获取查询字符串作为烧瓶上的函数参数

Question

有没有办法在flask上获取查询字符串作为函数参数？
例如，请求将是这样的。

http://localhost:5000/user?age=15&gender=Male

并希望代码与此类似。

@app.route("/用户")
def getUser（年龄，性别）：
...

Answer 1

如果你愿意写一个装饰器，一切皆有可能：

from functools import wraps

def extract_args(*names, **names_and_processors):
    user_args = ([{"key": name} for name in names] +
        [{"key": key, "type": processor}
            for (key, processor) in names_and_processors.items()])

    def decorator(f):
        @wraps(f)
        def wrapper(*args, **kwargs):
            final_args, final_kwargs = args_from_request(user_args, args, kwargs)
            return f(*final_args, **final_kwargs)
        return wrapper
    return decorator if len(names) < 1 or not callable(names[0]) else decorator(names[0])

def args_from_request(to_extract, provided_args, provided_kwargs):
    # Ignoring provided_* here - ideally, you'd merge them
    # in whatever way makes the most sense for your application
    results = {}
    for arg in to_extract:
        result[arg["key"]] = request.args.get(**arg)
    return provided_args, results

用法：

@app.route("/somewhere")
@extract_args("gender", age=int)
def somewhere(gender, age):
    return jsonify(gender=gender, age=age)