python - 如何将变量传递到 FLASK 中的另一条路线

Question

我试图将一个变量从一条路径传递到另一条路径，但我做不到。有人可以指导我这样做吗？我在最后一行收到错误。

@app.route("/search", methods=['GET', 'POST'])
def search():

    if request.method == 'GET':

        return render_template('search.html', navbar=True)

    else:

        query = request.form.get('query').lower()
        query_like = '%' + query + '%'

        books = db.execute('SELECT * FROM books WHERE (LOWER(isbn) LIKE :query) OR (LOWER(title) LIKE :query) '
                           'OR (LOWER(author) LIKE :query)',
                           {'query': query_like}).fetchall()

        if not books:
            return render_template('error.html', message='No Books were Found!', navbar=True)

        return render_template('books.html', query=query, books=books, navbar=True)

@app.route("/books", methods=['GET', 'POST'])
def books():

    return render_template('books.html', query=query, books=books)

Answer 1

问题在于您的代码组织方式。函数中的变量在函数内作用域，因此books 不能从第二条路径获得。除此之外，您还有一个命名冲突，其中 books=books 指的是函数本身(在模块范围内定义)。

如果你想在路由之间共享代码，把它放在一个单独的函数中:

def get_books(query, show_nav_bar=False):
    query = query.lower()
    query_like = '%' + query + '%'
    books = db.execute('SELECT * FROM books WHERE (LOWER(isbn) LIKE :query) OR (LOWER(title) LIKE :query) '
                    'OR (LOWER(author) LIKE :query)', {'query': query_like}).fetchall()

    if not books:
        return render_template('error.html', message='No Books were Found!', navbar=True)

    return render_template('books.html', query=query, books=books, navbar=show_nav_bar)


@app.route("/search", methods=['GET', 'POST'])
def search():
    if request.method == 'GET':
        return render_template('search.html', navbar=True)

    else:
        return get_books(request.form.get('query'), show_nav_bar=True)


@app.route("/books", methods=['GET', 'POST'])
def books():
    return get_books(request.form.get('query'))