RobotFramework之Collections
RobotFramework之Collections
背景
继续学习RobotFramework，此次看的是Collections的源码。html
Collections库是RobotFramework用来处理列表和字典的库，官方文档是这么介绍的python
A test library providing keywords for handling lists and dictionaries.app
Append To List
示例代码框架
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${l1} create list a b
append to list ${l1} 1 2
log ${l1}
执行结果：less
KEYWORD BuiltIn . Log ${l1}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170422 19:11:52.624 / 20170422 19:11:52.624 / 00:00:00.000
19:11:52.624 INFO [u'a', u'b', u'1', u'2']
源代码学习
def append_to_list(self, list_, *values):
for value in values:
list_.append(value)
Combine List
示例代码ui
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${l1} create list a b
${l2} create list 1 2
${l} combine lists ${l1} ${l2}
log ${l}
执行结果spa
KEYWORD ${l} = Collections . Combine Lists ${l1}, ${l2}
Documentation:
Combines the given lists together and returns the result.
Start / End / Elapsed: 20170422 19:15:52.086 / 20170422 19:15:52.087 / 00:00:00.001
19:15:52.087 INFO ${l} = [u'a', u'b', u'1', u'2']
源代码code
def combine_lists(self, *lists):
ret = []
for item in lists:
ret.extend(item)
return ret
说明regexp
源代码中的extend方法和append方法是不同的，append方法是追加一个元素，extend方法是追加一个列表的内容，若是用append方法，则结果会变成以下状况：
l1 = [1,2,3] l2 = [4,5,6] l1.append(l2) # ==>[1,2,3,[4,5,6]] l1.extend(l2) # ==>[1,2,3,,4,5,6]
因此合并列表的时候使用extend而不是append
Count Values In List
计算某一个值在列表中重复的次数
示例
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${l1} create list 1 2 3 4 3
${count} count values in list ${l1} 3
log ${count}
执行结果
KEYWORD ${int} = Collections . Count Values In List ${l1}, 3
Documentation:
Returns the number of occurrences of the given value in list.
Start / End / Elapsed: 20170422 19:30:33.807 / 20170422 19:30:33.807 / 00:00:00.000
19:30:33.807 INFO ${int} = 2
源代码
def count_values_in_list(self, list_, value, start=0, end=None):
return self.get_slice_from_list(list_, start, end).count(value)
def get_slice_from_list(self, list_, start=0, end=None):
start = self._index_to_int(start, True)
if end is not None:
end = self._index_to_int(end)
return list_[start:end]
def _index_to_int(self, index, empty_to_zero=False):
if empty_to_zero and not index:
return 0
try:
return int(index)
except ValueError:
raise ValueError("Cannot convert index '%s' to an integer." % index)
说明
这里方法是默认从第一位开始计算，若是有须要，能够输入一个区间，好比指定起始位置和结束位置，好比以下代码
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${l1} create list 1 2 3 4 3 3
${count} count values in list ${l1} 3 3 5
log ${count}
执行结果就是1
Dictionaries Should Be Equal
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${dict1} create dictionary a=1 b=2
${dict2} create dictionary a=1 b=3
dictionaries should be equal ${dict1} ${dict2}
执行结果
KEYWORD Collections . Dictionaries Should Be Equal ${dict1}, ${dict2}
Documentation:
Fails if the given dictionaries are not equal.
