使用 Flask-SQLAlchemy 优化 Flask 中的 ORM 查询

Question

我用 Django 和 Flask 编写了同一个项目。完整的代码可以在我的 Github 帐户上找到。该网站是一个基于问答的小测验网站（CTF格式，非常简单的问题）。以下是链接：

我的问题是关于在 SQLAlchemy 或 Flask-SQLAlchemy 中优化 ORM 查询。

我会尽量写出表的架构，以便更好地理解。

Teams (id, team_name, email, phone, password)
Questions (id, name, body, hint, answer, points, visible)
Submissions (id, team(fk), question(fk), timestamp)

如果您想查看实际代码，请看这里：
对于 Django - Question & Submission, Team
对于烧瓶 - Question, Team, Submission

对于两条路线，/submissions 和 /leaderboard，我必须使用 ORM 编写某些查询。这是页面的样子：

对于 Django，查询看起来很不错（或者至少我是这么认为的 :P ）

def submissions(request):
    all_submissions = Question.objects \
        .values('id', 'name') \
        .order_by('id') \
        .annotate(submissions=Count('submission'))

    print(all_submissions.query)

    return render(request, 'questions/submissions.html', {
        'submissions': all_submissions
    })

def leaderboard(request):
    team_scores = Team.objects \
        .values('team_name') \
        .order_by('team_name') \
        .annotate(score=Coalesce(Sum('submission__question__points'), 0)) \
        .order_by('-score')

    print(team_scores.query)

    return render(request, 'questions/leaderboard.html', {
        'team_scores': team_scores,
    })

原始 SQL 查询如下所示：

SELECT "teams_team"."team_name", COALESCE(SUM("questions_question"."points"), 0) AS "score" FROM "teams_team" LEFT OUTER JOIN "questions_submission" ON ("teams_team"."id" = "questions_submission"."team_id") LEFT OUTER JOIN "questions_question" ON ("questions_submission"."question_id" = "questions_question"."id") GROUP BY "teams_team"."team_name" ORDER BY "score" DESC

SELECT "questions_question"."id", "questions_question"."name", COUNT("questions_submission"."id") AS "submissions" FROM "questions_question" LEFT OUTER JOIN "questions_submission" ON ("questions_question"."id" = "questions_submission"."question_id") GROUP BY "questions_question"."id", "questions_question"."name" ORDER BY "questions_question"."id" ASC

这是对我的问题的冗长介绍。

我的问题从这里开始，我似乎无法使用 SQLAlchemy ORM 编写此查询或类似查询，并且 PyCharm 没有提供正确的代码 completion/suggestions一样。

对于 Flask，这些是我的函数的样子：

def get_team_score(team):
    team_submissions = Submission.query.filter_by(team_id=team.id)
    score = sum(
        submission.question.points
        for submission in team_submissions
    )

    return score

@question_blueprint.route('/submissions')
def submissions():
    all_submissions = [
        {
            'id': q.id,
            'name': q.name,
            'submissions': Submission.query.filter_by(question_id=q.id).count()
        }
        for q in Question.get()  # fetch all Question rows
    ]
    return render_template('submissions.html', **{
        'submissions': all_submissions
    })

@question_blueprint.route('/leaderboard')
def leaderboard():
    team_scores = [
        {
            'team_name': team.team_name,
            'score': get_team_score(team)
        }
        for team in Team.query.filter_by()
    ]
    return render_template('leaderboard.html', **{
        'team_scores': team_scores
    })

查询未优化，我想知道是否可以编写像 django-orm 这样优雅的查询而不必编写原始 SQL 语句。如果可能的话，我想对这个问题中提到的这两条路线进行一些很好的优化查询。

呼。

Answer 1

您可以像这样更改上面的代码。

@question_blueprint.route('/leaderboard')
def leaderboard():
    team_scores =\
        Team.query.join(
            Submission,
        ).join(
            Question,
        ).with_entities(
            Team.team_name,
            func.sum(Question.points).label("score")
        ).all()

    return render_template('leaderboard.html', **{
        'team_scores': team_scores
    })

在下面的这个片段中，你加入了团队 -> 提交 -> 问题，然后因为你只需要团队名称和基于每个提交的问题的总分，所以你只需要 with_entities。

@question_blueprint.route('/submissions')
def submissions():
    all_submissions = Question.query.join(
        Submission
    ).with_entities(
        Question.id,
        Question.name,
        func.count().label("submissions")
    ).all()

    return render_template('submissions.html', **{
        'submissions': all_submissions
    })

要通过提交查询问题，您可以像我在上面添加的那样进行类似的查询，但是您不想计算总和，而是只想计算行数，因此只需将其更改为 func.sum

Answer 2

看完 Kelvin 的回答后，我有点明白我需要如何运行在 SQLAlchemy 中查询。我不断修改 Kelvin 的答案以满足我的需要，这些是完全有效的查询。

@question_blueprint.route('/submissions')
def submissions():
    all_submissions = Question.query \
        .outerjoin(Submission) \
        .group_by(Question.id) \
        .order_by(Question.id) \
        .with_entities(
            Question.id,
            Question.name,
            func.count(Submission.id).label('submissions')) \
        .all()

    return render_template('submissions.html', **{
        'submissions': all_submissions
    })


@question_blueprint.route('/leaderboard')
def leaderboard():
    team_scores = Team.query \
        .outerjoin(Submission) \
        .outerjoin(Question) \
        .group_by(Team.team_name) \
        .with_entities(
            Team.team_name,
            func.coalesce(func.sum(Question.points), 0).label('score')) \
        .with_labels().all()

    return render_template('leaderboard.html', **{
        'team_scores': team_scores
    })

使用 Flask-SQLAlchemy 优化 Flask 中的 ORM 查询

Optimize ORM queries in Flask using Flask-SQLAlchemy

python

django

django-orm

flask

flask-sqlalchemy