Flask 在单击按钮时转到新页面

Question

我有一个带有索引页的小型 Flask 应用程序，用户可以在其中从下拉菜单中选择一个目录。一旦用户做出选择并点击网页上的“提交”，我想将他们发送到一个包含特定于该目录的信息的新页面。

我正在尝试通过在从索引页面获取输入后调用该函数来加载新页面。我的问题是在网页上单击提交会重新加载相同的索引页。

烧瓶代码：

app = Flask(__name__)
dirname = os.path.dirname(sys.argv[0])

run_path = ""

@app.route("/", methods = ["GET", "POST"])
def index():
    dir_loc = "/Users/kregan1/Documents"
    dir_data = list(os.listdir(dir_loc))

    if request.method == "POST":
        run_dir = request.form.get("run_list")
        run_path = dir_loc + "/" + run_dir
        run(run_path)

    return render_template('index.html', data = dir_data)


@app.route("/run", methods = ["GET", "POST"])
def run(run_path):
    if os.path.isdir(run_path) == True:
        file_lists = []
        for f in os.listdir(run_path):
            if f.startswith("pattern"):
                file_lists.append(run_path + "/" + f)

        projects = []
        for fl in file_lists:
            with open(fl) as f:
                for row in f:
                    if row.split(",")[0] not in projects:
                        projects.append(row.split(",")[0])
    else:
        projects = ["a","b","c"]

    return render_template('run.html', run = run_path, proj_data = projects )

index.html:

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8">
    <link rel="stylesheet" type="text/css" href="{{ url_for('static', filename='style.css') }}">
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
    <script src="{{url_for('static', filename='asp_javascript.js')}}"></script>
</head>
<body>

    <h2>Packaging Tool - Index</h2>
    <p>Select Run:</p>
    <form action="{{ url_for("index") }}" method="post">
        <select name="run_list">
            {% for d in data %}
            <option>{{d}}</option>
            {% endfor %}
        </select>
        <br>
        <input type="submit" value="submit">
    </form>

</body>
</html>

Answer 1

您可以使用 redirect(...) 而不仅仅是调用函数。

from flask import redirect
...
if request.method == "POST":
        run_dir = request.form.get("run_list")
        run_path = dir_loc + "/" + run_dir
        return redirect(url_for('run', run_path=run_path))

@app.route("/run/<run_path>", methods = ["GET", "POST"])
def run(run_path):
    if os.path.isdir(run_path) == True:
    ...

Flask 在单击按钮时转到新页面

Flask go to new page on button click

python

flask

python-3.x