小编shu*_*014的帖子

是否有任何有效的区别

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？

我在Google上搜索并找到答案,但这对我不起作用.我必须像JsonResponseDjango 一样发送一个列表,类似于:

list_to_json =[{"title": "hello there",
                "link": "www.domain.com",
                "date":   ...},
               {},{},{},...]

我通过应用StackOverflow 问题1和问题2将其转换为JSON,但它不适用于我.我收到以下错误:

为了允许非dict对象被序列化,请将safe参数设置为False

这是我的代码:

    def json_response(request):
        list_to_json=[{"title": ..., "link": ..., "date": ...},{...}]
        return JsonResponse(json.dumps(list_to_json) )

我尝试使用python在Django中添加搜索字段.以下是我使用过的代码.

# admin.py file

from django.db import models
from blog.models import Blog
from django.contrib import admin
admin.site.register(Blog)

class Blog(models.Model):
    title = models.CharField(max_length=60)
    body = models.TextField()
    created = models.DateTimeField("Date Created")
    updated = models.DateTimeField("Date Updated")
    def __unicode__(self):
        return self.title

class Comment(models.Model):
    body = models.TextField()
    author = models.CharField(max_length=60)
    created = models.DateTimeField("Date Created")
    updated = models.DateTimeField("Date Updated")
    post = models.ForeignKey(Blog)
    def __unicode__(self):
        return self.body

class CommentInline(admin.TabularInline):
    model = Comment


class BlogAdmin(admin.ModelAdmin):
    list_display = ('title','created', 'updated')
    search_fields = ['title','body']
    list_filter = ('Date Created','Date Updated')
    inlines …

我正在制作Google Chrome扩展程序,我想检查是否设置了密钥chrome.storage.sync.

示例:
我想检查密钥'links':

if (chrome.storage.sync.get('links',function(){
    // if already set it then nothing to do 
}));
else{
    // if not set then set it 
}

任何有用的建议将不胜感激.

我想在视图中生成带有静态URL的图像,然后在模板中呈现它.我mark_safe这样使用HTML不会被转义.但是,它仍然无法渲染图像.如何在Python中生成图像标记？

from django.shortcuts import render
from django.utils.safestring import mark_safe

def check(request):
    html = mark_safe('<img src="{% static "brand.png" %}" />')
    return render(request, "check.html", {"image": html})

在模板中渲染它:

{{ image }}

呈现img标记而不生成url:

<img src="{% static "brand.png" %} />

示例情况如下：

# models.py
class Form(models.Model):
    name = models.CharField()

class A(models.Model):
    form = models.ForeignKey(Form)

class B(A):
    name = models.CharField()

# view.py
form = Form.objects.get(id=1)
form.a_set.all()  # works
form.b_set.all()  # doesn't work

我想B通过父类A外键访问所有相关对象，但我似乎无法做到这一点。如果我通过访问它们，A我只会获得通用父类查询集。谢谢。

我最近在我的项目中添加了用于错误跟踪的哨兵，并根据此处的文档配置了 raven，但出现如下所示的错误。

设置.py

import raven


INSTALLED_APPS = [    
    # 3rd party app
    'raven.contrib.django.raven_compat',
]    

RAVEN_CONFIG = {
    'dsn': 'https://*****@sentry.io/224093',
    'release': raven.fetch_git_sha(os.path.dirname(os.pardir)),
}

错误日志

System check identified no issues (0 silenced).
October 02, 2017 - 11:31:58
Django version 1.10, using settings 'FoodCham.settings.development'
Starting development server at http://127.0.0.1:8000/
Quit the server with CONTROL-C.
Sending message of length 2240 to https://sentry.io/api/224093/store/
Internal Server Error: /
Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/django/core/handlers/exception.py", line 39, in inner
    response = get_response(request)
  File "/usr/local/lib/python2.7/dist-packages/django/core/handlers/base.py", …

class Test
{
    int x;
};
int main()
{
   cout << sizeof(Test) ;
   return 0;
}

输出:4
我只是想问,即使我没有创建任何类Test的对象为什么它打印4？

我想编写一个脚本，将用户信息（如姓名、联系电话等）以文件的形式保存为电子名片。有没有适合它的图书馆？