id	summary	reporter	owner	description	type	status	component	version	severity	resolution	keywords	cc	stage	has_patch	needs_docs	needs_tests	needs_better_patch	easy	ui_ux
25670	`dictsort` does not work when `arg` parameter is numeric	Andrew Kuchev	Andrew Kuchev	"According to `dictsort` documentation, it orders given list of dictionaries, using `arg` as property in each dictionary:

{{{
@register.filter(is_safe=False)
def dictsort(value, arg):
    """"""
    Takes a list of dicts, returns that list sorted by the property given in
    the argument.
    """"""
    try:
        return sorted(value, key=Variable(arg).resolve)
    except (TypeError, VariableDoesNotExist):
        return ''
}}}

However, it is not possible to order list of dictionaries by a numeric key. Let's consider the following test case:

{{{
def test_sort_list_of_tuple_like_dicts(self):
    data = [{'0': 'a', '1': '42'},
            {'0': 'c', '1': 'string'},
            {'0': 'b', '1': 'foo'}]

    sorted_data = dictsort(data, '0')

    self.assertEqual([{'0': 'a', '1': '42'},
                      {'0': 'b', '1': 'foo'},
                      {'0': 'c', '1': 'string'}], sorted_data)
}}}

This test fails with the following message:

{{{
Traceback (most recent call last):
  File "".../django/tests/template_tests/filter_tests/test_dictsort.py"", line 50, in test_sort_list_of_tuple_like_dicts
    {'0': 'c', '1': 'string'}], sorted_data)
AssertionError: Lists differ: [{'0': 'a', '1': '42'}, {'0': 'b', '1': 'foo'}, {'0': 'c', '1': 'string'}] != [{'0': 'a', '1': '42'}, {'0': 'c', '1': 'string'}, {'0': 'b', '1': 'foo'}]

First differing element 1:
{'0': 'b', '1': 'foo'}
{'0': 'c', '1': 'string'}

- [{'0': 'a', '1': '42'}, {'0': 'b', '1': 'foo'}, {'0': 'c', '1': 'string'}]
+ [{'0': 'a', '1': '42'}, {'0': 'c', '1': 'string'}, {'0': 'b', '1': 'foo'}]
}}}

The `dictsort` uses `sorted` function with `key=Variable(arg).resolve`. When `arg` is `'0'`, `key` function should behave like `operator.itemgetter('0')`, but `Variable('0').resolve(context)` returns `0` regardless of given `context`.

There are five usages of `dictsort` with `""0""` as `arg` in [https://github.com/django/django/blob/master/django/views/debug.py#L816 debug.py]

As [https://code.djangoproject.com/ticket/25646#comment:4 mentioned] by ''bmispelon'', this may be some kind of regression in Django 1.3. "	Bug	closed	Template system	1.8	Normal	fixed	template filter dictsort		Accepted	1	0	0	0	0	0