models.py:
class UserProfile(models.Model):
photo = models.ImageField(upload_to = get_upload_file_name,
storage = OverwriteStorage(),
default = os.path.join(settings.STATIC_ROOT,'images','generic_profile_photo.jpg'),
height_field = 'photo_height',
width_field = 'photo_width')
photo_height = models.PositiveIntegerField(blank = True, default = 0)
photo_width = models.PositiveIntegerField(blank = True, default = 0)
views.py:
def EditProfile(request):
register_generator()
source_file = UserProfile.objects.get(user = request.user).photo
args = {}
args.update(csrf(request))
args.update({'source_file' : source_file})
在我的模板中的某个地方:
{% generateimage 'user_profile:thumbnail' source=source_file %}
我收到一个错误:UserProfile匹配查询不存在.
在这一行:
source_file = UserProfile.objects.get(user = request.user).photo
问题是ImageField的默认属性不起作用.因此,对象不是在我的模型中创建的.如何正确使用此属性？如果我省略此属性,则创建对象时没有错误.我需要通过绝对路径还是相对路径？我使用django-imagekit在显示之前调整图像大小:http://django-imagekit.readthedocs.org/en/latest/
以下代码正常工作.它画了一个圆圈.我想要的是在这个圆圈里面绘制一个bootstrap glyphicon:<span class="glyphicon glyphicon-zoom-in"></span>
d3.json("relations", function(error, graph) {
var links = svg.selectAll()
.data(graph.links).enter();
links
.append("circle")
.attr("class", "circle")
.attr("r", circle_r)
.attr("cx", function(d) { return d.x; })
.attr("cy", function(d) { return d.y; });
我试图做以下继续前面的代码:
links
.append("span")
.attr("class", "glyphicon glyphicon-zoom-in")
但符号不呈现.我想有一些关于定位它的东西,因为使用firebug/inspector我可以看到已经<span>解决了.
有谁知道怎么做？谢谢
我有这个型号:
class UserProfile(models.Model):
#..........
photo = models.ImageField(upload_to = get_upload_file_name,
storage = OverwriteStorage(),
blank = True, null = True,
height_field = 'photo_height',
width_field = 'photo_width')
这是我的存储功能:
class OverwriteStorage(FileSystemStorage):
def _save(self, name, content):
self.delete(r'.*')
return super(OverwriteStorage, self)._save(name, content)
def get_available_name(self, name):
return name
我该怎么做以下两件事:
每当用户上传文件(即图像)时,我想删除旧文件,无论名称是否相同.我试图删除任何与上述正则表达式匹配的内容,但这不起作用.
如果用户上传名为"me.jpg"的图像,我想以不同的方式重命名,具体取决于用户的用户名.所以我会做一些像这样做的事情return super(OverwriteStorage, self)._save(SOMETHING_ELSE_HERE, content)？我可以将其他参数传递给OverwriteStorage函数吗？
还有第三个问题:我为这个表单创建了一个ModelForm.因此用户可以上传图像.因此,当有人按下"选择文件"时,会弹出一个窗口窗口以浏览和选择照片.我怎么才能在这里显示某些文件？(例如,只有.jpg和.jpeg文件)
谢谢!
编辑:该get_upload_file_name功能
def get_upload_file_name(instance, filename):
return "%s/%s/profile_photo/%s" % (instance.user.username[0].lower(), instance.user.username, filename)
编辑2:我已经包含了我的models.py
import datetime
import os
import urllib2, urlparse
import re
from django.db import models
from …我读过其他类似的帖子,但我不能让我的工作.我想让点击的标签处于活动状态
所以这是我的bootstrap导航栏在html中的样子:
<div class="navbar navbar-default" role="navigation">
<div class="container">
<div class="navbar-header">
<button type="button" class="navbar-toggle" data-toggle="collapse" data-target=".navbar-collapse">
<span class="sr-only">Toggle navigation</span>
<span class="icon-bar"></span>
<span class="icon-bar"></span>
<span class="icon-bar"></span>
</button>
<a href="/">
<img src="/static/images/logo.png" alt="Prime Solutions"/>
</a>
</div>
<div class="navbar-collapse collapse">
<ul class="nav navbar-nav">
<li class="active"><a href="/"><strong>Acasa</strong></a></li>
<li><a href="/software"><strong>Software</strong></a></li>
<li><a href="/hardware"><strong>Hardware</strong></a></li>
<li><a href="/about"><strong>Despre Noi</strong></a></li>
<li><a href="/contact"><strong>Contact</strong></a></li>
<li class="dropdown">
<a href="#" class="dropdown-toggle" data-toggle="dropdown"><strong>Servicii</strong><b class="caret"></b></a>
<ul class="dropdown-menu">
<li><a href="/servicii/servicii">Servicii IT&C </a></li>
<li><a href="/servicii/consultanta">Consultanta IT&C </a></li>
<li class="divider"></li>
<li class="dropdown-header">Parteneri</li>
<li><a href="/parteneri">Partenerii nostri</a></li>
</ul>
</li>
</ul>
<ul class="nav …在我的某个地方，我抛出了一个错误，尤其是这个错误：
views.py
from xmlrpclib import Fault
def some_function(request):
if ....:
return Fault(-1, 'foo')
然后，也在views.py中，我有自定义500处理程序来捕获服务器错误：
def my_custom_500(request):
context = {...}
### Here is where I need to catch `'foo'`
### in order to put it in the context and pass it to the template
render(request, '500.html', context)
无论如何，我可以访问该错误消息吗？谢谢
这是我的/home/ubuntu/project/nginx.conf样子:
http {
# configuration of the server
server {
# the port your site will be served on
listen 80;
# the domain name it will serve for
server_name ec2-xxx-xxx-xxx-xxx.eu-central-1.compute.amazonaws.com; # substitute your machine's IP address or FQDN
charset utf-8;
# max upload size
client_max_body_size 75M; # adjust to taste
# Django media
location /media {
alias /home/ubuntu/project/media; # your Django project's media files - amend as required
}
location /static {
alias /home/ubuntu/project/static; # your Django …