Django：单击图像并在新页面中打开它答案

【问题标题】：Django: click image and have it open in a new pageDjango：单击图像并在新页面中打开它
【发布时间】：2015-05-24 03:25:56
【问题描述】：

django 和 python 的新手，所以请用非常简单的术语给我你的建议。

基本上我有一个图像托管网站，我希望能够单击任何图像并让它打开一个显示图像的新页面。我该怎么做呢？我想我必须获取图片网址，但我不确定如何获取。这也必须是自动的。

型号：

class UploadImages(models.Model):

    category_choice=(('Full','Full Beard'),
                ('tashe', 'Moustache'),
                ('goat','Goatee'))

    image = models.ImageField(upload_to='rate', blank=True)
    name =  models.CharField(max_length=128, default=0)
    user = models.ForeignKey(User)
    date = models.DateTimeField(default=now())
    #category = models.ForeignKey(Categories)
    category = models.CharField(max_length=30, choices=category_choice)
    rating = models.IntegerField(default=0)

    def __unicode__(self):
        return self.name

观看次数：

def index(request):

    if request.method == 'GET':

        #images_list = UploadImages.objects.order_by('-date')
        images_list = UploadImages.objects.filter(category='tashe')
        category_list = Category.objects.order_by('-likes')[:5]
        page_list = Page.objects.order_by('-views')[:5]

        context_dict = {'images':images_list}
        response = render(request,'RateMyBeard/index.html', context_dict)
        return response

    else:
        rating = request.POST['submit']
        #images_list = UploadImages.objects.order_by('-date')
        images_list = UploadImages.objects.filter(category='tashe')
        print rating
        #print images_list

        return render(request, 'RateMyBeard/index.html', images_list)

HTML

{% extends 'base.html' %}

{% load staticfiles %} <!-- New line -->



{% block body_block %}

<p> 'check our images'</p>

             {% for image in images %}
                 <a href="upload/">
                 <img src="{{image.image.url}}" alt="{{image.name}}" >
                    </a>

              {% endfor %}

  {% csrf_token %}
{% endblock %}

【问题讨论】：

标签： django image hyperlink

【解决方案1】：

将target="_blank" 属性添加到<a> 标签，链接将在新窗口/标签中打开：

<a href="{{image.image.url}}" target="_blank">
    <img src="{{image.image.url}}" alt="{{image.name}}">
</a>

【讨论】：