[英]Cant access value of JSON object
Im doing a little project with flask. 我正在用烧瓶做一个小项目。 Right now, im trying to add options to a select from a JSON object.Despite being able to insert options, i cant access their value
现在,我试图将选项添加到JSON对象的选择中。尽管能够插入选项,但我无法访问它们的值
This my flask code: 这是我的烧瓶代码:
for dealership in session.query(Dealership).filter_by(ownerID = userId).all():
dealership_list.append({'Name' : dealership.name})
return jsonify(names = dealership_list)
Ajax code: Ajax代码:
success:function(result){
$.each(result.names, function(Name, value) {
$('#dealerships').append($("<option/>", {
value: Name,
text: value
}));
});
},
each item in result.names
is a dealership (as defined by your view function). result.names
中的每个项目都是经销result.names
(由您的查看功能定义)。
each dealership is a dict consisting of {"Name":the_name_of_this_dealership}
每个经销商都是由
{"Name":the_name_of_this_dealership}
组成的字典
so $.each(result.names,function(data){console.log(data)}
should print each object which is {"Name":some_dealership_name}
因此
$.each(result.names,function(data){console.log(data)}
应打印每个对象,这些对象为{"Name":some_dealership_name}
so just change it to 所以只需将其更改为
success:function(result){
$.each(result.names, function(dealership) {
$('#dealerships').append($("<option/>", {
value: dealership.Name,
text: dealership.Name
}));
});
},
better yet just change your python code to just return a list of names since thats the only data you are using 更好的方法是更改python代码以返回名称列表,因为多数民众赞成在使用唯一的数据
dealerships = session.query(Dealership).filter_by(ownerID = userId).all()
names = [dealership.name for dealership in dealerships]
return jsonify(names = names)
and just change your js to 只需将您的js更改为
success:function(result){
$.each(result.names, function(index,a_dealer_name) {
$('#dealerships').append($("<option/>", {
value: a_dealer_name,
text: a_dealer_name
}));
});
}
@JoranBeasley Awesome!I used the second solution, but added/change the following: @JoranBeasley太棒了!我使用了第二个解决方案,但添加/更改了以下内容:
$.each(result.names, function(key,name) {
$('#dealerships').append($("<option/>", {
value: name,
text: name
}));
});
This because your solution was printing the index instead of value.Thanks for the help! 这是因为您的解决方案正在打印索引而不是值。感谢您的帮助!
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