【发布时间】:2017-11-22 15:15:25
【问题描述】:
我开始阅读一些烧瓶应用程序编程,我一直试图让下拉菜单工作,但到目前为止我没有运气。我想要做的是,当用户从第一个下拉列表中选择一种食物类型时,它应该从数据库中获取相应的列表并填充第二组下拉列表。一旦做出选择,我不知道如何让它发送一个快速请求。我真的不明白这里应该做什么。
<body>
<div>
<form action="{{ url_for('test') }}" method="POST">
<div>
<label>Food:</label>
<select id="food" name="food" width="600px">
<option SELECTED value='0'>Choose your fav food</option>
{% for x in food %}
<option value= '{{ x }}'>{{x}}</option>
{% endfor %}
</select>
<!-- After a selection is made, i want it to go back to the database and fectch the results for the below drop box based on above selection -->
</div>
<div>
<label>Choose Kind of Food:</label>
<select id="foodChoice" name="foodChoice" width="600px">
<option selected value='0'>Choose a kind</option>
{% for x in foodChoice %}
<option value= '{{ x }}'>{{x}}</option>
{% endfor %}
</select>
</div>
<div>
<input type="submit">
</div>
</form>
</div>
app.html
@app.route('/', method = ['GET', 'POST'])
def index():
foodList = [ i.type for i in db.session.query(FoodType)]
return render_template('main.html', food=foodList)
@app.route(/foodkind', method = ['GET', 'POST'])
def foodkind():
selection = request.form['foodChoice']
foodKind = [ i.kind for i in db.session.query(FoodType).filter(FoodKind == selection)]
return render_template('main.html', foodChoice = foodKind)
我查看了很多问题,但还没有找到任何简单的可以帮助我的问题。如果有人可以为我演示一段代码,那将是非常棒的,这样我就可以查看并从中学习。
【问题讨论】:
标签: javascript jquery python html flask