小编Zol*_*oli的帖子

我试图将一些数据发布到服务器,但我不知道如何取回响应数据.

我有以下代码:

fetch(url, {
  method: 'POST',
  headers: {
    'Accept': 'application/json',
    'Content-Type': 'application/json'
  },
  body: JSON.stringify({
    email: login,
    password: password,
  })
}).then(function(a){
  console.log(a);
})

它打印出一个Response包含body(ReadableByteStream),bodyUsed(false),ok(true),status(200)等数据的数据,但是我找不到我回来的数据,无处可去.当我打开chrome开发者控制台 - 网络时,我看到那里的响应数据.

我究竟做错了什么？

我一直在寻找一些资源如何获取,承诺,...工作,但我找不到任何写得好.

我是单元测试的新手,所以我不知道我做错了什么.我在Django1.8中使用python2.7

我跑的时候

python manage.py test myapp --keepdb

我明白了

======================================================================
ERROR: test_view_content (myproject.news.tests.test_views.EntryTestCase)
----------------------------------------------------------------------
Traceback (most recent call last):
  File "/home/zoli/projects/project_dict/myproject/news/tests/test_views.py", line 27, in test_view_content
    response = client.get(reverse('news_list', kwargs={'page': 1}))
  File "/home/zoli/.virtualenvs/project_dict/local/lib/python2.7/site-packages/django/test/client.py", line 500, in get
    **extra)
  File "/home/zoli/.virtualenvs/project_dict/local/lib/python2.7/site-packages/django/test/client.py", line 303, in get
    return self.generic('GET', path, secure=secure, **r)
  File "/home/zoli/.virtualenvs/project_dict/local/lib/python2.7/site-packages/django/test/client.py", line 379, in generic
    return self.request(**r)
  File "/home/zoli/.virtualenvs/project_dict/local/lib/python2.7/site-packages/django/test/client.py", line 466, in request
    six.reraise(*exc_info)
  File "/home/zoli/.virtualenvs/project_dict/local/lib/python2.7/site-packages/django/core/handlers/base.py", line 108, in get_response
    response = middleware_method(request)
  File "/home/zoli/projects/project_dict/myproject/middleware/multihostname.py", line 18, in process_request
    host = request.META['HTTP_HOST'].split(':')[0] …

我很难从django形式获取图像url.

模型:

class Sponsor(models.Model):
    image = ProcessedImageField(upload_to='sponsors/', 
                                processors=[SmartResize(300, 120, upscale=False)],
                                format='JPEG', 
                                options={'quality': 100}, 
                                null=True)

形成:

class SponsorForm(forms.ModelForm):

    class Meta:
        model = Sponsor
        fields = ('image',)

我使用时渲染它:

{{ sponsorform.image }}

它打印:

Currently: <a href="/media/sponsors/random1.jpg">sponsors/random1.jpg</a> <br>Change: <input id="id_sponsor-image" name="sponsor-image" type="file">

所以我的问题是,如何在模板中获取该URL？我试过了:

{{ sponsorform.image.url }}
{{ sponsorform.image.path }}
{{ sponsorform.image.href }}

但似乎没有任何效果,任何建议？

如何合并按键和点击？我的意思是当用户按下回车并在同一时间点击某处我需要调用一个功能.

$(document).keypress(function(e) {
    var code = e.keyCode || e.which;
    if(code == 13) {
        alert('keypress');
    }
});
$(document).on( "click", function() {
    alert('click');
});

我有这个代码,但我无法合并它(通常我不使用jQuery/javascript).

我有以下格式的网址。

http://127.0.0.1:8000/accounts/login/?next=/event/contract-risk-management/review/

我需要从模板中“解析”“ / event / contract-risk-management / review /”部分。问题是我不知道如何在问号后得到零件。

我尝试了request.path，但它只返回了网址的第一部分。（无域）。

有谁知道我该怎么用？谢谢。