Start / End / Elapsed: 20170422 19:49:36.865 / 20170422 19:49:36.866 / 00:00:00.001
19:49:36.866 FAIL Following keys have different values:
Key b: 2 != 3
源代码
def dictionaries_should_be_equal(self, dict1, dict2, msg=None, values=True):
keys = self._keys_should_be_equal(dict1, dict2, msg, values)
self._key_values_should_be_equal(keys, dict1, dict2, msg, values)
def _keys_should_be_equal(self, dict1, dict2, msg, values):
keys1 = self.get_dictionary_keys(dict1)
keys2 = self.get_dictionary_keys(dict2)
miss1 = [unic(k) for k in keys2 if k not in dict1]
miss2 = [unic(k) for k in keys1 if k not in dict2]
error = []
if miss1:
error += ['Following keys missing from first dictionary: %s'
% ', '.join(miss1)]
if miss2:
error += ['Following keys missing from second dictionary: %s'
% ', '.join(miss2)]
_verify_condition(not error, '\n'.join(error), msg, values)
return keys1
def _key_values_should_be_equal(self, keys, dict1, dict2, msg, values):
diffs = list(self._yield_dict_diffs(keys, dict1, dict2))
default = 'Following keys have different values:\n' + '\n'.join(diffs)
_verify_condition(not diffs, default, msg, values)
def _yield_dict_diffs(self, keys, dict1, dict2):
for key in keys:
try:
assert_equal(dict1[key], dict2[key], msg='Key %s' % (key,))
except AssertionError as err:
yield unic(err)
说明
原本Python中比较两个字典是否相等只要用==就好了，这里为了可以精确的报错，遍历了全部的key和value来作比较。
Dictionary Should Contain Item
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${dict1} create dictionary a=1 b=2
dictionary should contain item ${dict1} a 1
结果
KEYWORD Collections . Dictionary Should Contain Item ${dict1}, a, 1
Documentation:
An item of key``/``value must be found in a `dictionary`.
Start / End / Elapsed: 20170422 19:58:37.086 / 20170422 19:58:37.087 / 00:00:00.001
源代码
def dictionary_should_contain_item(self, dictionary, key, value, msg=None):
self.dictionary_should_contain_key(dictionary, key, msg)
actual, expected = unic(dictionary[key]), unic(value)
default = "Value of dictionary key '%s' does not match: %s != %s" % (key, actual, expected)
_verify_condition(actual == expected, default, msg)
def dictionary_should_contain_key(self, dictionary, key, msg=None):
default = "Dictionary does not contain key '%s'." % key
_verify_condition(key in dictionary, default, msg)
说明
判断一个k-v是否在一个字典里，一样是经过遍历原有字典的方式来处理，只不过这里传值必须是按照示例代码中那样传递，不能传入a=1或者一个字典。
Dictionary Should Contain Sub Dictionary
示例
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${dict1} create dictionary a=1 b=2
${dict2} create dictionary a=1
Dictionary Should Contain Sub Dictionary ${dict1} ${dict2}
结果
KEYWORD Collections . Dictionary Should Contain Sub Dictionary ${dict1}, ${dict2}
Documentation:
Fails unless all items in dict2 are found from dict1.
Start / End / Elapsed: 20170422 20:09:41.864 / 20170422 20:09:41.865 / 00:00:00.001
源代码
def dictionary_should_contain_sub_dictionary(self, dict1, dict2, msg=None,
values=True):
keys = self.get_dictionary_keys(dict2)
diffs = [unic(k) for k in keys if k not in dict1]
default = "Following keys missing from first dictionary: %s" \
% ', '.join(diffs)
_verify_condition(not diffs, default, msg, values)
self._key_values_should_be_equal(keys, dict1, dict2, msg, values)
说明
上一个方法只能传入单一的k-v，使用的局限性比较大，这个方法就能够支持传入一个字典
Get Dictionary Items
示例
*** Settings ***
Library Collections
*** Test Cases ***
TestCase001
${dict1} create dictionary a=1 b=2
${dict2} create dictionary a=1
${dict3} Get Dictionary Items ${dict1}
执行结果
KEYWORD ${dict3} = Collections . Get Dictionary Items ${dict1}
Documentation:
Returns items of the given dictionary.
