;
private String OPTB;
private String OPTC;
private String OPTD;
private String ANSWER;

public Question() {
    ID=0;
    QUESTION="";
    OPTA="";
    OPTB="";
    OPTC="";
    OPTD="";
    ANSWER="";
}

public Question(String question, String opta, String optb, String optc, String optd, String answer) {
    QUESTION = question;
    OPTA = opta;
    OPTB = optb;
    OPTC = optc;
    OPTD = optd;
    ANSWER = answer;
}

public int getID() { return ID; }
public String getQUESTION() { return QUESTION; }
public String getOPTA() { return OPTA; }
public String getOPTB() { return OPTB; }
public String getOPTC() {
    return OPTC;
}
public String getOPTD() {
    return OPTD;
}
public String getANSWER() { return ANSWER; }

public void setID(int id)
{
    ID=id;
}
public void setQUESTION(String question) {
    QUESTION = question;
}
public void setOPTA(String opta) {
    OPTA = opta;
}
public void setOPTB(String optb) {
    OPTB = optb;
}
public void setOPTC(String optc) { OPTC = optc; }
public void setOPTD(String optd) { OPTD = optd; }
public void setANSWER(String answer) { ANSWER = answer; }
}

我的方法不成功

private void addQuestions() {
    for (int i = 0; i <= 30; i++) {
        // ACE Questions
        String aceQuestioni = MainActivity.getContext().getResources().getString(R.string."ace_Question" + i);
        String aceQuestionia = MainActivity.getContext().getResources().getString(R.string."ace_Question" + i + "a");
        String aceQuestionib = MainActivity.getContext().getResources().getString(R.string."ace_Question" + i + "b");
        String aceQuestionic = MainActivity.getContext().getResources().getString(R.string."ace_Question" + i + "c");
        String aceQuestionid = MainActivity.getContext().getResources().getString(R.string."ace_Question" + i + "d");
        Question ace1 = new Question(aceQuestion1, aceQuestion1a, aceQuestion1b, aceQuestion1c, aceQuestion1d, aceQuestion1b);
        this.addACEQuestion(ace1);
        Question ace2 = new Question(aceQuestion2, aceQuestion2a, aceQuestion2b, aceQuestion2c, aceQuestion2d, aceQuestion2d);
        this.addACEQuestion(ace2);
    }
}

Answer 1

问题在于，R.string.ace_Question2是Java标识符(最可能用于字符串)，问题和答案有独立的标识符(我猜)。

将R.string中的常数分解为两个enums

class R {
  enum QandAkey {
   KEY_QUESTION, KEY_ANSWER_A, KEY_ANSWER_B /*, ...*/;
  }
  enum QuestionKey { // string is not a good name!!
    ace_Question1,ace_Question1,androgens_Question1,androgens_Question2 /*,...*/;
  }
}

然后，您可以将代码简化为循环：

for(QuestionKey qk : R.QuestionKey.values()) {
  String[] questionAndAnswers = new String[R.QandA.values().length];
  for(QandAkey qak : R.QandAkey.values()){
     questionAndAnswers[qak.ordinal()]
       = MainActivity.getContext().getResources()
               .getString(qk.name()+"."+qak.name());
  }
  // let the Question-constructor deal with the strings array itself 
  addAndrogensQuestion( new Question(questionAndAnswers)); 
}

取消，因为您必须更改Resource属性中条目的键以匹配新模式：

ace_Question1.KEY_QUESTION = dflafDSTDJWEfd
ace_Question1.KEY_ANSWER_A = dflafDSTDJWEfd
ace_Question1.KEY_ANSWER_B = dflafDSTDJWEfd

如果您想要添加更多的问题，您只需要在QuestionKey中为每个问题添加一个新的常量，并在资源文件中添加与QandAkey实体相同的条目。

Answer 2

方法中的另一个问题是在运行时选择正确的答案。

应该将答案保存为参考资源中的正确答案，而不是在运行时确定：

ace_Question1.KEY_QUESTION = dflafDSTDJWEfd
ace_Question1.KEY_CORRECT = KEY_ANSWER_A 
ace_Question1.KEY_ANSWER_A = dflafDSTDJWEfd
ace_Question1.KEY_ANSWER_B = dflafDSTDJWEfd

ace_Question2.KEY_QUESTION = dflafDSTDJWEfd
ace_Question2.KEY_CORRECT = KEY_ANSWER_B 
ace_Question2.KEY_ANSWER_A = dflafDSTDJWEfd
ace_Question2.KEY_ANSWER_B = dflafDSTDJWEfd

这需要第一个枚举中的第二个默认条目：

  enum QandAkey {
   KEY_QUESTION, KEY_CORRECT, KEY_ANSWER_A, KEY_ANSWER_B /*, ...*/;
   // the order of entries here does not need to reflect
   // the order of entries in the resource file
   // as long as we can rely on String type keys...
  }

然后，我们必须更改问题构造函数，以处理我的第一个答案(解析Error:(422, 28) error: no suitable constructor found for Question(String[]))中的新方法，以及这个方法中的更改：

public Question(String... questionAndAnswers) {
    QUESTION = questionAndAnswers[QandAkey.KEY_QUESTION.ordinal()];
    OPTA =questionAndAnswers[QandAkey.KEY_ANSWER_A.ordinal();
    OPTB = questionAndAnswers[QandAkey.KEY_ANSWER_B.ordinal();
    OPTC = questionAndAnswers[QandAkey.KEY_ANSWER_C.ordinal();
    OPTD = questionAndAnswers[QandAkey.KEY_ANSWER_A.ordinal();
    ANSWER = questionAndAnswers[QandAkey.valueOf(
                  questionAndAnswers[QandAkey.KEY_CORRECT.ordinal()]
                ).ordinal()
             ];
}

这是对Question类的租约破坏性更改。

当然，更好的解决方案是将ANSWER更改为int并存储索引，但这需要在类中进行更多的更改。

Answer 3

由于您的问题和答案存储在XML中，所以可以使用JaxB创建问题的集合。https://docs.oracle.com/javase/tutorial/jaxb/intro/examples.html

首先，您必须使用如下的Question注释增强您的JaxB类：

@XmlElement
public class Question {
    private int ID;
    @XmlElement(name="TheQuestion") // because we need a different Name in XML
    private String QUESTION;
    @XmlElement
    private String OPTA;
    @XmlElement
    private String OPTB;
    @XmlElement
    private String OPTC;
    @XmlElement
    private String OPTD;
    @XmlAttribute(name="answer") // because of case change
    private int ANSWER;

    // could be removed completely...
    public Question() {
    }

然后，您必须使用本教程中提到的jaxb编译器创建映射代码。

您的XML必须从现在的样子更改为(或类似的)

<Questions> <!-- the root element -->
  <Question answer="2">
    <TheQuestion>flaiufdgl</TheQuestion>
    <OPTA>flaiufdgl</OPTA>
    <OPTB>flaiufdgl</OPTB>
    <OPTC>flaiufdgl</OPTC>
    <OPTD>flaiufdgl</OPTD>
  </Question>
</Questions>