python - 在单个 View 方法中接受多种请求类型的更好方法？

Question

我正在尝试向各种请求方法(GET、url x-www-form-urlencoded POST 和 json POST)公开 API:

@app.route('/create', methods=['GET', 'POST'])
def create_file():
    if request.method == 'GET':
        n = request.args.get('n')
        t = request.args.get('t')
    if request.method == 'POST':
        if request.json:
            n = request.json['n']
            t = request.json['t']
        else:
            n = request.form['n']
            t = request.form['t']
    try:
        n = int(n)
    except:
        n = 1
    ...

以上内容显得过于冗长。有没有更简单或更好的写法？谢谢。

Answer 1

这样看起来更好吗？在我看来，如果您可以接受将 JSON POST 请求移动到不同的路由(无论如何您确实应该这样做)，它会更简洁一些。

def _create_file(n, t):
    try:
        n = int(n)
    except:
        n = 1
    ...

@app.route('/create')
def create_file():
    n = request.args.get('n')
    t = request.args.get('t')
    return _create_file(n, t)

@app.route('/create', methods = ['POST'])
def create_file_form():
    n = request.form.get('n')
    t = request.form.get('t')
    return _create_file(n, t)

@app.route('/api/create', methods = ['POST'])
def create_file_json():
    if not request.json:
        abort(400); # bad request
    n = request.json.get('n')
    t = request.json.get('t')
    return _create_file(n, t)