python - 如何使新页面在同一选项卡中打开。表单POST方法。 flask

Question

当我尝试使用表单打开具体页面时，出现以下问题，当我按下表单按钮时，页面的新实例将在新选项卡中打开。但我希望它在同一个选项卡中打开。我该怎么做？

有趣的事实是，当我转到新选项卡并按同一表单中的同一按钮时，会在同一选项卡中打开一个新页面。

我有以下 Flask 代码:

@app.route('/wifi', methods=["GET", "POST"])
def wifi():
    result_dict = None
    pic_end = ""
    if request.method == 'POST':
        start_date = request.form['calendar1']
        end_date = request.form["calendar2"]
        period = request.form["period"]
        # do some work here ...

    return render_template("test_wifi.html", result_dict=result_dict, wifipic="wifi" + pic_end)

我有以下 html 页面:

<form action="" class="ordinary_form" id="wifi_form">

    <label for="calendar1"> Choose period start: </label>
    <input type="date" name="calendar1" id="calendar1" required> <br> <br>
    <label for="calendar2"> Choose period end: </label>
    <input type="date" name="calendar2" id="calendar2" required></p>
    <label for="period"> Step size: </label>
    <select required name="period" id="period">
        <option value="0">Day</option>
        <option value="1">Week</option>
        <option value="2">Month</option>
    </select>
    <br><br>


    <input target="_self" type="submit" value="SEND" formtarget="{{url_for('wifi')}}" formmethod="post" onclick="loading();">
</form>

Answer 1

解决办法如下:

我应该在表单行中进行更改:

<form action="" method="POST"  class="ordinary_form" id="wifi_form">

并在输入提交部分:

<input type="submit" value="SEND" onclick="loading();">