Warning: file_get_contents(/data/phpspider/zhask/data//catemap/2/django/23.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Django:无法显示数据_Django_Django Templates_Django Views - Fatal编程技术网

Django:无法显示数据

Django:无法显示数据,django,django-templates,django-views,Django,Django Templates,Django Views,我的操作非常简单,只需尝试获取单个对象并按如下方式显示它: 查看: def sample(request,faculty): faculty_details = Faculty.objects.get(username=faculty) faculty_json=serializers.serialize("json",[faculty_details]) faculty_json = faculty_json.strip("[]") return HttpRes

我的操作非常简单,只需尝试获取单个对象并按如下方式显示它:

查看:

def sample(request,faculty):
    faculty_details = Faculty.objects.get(username=faculty)
    faculty_json=serializers.serialize("json",[faculty_details])
    faculty_json = faculty_json.strip("[]")
    return HttpResponse(json.dumps({'faculty':faculty_json}), mimetype="application/json")
$.getJSON(url, function(data) {
    faculty=data['faculty'];
    alert(faculty.fields['username']);    //this does not work!!
}
模板:

def sample(request,faculty):
    faculty_details = Faculty.objects.get(username=faculty)
    faculty_json=serializers.serialize("json",[faculty_details])
    faculty_json = faculty_json.strip("[]")
    return HttpResponse(json.dumps({'faculty':faculty_json}), mimetype="application/json")
$.getJSON(url, function(data) {
    faculty=data['faculty'];
    alert(faculty.fields['username']);    //this does not work!!
}

我在这里浪费了很多时间。我想知道我做错了什么?

好吧,
feculture\u json
已经是json了,因为它是由json序列化程序创建的。所以我不明白为什么在return语句中再次将其转储到JSON。这将简单地对引号进行双重编码,因此Javascript现在将其视为字符串,而不是实际的JSON对象

删除
json.dumps
,一切都会好起来