我是Flask的新手,我想在我的网站上创建一个开关按钮。我想知道这是否可能,以及如何包括动态标签。下图显示了我的想法:

我正在考虑使用wtfforms SubmitField,但我不太知道如何在我的routes.py文件和html模板之间实现这种动态行为。我是这样想的:
forms.py:
from flask_wtf import FlaskForm
from wtforms import SubmitField
class PowerSwitchForm(FlaskForm):
power_switch = SubmitField("ON")routes.py:
from flask import render_template, flash, redirect, url_for
from app import app
from app.forms import PowerSwitchForm
@app.route('/power', methods=['GET', 'POST'])
def power():
power_switch = PowerSwitchForm()
if power_switch.power_switch.label.text == "ON" and power_switch.validate():
flash("Power has been turned ON")
power_switch.power_switch.label.text = "OFF"
return redirect(url_for('power')
elif power_switch.power_switch.label.text == "OFF" and power_switch.validate():
flash("Power has been turned OFF")
power_switch.power_switch.label.text = "ON"
return redirect(url_for('power')
return render_template('power.html', form0=power_switch)power.html:
<!DOCTYPE html>
{% extends "base.html" %}
{% block content %}
<h2>Power switch</h2>
<form action="" method="post" novalidate>
{{ form0.hidden_tag() }}
{{ form0.power_switch() }}
</form>
{% endblock %}发布于 2018-06-27 22:43:42
单击切换按钮时,可以使用jquery处理所需的操作。此外,如果在切换按钮时需要执行后端进程,则可以使用ajax。这个答案说明了这两个问题。bootstrap-toggle是一个库,可以实现简单的切换。要使用,请将header标记值复制到下面:
显示"toggled“或”untoggled“的简单切换:
<html>
<body>
<head>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-beta.2/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-beta.2/js/bootstrap.min.js"></script>
<link href="https://gitcdn.github.io/bootstrap-toggle/2.2.2/css/bootstrap-toggle.min.css" rel="stylesheet">
<script src="https://gitcdn.github.io/bootstrap-toggle/2.2.2/js/bootstrap-toggle.min.js"></script>
</head>
<input type="checkbox" class='toggle' checked data-toggle="toggle">
<div class='status'>Toggled</div>
</body>
<script>
$(document).ready(function() {
$('.toggle').click(function() {
var current_status = $('.status').text();
if (current_status === 'Untoggled'){
$('.status').html('Toggled');
}
else{
$('.status').html('Untoggled');
}
});
});
</script>
</html>

切换触发"toggled“或”untoggled“的后端脚本:
在模板中,稍微更改script
<script>
$(document).ready(function() {
$('.toggle').click(function() {
var current_status = $('.status').text();
$.ajax({
url: "/get_toggled_status",
type: "get",
data: {status: current_status},
success: function(response) {
$(".status").html(response);
},
error: function(xhr) {
//Do Something to handle error
}
});
});
});
</script>然后,在您的应用程序中创建路径/get_toggled_status
@app.route('/get_toggled_status')
def toggled_status():
current_status = flask.request.args.get('status')
return 'Toggled' if current_status == 'Untoggled' else 'Untoggled'此示例与纯html/jquery解决方案做相同的事情,但是,它确实演示了在使用切换时如何与后端通信。
发布于 2021-04-20 13:09:02
我也是第一次接触Flask。下面是我尝试过的纯python代码和flask。看起来起作用了。
在templates/demo.html中:
{% extends "bootstrap/base.html" %}
{% import "bootstrap/wtf.html" as wtf %}
{% block content %}
<div class="page-header">
{{ wtf.quick_form(form) }}
</div>
{% endblock %}在demo.py中:
from flask import Flask, render_template, redirect, url_for
from flask_bootstrap import Bootstrap
from flask_wtf import FlaskForm
from wtforms import SubmitField
class PowerState(FlaskForm) :
state = SubmitField('OFF')
app = Flask(__name__)
Bootstrap(app)
app.config['SECRET_KEY'] = 'YOUR SECRET KEY'
@app.route('/', methods=['GET', 'POST'])
def home() :
form = PowerState()
if form.validate_on_submit() :
if form.state.label.text == 'OFF' :
PowerState.state = SubmitField('ON')
elif form.state.label.text == 'ON' :
PowerState.state = SubmitField('OFF')
return redirect(url_for('home'))
return render_template('demo.html', form=form)然后运行: flask run
致敬,Alex.Wu
https://stackoverflow.com/questions/51057966
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