view test/test_hypothesis.py @ 7853:03c1b7ae3a68

issue2551328/issue2551264 unneeded next link and total_count incorrect Fix: issue2551328 - REST results show next link if number of results is a multiple of page size. (Found by members of team 3 in the UMass-Boston CS682 Spring 2024 class.) issue2551264 - REST X-Total-Count header and @total_size count incorrect when paginated These issues arose because we retrieved the exact number of rows from the database as requested by the user using the @page_size parameter. With this changeset, we retrieve up to 10 million + 1 rows from the database. If the total number of rows exceeds 10 million, we set the total_count indicators to -1 as an invalid size. (The max number of requested rows (default 10 million +1) can be modified by the admin through interfaces.py.) By retrieving more data than necessary, we can calculate the total count by adding @page_index*@page_size to the number of rows returned by the query. Furthermore, since we return more than @page_size rows, we can determine the existence of a row at @page_size+1 and use that information to determine if a next link should be provided. Previously, a next link was returned if @page_size rows were retrieved. This change does not guarantee that the user will get @page_size rows returned. Access policy filtering occurs after the rows are returned, and discards rows inaccessible by the user. Using the current @page_index/@page_size it would be difficult to have the roundup code refetch data and make sure that a full @page_size set of rows is returned. E.G. @page_size=100 and 5 of them are dropped due to access restrictions. We then fetch 10 items and add items 1-4 and 6 (5 is inaccessible). There is no way to calculate the new database offset at: @page_index*@page_size + 6 from the URL. We would need to add an @page_offset=6 or something. This could work since the client isn't adding 1 to @page_index to get the next page. Thanks to HATEOAS, the client just uses the 'next' url. But I am not going to cross that bridge without a concrete use case. This can also be handled client side by merging a short response with the next response and re-paginating client side. Also added extra index markers to the docs to highlight use of interfaces.py.
author John Rouillard <rouilj@ieee.org>
date Mon, 01 Apr 2024 09:57:16 -0400
parents ce740d9a7d8d
children 9ff94a2e8c82
line wrap: on
line source

import unittest

import pytest

pytest.importorskip("hypothesis")

# ruff: noqa: E402
from hypothesis import example, given, settings
from hypothesis.strategies import binary, none, one_of, sampled_from, text

from roundup.anypy.strings import b2s, s2b, s2u, u2s
# ruff: noqa: I001  - yes I know I am using \ to continue the line...
from roundup.password import PasswordValueError, encodePassword, \
     h64decode, h64encode


def Identity(x):
    return x


_max_examples = 1000


class HypoTestStrings(unittest.TestCase):

    @given(text())
    @settings(max_examples=_max_examples)
    def test_b2s(self, utf8_bytes):
        self.assertEqual(b2s(utf8_bytes.encode("utf-8")), utf8_bytes)

    @given(text())
    @settings(max_examples=_max_examples)
    def test_s2b(self, s):
        self.assertTrue(isinstance(s2b(s), bytes))

    @given(text())
    @settings(max_examples=_max_examples)
    @example("\U0001F600 hi there")  # smiley face emoji
    def test_s2u_u2s_invertable(self, s):
        self.assertEqual(u2s(s2u(s)), s)


class HypoTestPassword(unittest.TestCase):

    @given(binary())
    @example(b"")
    @settings(max_examples=_max_examples)
    def test_h64encode_h64decode(self, s):

        self.assertEqual(h64decode(h64encode(s)), s)

    @given(one_of(none(), text()),
           sampled_from(("PBKDF2S5", "PBKDF2", "SSHA",
                         "SHA", "MD5", "crypt", "plaintext",
                         "zot")))
    @example("asd\x00df", "crypt")
    @settings(max_examples=_max_examples)
    def test_encodePassword(self, password, scheme):

        if	scheme == "crypt" and password and "\x00" in password:
            with self.assertRaises(ValueError) as e:
                encodePassword(password, scheme)
            self.assertEqual(e.exception.args[0],
                             "embedded null character")
        elif scheme == "plaintext":
            if password is not None:
                self.assertEqual(encodePassword(password, scheme), password)
            else:
                self.assertEqual(encodePassword(password, scheme), "")
        elif scheme == "zot":
            with self.assertRaises(PasswordValueError) as e:
                encodePassword(password, scheme)
            self.assertEqual(e.exception.args[0],
                             "Unknown encryption scheme 'zot'")
        else:
            # it shouldn't throw anything.
            pw = encodePassword(password, scheme)

            # verify format
            if scheme in ["PBKDF2S5", "PBKDF2"]:
                # 1000$XbSsijELEQbZZb1LlD7CFuotF/8$DdtssSlm.e
                self.assertRegex(pw, r"^\d{4,8}\$.{27}\$.*")
            elif scheme == "SSHA":
                # vqDbjvs8rhrS1AJxHYEGGXQW3x7STAPgo7uCtnw4GYgU7FN5VYbZxccQYCC0eXOxSipLbtgBudH1vDRMNlG0uw==
                self.assertRegex(pw, r"^[^=]*={0,3}$")
            elif scheme == "SHA":
                # da39a3ee5e6b4b0d3255bfef95601890afd80709'
                self.assertRegex(pw, r"^[a-z0-9]{40}$")
            elif scheme == "MD5":
                # d41d8cd98f00b204e9800998ecf8427e'
                self.assertRegex(pw, r"^[a-z0-9]{32}$")
            elif scheme == "crypt":
                # WqzFDzhi8MmoU
                self.assertRegex(pw, r"^[A-Za-z0-9./]{13}$")
            else:
                self.assertFalse("Unknown scheme: %s, val: %s" % (scheme, pw))

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