Question

让我们举个例子：我有一个表单，其中有几个部分，每个部分都有问题。侧面，我有答案映射到问题，他们有另一列，我想在查询时过滤：

所以我有以下实体：

@Entity(tableName = "sections")
public class Section {
    @PrimaryKey
    public long id;
    public String title;
}
@Entity(tableName = "questions")
public class Question {
    @PrimaryKey
    public long id;
    public String title;
    public long sectionId;
}
@Entity(tableName = "answers")
public class Answer {
    @PrimaryKey
    public long id;
    public long questionId;
    public int otherColumn;
}

在DAO部分，我想要检索所有这些。

以下是我想通过此查询填写的POJO：

class SectionWithQuestions {
    @Embedded
    public Section section;

    @Relation(parentColumn = "id", entityColumn = "sectionId", entity = Question.class)
    public List<QuestionWithAnswer> questions;

    public static class QuestionWithAnswer {
        @Embedded
        public Question question;

        @Relation(parentColumn = "id", entityColumn = "questionId", entity = Answer.class)
        List<Answer> answers;
    }
}

在另一个应用程序中，查询将是：

SELECT s.*, q.*, a.*
FROM sections s
LEFT JOIN questions q ON q.sectionId = s.id
LEFT JOIN answers a ON a.questionId = q.id
WHERE s.id = :sectionId and a.otherColumn = :otherColumn

然而在Room中我发现如果你想要一个对象及其关系（如示例中的用户及其宠物），则只选择该对象，并在第二个查询中查询关系。那将是：

@Query("SELECT * FROM sections WHERE id = :sectionId")

然后在生成的代码中会有（伪代码）：

sql = "SELECT * FROM sections WHERE id = :sectionId" // what's inside @Query
cursor = query(sql)
int indexColumn1 = cursor.getColumnIndex(col1)
int indexColumn2
... etc
while (cursor.moveToNext) {
    masterObject = new object()
    masterObject.property1 = cursor.get(indexColumn1)
    ... etc

    __fetchRelationshipXXXAsYYY(masterObject.relations) // fetch the child objects
}

这个__fetch XXX as YYY方法如下：

sql = "SELECT field1, field2, ... FROM a WHERE foreignId IN (...)"
similar algo as previously: fetch column indices, and loop through the cursor

所以基本上它创建了2个查询：一个用于主对象，一个用于关系。第二个查询是自动创建的，我们无法控制它。

要回到我想要关系但又过滤子列的问题，我就陷入了困境：

在第一个查询中，我无法引用otherColumn列，因为它不存在

@Relation

，因为此注释的唯一属性是连接列和实体定义

这可以在Room中进行，还是我必须自己制作子查询？

奖金问题：为什么他们不在一个查询中加入表格而是创建2个查询？这是出于性能原因吗？

编辑以澄清我的期望：

这就是我期望写的内容：

@Query("SELECT s.*, q.*, a.* " +
       "FROM sections s " +
       "LEFT JOIN questions q ON q.sectionId = s.id " +
       "LEFT JOIN answers a ON a.questionId = q.id " +
       "WHERE s.id = :sectionId and a.otherColumn = :additionalIntegerFilter")
SectionWithQuestionsAndAnswers fetchFullSectionData(long sectionId);

static class SectionWithQuestionsAndAnswers {
    @Embedded Section section;
    @Relation(parentColumn = "id", entityColumn = "sectionId", entity = Question.class)
    List<QuestionWithAnswers> questions;
}
static class QuestionWithAnswers {
    @Embedded Question question;
    @Relation(parentColumn = "id", entityColumn = "questionId", entity = Answer.class)
    Answer answer; // I already know that @Relation expects List<> or Set<> which is
                   // not useful if I know I have zero or one relation (ensured
                   // through unique keys)
}

我想象由Room实现的伪代码作为生成的代码：

function fetchFullSectionData(long sectionId, long additionalIntegerFilter) {
    query = prepare(sql); // from @Query
    query.bindLong("sectionId", sectionId);
    query.bindLong("additionalIntegerFilter", additionalIntegerFilter);
    cursor = query.execute();
    Section section = null;
    long prevQuestionId = 0;
    Question question = null;
    while (cursor.hasNext()) {
        if (section == null) {
            section = new Section();
            section.questions = new ArrayList<>();
            section.field1 = cursor.get(col1); // etc for all fields
        }
        if (prevQuestionId != cursor.get(questionIdColId)) {
            if (question != null) {
                section.questions.add(question);
            }
            question = new Question();
            question.fiedl1 = cursor.get(col1); // etc for all fields
            prevQuestionId = question.id;
        }
        if (cursor.get(answerIdColId) != null) { // has answer
            Answer answer = new Answer();
            answer.field1 = cursor.get(col1); // etc for all fields
            question.answer = answer;
        }
    }
    if (section !=null && question != null) {
        section.questions.add(question);
    }
    return section;
}

这是一个查询，我的所有对象都已获取。

如何在Room中过滤嵌套关系？

0 个答案: