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[minor] fetch dictory value using get method instead of indexing (#4324)

version-14
Makarand Bauskar пре 7 година
committed by Rushabh Mehta
родитељ
комит
fe75ffb8da
1 измењених фајлова са 54 додато и 34 уклоњено
  1. +54
    -34
      frappe/utils/password_strength.py

+ 54
- 34
frappe/utils/password_strength.py Прегледај датотеку

@@ -10,7 +10,9 @@ from frappe import _
def test_password_strength(password, user_inputs=None):
'''Wrapper around zxcvbn.password_strength'''
result = zxcvbn(password, user_inputs)
result['feedback'] = get_feedback(result['score'], result['sequence'])
result.update({
"feedback": get_feedback(result.get('score'), result.get('sequence'))
})
return result

# NOTE: code modified for frappe translations
@@ -33,23 +35,27 @@ default_feedback = {
],
}

def get_feedback (score, sequence):
def get_feedback(score, sequence):
"""
Returns the feedback dictionary consisting of ("warning","suggestions") for the given sequences.
"""
minimum_password_score = int(frappe.db.get_single_value("System Settings", "minimum_password_score"))

global default_feedback
minimum_password_score = int(frappe.db.get_single_value("System Settings", "minimum_password_score") or 2)

# Starting feedback
if len(sequence) == 0:
return default_feedback

# No feedback if score is good or great
if score >= minimum_password_score:
return dict({"warning": "","suggestions": []})
return dict({"warning": "", "suggestions": []})

# Tie feedback to the longest match for longer sequences
longest_match = max(sequence, key=lambda x: len(x['token']))
longest_match = max(sequence, key=lambda seq: len(seq.get('token', '')))

# Get feedback for this match
feedback = get_match_feedback(longest_match, len(sequence) == 1)

# If no concrete feedback returned, give more general feedback
if not feedback:
feedback = {
@@ -64,50 +70,55 @@ def get_match_feedback(match, is_sole_match):
"""
Returns feedback as a dictionary for a certain match
"""
# Define a number of functions that are used in a look up dictionary
def fun_bruteforce():
# Define a number of functions that are used in a look up dictionary
return None

def fun_dictionary():
# If the match is of type dictionary, call specific function
return get_dictionary_match_feedback(match, is_sole_match)

def fun_spatial():
if match["turns"] == 1:
feedback ={
feedback = {
"warning": _('Short keyboard patterns are easy to guess'),
"suggestions":[
_("Make use of longer keyboard patterns")
],
}

if match.get("turns") == 1:
feedback = {
"warning": _('Straight rows of keys are easy to guess'),
"suggestions":[
_("Try to use a longer keyboard pattern with more turns")
],
}
else:
feedback ={
"warning": _('Short keyboard patterns are easy to guess'),
"suggestions":[
_("Make use of longer keyboard patterns")
],
}

return feedback

def fun_repeat():
feedback ={
"warning": _('Repeats like "abcabcabc" are only slightly harder to guess than "abc"'),
"suggestions":[
_("Try to avoid repeated words and characters")
],
}
if len(match["repeated_char"]) == 1:
feedback ={
feedback = {
"warning": _('Repeats like "aaa" are easy to guess'),
"suggestions":[
_("Let's avoid repeated words and characters")
],
}
else:
feedback ={
"warning": _('Repeats like "abcabcabc" are only slightly harder to guess than "abc"'),
"suggestions":[
_("Try to avoid repeated words and characters")
],
}
return feedback

def fun_sequence():
return {
"suggestions":[
_("Avoid sequences like abc or 6543 as they are easy to guess")
],
}

def fun_regex():
if match["regex_name"] == "recent_year":
return {
@@ -117,6 +128,7 @@ def get_match_feedback(match, is_sole_match):
_("Avoid years that are associated with you.")
],
}

def fun_date():
return {
"warning": _("Dates are often easy to guess."),
@@ -138,7 +150,7 @@ def get_match_feedback(match, is_sole_match):
}
pattern_fn = patterns.get(match['pattern'])
if pattern_fn:
return(pattern_fn())
return (pattern_fn())

def get_dictionary_match_feedback(match, is_sole_match):
"""
@@ -146,35 +158,43 @@ def get_dictionary_match_feedback(match, is_sole_match):
"""
warning = ""
suggestions = []

# If the match is a common password
if match["dictionary_name"] == "passwords":
if is_sole_match and not match["l33t_entropy"]:
if match["rank"] <= 10:
if match.get("dictionary_name") == "passwords":
if is_sole_match and not match.get("l33t_entropy"):
if match.get("rank") <= 10:
warning = _("This is a top-10 common password.")
elif match["rank"] <= 100:
elif match.get("rank") <= 100:
warning = _("This is a top-100 common password.")
else:
warning = _("This is a very common password.")
else:
warning = _("This is similar to a commonly used password.")

# If the match is a common english word
elif match["dictionary_name"] == "english":
elif match.get("dictionary_name") == "english":
if is_sole_match:
warning = _("A word by itself is easy to guess.")

# If the match is a common surname/name
elif match["dictionary_name"] in ["surnames", "male_names", "female_names"]:
elif match.get("dictionary_name") in ["surnames", "male_names", "female_names"]:
if is_sole_match:
warning = _("Names and surnames by themselves are easy to guess.")
else:
warning = _("Common names and surnames are easy to guess.")
word = match["token"]

word = match.get("token")
# Variations of the match like UPPERCASES
if re.match(scoring.START_UPPER, word):
suggestions.append(_("Capitalization doesn't help very much."))
elif re.match(scoring.ALL_UPPER, word):
suggestions.append(_("All-uppercase is almost as easy to guess as all-lowercase."))

# Match contains l33t speak substitutions
if match["l33t_entropy"]:
if match.get("l33t_entropy"):
suggestions.append(_("Predictable substitutions like '@' instead of 'a' don't help very much."))
return {"warning": warning, "suggestions": suggestions}

return {
"warning": warning,
"suggestions": suggestions
}

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