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Problem :

I have below error message on the django 1.4:

“Dictionary update sequence element #0 has length 1; 2 is required”

It happened to me when I tried using a template tag like: `{% for v in values %}

It happens too when I try to access on the hstore queryset

My code is as below:

mytmp = Item.objects.where(HE("kv").contains({'key':value}))
if mytmp.count() > 0:
item_id = mytmp[0].id,

I am just trying to access a value. I don't understand a "update sequence" message. When I use the cursor instead of the hstore queryset, a function works. A error comes on template rendering too. I just restarted the uwsgi and everything works well, but above error comes back later.

Has someone an idea to fix this error?

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2 Answers

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Solution :

Just ran into your problem. The root cause was because I just forgot to put the name= on a last argument of a url (or a path in Django 2.0+) function call.

For example, the below functions throw a error from your question:

url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), 'foo')

path('foo/{slug:bar}/', views.FooBar, 'foo')

But below one actually work:

url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), name='foo')

path('foo/{slug:bar}/', views.FooBar, name='foo')

So a reason why a traceback is unhelpful is as internally, Django wants to parse a given positional argument as a keyword argument kwargs, and since the string is the iterable, the atypical code path begins to unfold. So always use a name= on your urls!

9 7 4
38,600 points
0 votes


Just ran into this problem. I don't know if it's the same thing that hit your code, but for me the root cause was because I forgot to put name= on the last argument of the url (or path in Django 2.0+) function call.

the following functions throw the error from the question:

url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), 'foo')
path('foo/{slug:bar}/', views.FooBar, 'foo')

However it will work,

url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), name='foo')
path('foo/{slug:bar}/', views.FooBar, name='foo')

The reason why the traceback is unhelpful is because internally, Django wants to parse the given positional argument as the keyword argument kwargs, and since a string is an iterable, an atypical code path begins to unfold. Always use name= on your urls!

 get dict from string is:

dic2 = eval(str1)
{'taras': 'vaskiv', 'iruna': 'vaskiv'}

Or in matter of security we can use literal_eval

from ast import literal_eval

pass a keyword argument name to url() function

url(r"^testing/$", views.testing, name="testing")


Another way of solve this issue

Pass a keyword argument name with value as your view name e.g home or home-view etc. to url() function.


Throws Error:

url(r'^home$', 'common.views.view1', 'home'),


url(r'^home$', 'common.views.view1', name='home'),

You are sending one parameter incorrectly; it should be a dictionary object:

Wrong: func(a=r)

Correct: func(a={'x':y})

I got the same issue and found that it was due to wrong parameters. In views.py, I used:

return render(request, 'demo.html',{'items', items})  

However, I found the issue: {'items', items}. Changing to {'items': items} resolved the issue.

10 6 4
31,120 points

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