flask中view怎么return给js数据

Question

JS提交代码

    //提交
    $('#submit').click(function(){
        var array = new Array();
        $("#select2 option").each(function(){
            var txt = $(this).text();
            array.push(txt);
         });
        alert(array);
        $.post("",
                {
                    'host_list':array
                },
        function(status){
                alert(status);
        });
    });

flask view函数

@main.route('/userhost/<int:userid>', methods=['GET', 'POST'])
@login_required
def userhost(userid):
    if current_user.role == '0':
        if request.method == 'POST':
            try:
                host_name_list = request.form['host_list[]'].split(',')
                return  jsonify({'status': u'更新成功!'})
            except Exception, e:
                print e
                return jsonify({'status': u'更新失败!'})

我在view里面接收到数据之后，如何返回一个状态给js，并让他弹出提示呢？
上面的js那样写弹不出来

Answer 1

这应该属于JS方面的问题, 另外, 你确定你的提交上去了吗

Answer 2

Sorry, my IME is broken, so can't write the answer in Chinese now. I think what you want to do is a restful API in flask, please check http://flask-restful.readthedocs.org/en/0.3.4/. It is flask plugin, you can get what you want.