Start / End / Elapsed: 20170424 00:39:49.766 / 20170424 00:39:49.766 / 00:00:00.000
00:39:49.766 INFO ${dict3} = [u'a', u'1', u'b', u'2']
源代码
def get_dictionary_items(self, dictionary):
ret = []
for key in self.get_dictionary_keys(dictionary):
ret.extend((key, dictionary[key]))
return ret
说明
这个方法就是把字典里面的成员以列表的形式返回出来，具体应用场景还不是很肯定
Get Dictionary Values
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${dict} create dictionary a=1 b=2
${rst} get from dictionary ${dict} a
log ${rst}
执行结果
Documentation: Logs the given message with the given level. Start / End / Elapsed: 20170428 19:45:59.199 / 20170428 19:45:59.200 / 00:00:00.001 19:45:59.200 INFO 1
源代码
def get_from_dictionary(self, dictionary, key):
try:
return dictionary[key]
except KeyError:
raise RuntimeError("Dictionary does not contain key '%s'." % key)
说明
源代码其实很是简单，传入一个字典和一个key，而后就返回这个key对应的值
Get From List
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst} create list 1 2 3
${new_lst} get from list ${lst} 0
log ${new_lst}
执行结果
KEYWORD BuiltIn . Log ${new_lst}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 19:50:42.533 / 20170428 19:50:42.533 / 00:00:00.000
19:50:42.533 INFO 1
源代码
def get_from_list(self, list_, index):
try:
return list_[self._index_to_int(index)]
except IndexError:
self._index_error(list_, index)
说明
执行方法就是获取列表和索引，而后返回列表的索引，值得一提的是，不论你传入的值是str类型仍是int类型，都会被python代码转成int类型，因此源代码中${new_lst} get from list ${lst} 0这样写,执行结果也是同样的。附上转换的源代码。
def _index_to_int(self, index, empty_to_zero=False):
if empty_to_zero and not index:
return 0
try:
return int(index)
except ValueError:
raise ValueError("Cannot convert index '%s' to an integer." % index)
Get Index From List
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst} create list 1 2 3 4 5 4
${rst} get index from list ${lst} 3
log ${rst}
执行结果
KEYWORD BuiltIn . Log ${rst}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 19:59:05.532 / 20170428 19:59:05.532 / 00:00:00.000
19:59:05.532 INFO 2
源代码
def get_index_from_list(self, list_, value, start=0, end=None):
if start == '':
start = 0
list_ = self.get_slice_from_list(list_, start, end)
try:
return int(start) + list_.index(value)
except ValueError:
return -1
说明
源代码中的self.get_slice_from_list方法是一个切片方法，在Get Index From List中还能够传入两个参数，start表示列表的开始索引，end表示列表结束的索引，若是没有给，默认就是整个列表来取索引结果。
Get Match Count
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${count} get match count aaabbcc a
log ${count}
执行结果
KEYWORD BuiltIn . Log ${count}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 20:08:08.699 / 20170428 20:08:08.700 / 00:00:00.001
20:08:08.700 INFO 3
源代码
def get_match_count(self, list, pattern, case_insensitive=False,
whitespace_insensitive=False):
return len(self.get_matches(list, pattern, case_insensitive,
whitespace_insensitive))
def _get_matches_in_iterable(iterable, pattern, case_insensitive=False,
whitespace_insensitive=False):
if not is_string(pattern):
raise TypeError("Pattern must be string, got '%s'." % type_name(pattern))
regexp = False
if pattern.startswith('regexp='):
pattern = pattern[7:]
regexp = True
elif pattern.startswith('glob='):
pattern = pattern[5:]
matcher = Matcher(pattern,
caseless=is_truthy(case_insensitive),
spaceless=is_truthy(whitespace_insensitive),
regexp=regexp)
return [string for string in iterable
if is_string(string) and matcher.match(string)]
说明
这个方法是返回一个字符在字符串中重复的次数。
源代码中的第二个方法才是真正的方法，Get Match Count方法是很是强大的，能够作普通匹配，也能够作正则匹配，首先，传入的必须是字符串，不然报错，而后断定传入的条件是否带有正则，若是传入的条件是regexp=开头的，表示要使用正则匹配，不然就是普通查找。
第三个参数表示是否忽略大小写，第四个参数表示是否忽略空格。
Get Matchs
此方法与上面一个方法调用的是同样的底层方法，请参考上面的内容。
