根据html选项修改python中的URL

Question

我将以下HTML代码用于表单选项：

<form action="/submit" method="POST">

        <div class="form-group">
          <h3>Section</h3>
          <select name="articles">
            <option value="">Select Section</option>
            <option value="Business">Business</option>
            <option value="Sport">Sports</option>
            <option value="Culture">Culture</option>
            <option value="Technology">Technology</option>

          </select>
        </div>

以及以下python代码：

url = 'https://api.nytimes.com/svc/topstories/v2/{section}.json?api-XXXX'

response1 = urllib.request.urlopen(url).read()
data = json.loads(response1.decode('utf-8').replace('\n', ''))

我正在使用Jinja2和Flask，希望在提交之前根据用户选择的选项填充{section}

谢谢

Answer 1

与此问题类似：Getting value from select tag using flask

使用flask的内置request.form.get方法获取select元素的结果。从这里开始，使用上面发布的URL字符串将值传递给f字符串，前面的f除外

见https://docs.python.org/3/reference/lexical_analysis.html#formatted-string-literals

section = flask.request.form.get('articles')
url = f'https://api.nytimes.com/svc/topstories/v2/{section}.json?api-XXXX'