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快速请求。我真的不明白应该在这里做些什么。
<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)
我看了很多问题,但我还没有发现任何可以帮助我的简单问题。如果有人可以为我演示代码,那么我可以从中查看和学习,这将是很棒的。
答案 0 :(得分:5)
您需要在这里使用Ajax来检索食物清单,具体取决于您选择的食物类型。因此,在您的模板中,您需要包含以下内容:
<html>
<head>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"
integrity="sha256-hwg4gsxgFZhOsEEamdOYGBf13FyQuiTwlAQgxVSNgt4="
crossorigin="anonymous">
</script>
<script>
$(document).ready(function() {
$('#foodkind').change(function() {
var foodkind = $('#foodkind').val();
// Make Ajax Request and expect JSON-encoded data
$.getJSON(
'/get_food' + '/' + foodkind,
function(data) {
// Remove old options
$('#food').find('option').remove();
// Add new items
$.each(data, function(key, val) {
var option_item = '<option value="' + val + '">' + val + '</option>'
$('#food').append(option_item);
});
}
);
});
});
</script>
</head>
<body>
<form>
{{ form.foodkind() }}
{{ form.food() }}
</form>
</body>
</html>
此代码将对JSON编码数据进行简写Ajax请求。此数据包含食物选择框的值列表。
为此,您需要在Flask视图中添加端点/get_food/<foodkind>
:
food = {
'fruit': ['apple', 'banana', 'cherry'],
'vegetables': ['onion', 'cucumber'],
'meat': ['sausage', 'beef'],
}
@app.route('/get_food/<foodkind>')
def get_food(foodkind):
if foodkind not in food:
return jsonify([])
else:
return jsonify(food[foodkind])