Get Slice From List
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst} create list 1 2 3 4 5 6 7 8
${slice_list} get slice from list ${lst} 3 6
log ${slice_list}
执行结果
KEYWORD BuiltIn . Log ${slice_list}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 20:20:00.780 / 20170428 20:20:00.780 / 00:00:00.000
20:20:00.780 INFO [u'4', u'5', u'6']
源代码
def get_slice_from_list(self, list_, start=0, end=None):
start = self._index_to_int(start, True)
if end is not None:
end = self._index_to_int(end)
return list_[start:end]
说明
源代码仍是比较容易看懂的，若是开始和结束不传值，那么就是返回原来的列表，不然就返回切片的列表。
Insert into list
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst} create list 1 2 3 4 5 6 7 8
insert into list ${lst} 0 a
log ${lst}
执行结果
KEYWORD BuiltIn . Log ${lst}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 20:27:02.736 / 20170428 20:27:02.737 / 00:00:00.001
20:27:02.737 INFO [u'a', u'1', u'2', u'3', u'4', u'5', u'6', u'7', u'8']
源代码
def insert_into_list(self, list_, index, value):
list_.insert(self._index_to_int(index), value)
说明
调用的就是Python中列表的insert方法。若是传入的索引大于列表的长度，那么值会默认插入到列表的最后一位，若是传入的是负数，好比传入-1，那么就会在列表的倒数第二位加入该元素。
根据实现的Python代码能够知道，传入的索引是int或者str类型均可以，实现的时候会强制转为int类型。
关于插入索引大于列表长度的问题，能够参考如下代码
Python 2.7.10 (default, Feb 6 2017, 23:53:20) [GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.34)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> a = [1,2,3,4,5] >>> a [1, 2, 3, 4, 5] >>> a.insert(9, 'a') >>> a [1, 2, 3, 4, 5, 'a']
Keep In Dictionary
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${dict} create dictionary a=1 b=2 c=3 d=4
keep in dictionary ${dict} a b
log ${dict}
执行结果
KEYWORD BuiltIn . Log ${dict}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 20:47:47.528 / 20170428 20:47:47.529 / 00:00:00.001
20:47:47.529 INFO {u'a': u'1', u'b': u'2'}
源代码
def keep_in_dictionary(self, dictionary, *keys):
remove_keys = [k for k in dictionary if k not in keys]
self.remove_from_dictionary(dictionary, *remove_keys)
def remove_from_dictionary(self, dictionary, *keys):
for key in keys:
if key in dictionary:
value = dictionary.pop(key)
logger.info("Removed item with key '%s' and value '%s'." % (key, value))
else:
logger.info("Key '%s' not found." % key)
说明
该方法是用来保留字典里指定的key，源代码也比较好理解，不过此处的注释有问题，
Keeps the given keys in the dictionary and removes all other.
If the given key cannot be found from the dictionary, it is ignored.
第二句是说明若是传入的key若是不在这个字典里，那么这个操做就会被忽略。可是实际上咱们能够看到代码执行时并非这样的，第一个方法会把字典中与传入的keys不匹配的key所有拿出来，调用第二个方法来所有从字典里移除，也就是说传入的key若是在字典里不存在，那么原字典就会变为一个空字典。
实际上的执行结果就是这样
KEYWORD ${dict} = BuiltIn . Create Dictionary a=1, b=2, c=3, d=4
00:00:00.001KEYWORD Collections . Keep In Dictionary ${dict}, e
Documentation:
Keeps the given keys in the dictionary and removes all other.
Start / End / Elapsed: 20170428 20:52:04.051 / 20170428 20:52:04.052 / 00:00:00.001
20:52:04.052 INFO Removed item with key 'a' and value '1'.
20:52:04.052 INFO Removed item with key 'b' and value '2'.
20:52:04.052 INFO Removed item with key 'c' and value '3'.
20:52:04.052 INFO Removed item with key 'd' and value '4'.
00:00:00.000KEYWORD BuiltIn . Log ${dict}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 20:52:04.052 / 20170428 20:52:04.052 / 00:00:00.000
20:52:04.052 INFO {}
List Should Contain Sub List
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst1} create list 1 2 3 4
${lst2} create list 2 3
list should contain sub list ${lst1} ${lst2}
执行结果
KEYWORD Collections . List Should Contain Sub List ${lst1}, ${lst2}
Documentation:
Fails if not all of the elements in list2 are found in list1.
源代码
def list_should_contain_sub_list(self, list1, list2, msg=None, values=True):
diffs = ', '.join(unic(item) for item in list2 if item not in list1)
default = 'Following values were not found from first list: ' + diffs
_verify_condition(not diffs, default, msg, values)
List Should Not Contain Duplicates
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst1} create list 1 2 3 4 2
list should not contain duplicates ${lst1} 2
执行结果
KEYWORD Collections . List Should Not Contain Duplicates ${lst1}, 2
Documentation:
Fails if any element in the list is found from it more than once.
Start / End / Elapsed: 20170428 21:08:59.719 / 20170428 21:08:59.720 / 00:00:00.001
21:08:59.719 INFO '2' found 2 times.
21:08:59.720 FAIL 2
源代码
def list_should_not_contain_duplicates(self, list_, msg=None):
if not isinstance(list_, list):
list_ = list(list_)
dupes = []
for item in list_:
if item not in dupes:
count = list_.count(item)
if count > 1:
logger.info("'%s' found %d times." % (item, count))
dupes.append(item)
if dupes:
raise AssertionError(msg or
'%s found multiple times.' % seq2str(dupes))
说明
该方法用于断言某个元素在列表中只会出现一次，若是出现屡次则报错。
Pop From Dictionary
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${dict} create dictionary a=1 b=2 c=3
${newdict} pop from dictionary ${dict} a
log ${newdict}
log ${dict}
执行结果
KEYWORD BuiltIn . Log ${dict}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 21:21:05.523 / 20170428 21:21:05.523 / 00:00:00.000
21:21:05.523 INFO {u'b': u'2', u'c': u'3'}
源代码
def pop_from_dictionary(self, dictionary, key, default=NOT_SET):
if default is NOT_SET:
self.dictionary_should_contain_key(dictionary, key)
return dictionary.pop(key)
return dictionary.pop(key, default)
说明
这里的方法有一个可选参数default，是用来处理不存在的键值，若是pop的键值在字典里不存在，那么框架是会报错处理的，可是若是给了一个default，那么传入的键值不存在时，会返回default的值。
Remove Duplicates
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst} create list 1 2 3 4 1 2
${rst} remove duplicates ${lst}
执行结果
KEYWORD ${rst} = Collections . Remove Duplicates ${lst}
Documentation:
Returns a list without duplicates based on the given list.
Start / End / Elapsed: 20170428 21:52:22.522 / 20170428 21:52:22.522 / 00:00:00.000
21:52:22.522 INFO 2 duplicates removed.
21:52:22.522 INFO ${rst} = [u'1', u'2', u'3', u'4']
源代码
def remove_duplicates(self, list_):
ret = []
for item in list_:
if item not in ret:
ret.append(item)
removed = len(list_) - len(ret)
logger.info('%d duplicate%s removed.' % (removed, plural_or_not(removed)))
return ret
说明
该方法是一个除重方法，能够吧列表中重复的元素移除。
Sort List
示例代码
*** Settings ***
Library Collections
*** Test Cases ***
test1
${lst} create list 1 2 3 4 1 2 and dib
sort list ${lst}
log ${lst}
执行结果
KEYWORD BuiltIn . Log ${lst}
Documentation:
Logs the given message with the given level.
Start / End / Elapsed: 20170428 22:13:57.288 / 20170428 22:13:57.288 / 00:00:00.000
22:13:57.288 INFO [u'1', u'1', u'2', u'2', u'3', u'4', u'and', u'dib']
源代码
def sort_list(self, list_):
list_.sort()
说明
该方法调用了列表的sort方法，须要注意的是，若是原始列表还须要使用的话，最好先保留一份，不然原始数据就会被破坏。
Note that the given list is changed and nothing is returned. Use
Copy Listfirst, if you need to keep also the original order.
总结
Collections库是一个很是简单的库，不过里面包含了大部分的字典和列表的操做，本文仅仅覆盖了一些关键字，还有一些重复的，或者一眼就能看出来怎么用的关键字我就没有写了。经过这些示例代码，基本上可以了解Robot Framework框架的使用方法，也可以对Robot Framework框架的编写方式和编写思想有了必定的认识。
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每一个你不满意的现在，都有一个你没有努力的曾经。
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