RobotFramework之Collections
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背景
继续学习RobotFramework，这次看的是Collections的源码。
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
A test library providing keywords for handling lists and dictionaries.
Append To List
示例代码
*** Settings ***Library Collections*** Test Cases ***TestCase001 ${l1} create list a b append to list ${l1} 1 2 log ${l1}执行结果：
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 / 00:00:00.00019:11:52.624 INFO [u'a', u'b', u'1', u'2']源代码
def append_to_list(self, list_, *values): for value in values: list_.append(value)Combine List
示例代码
*** 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
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 / 00:00:00.00119:15:52.087 INFO ${l} = [u'a', u'b', u'1', u'2']源代码
def combine_lists(self, *lists): ret = [] for item in lists: ret.extend(item) return ret说明
源代码中的extend方法和append方法是不一样的，append方法是追加一个元素，extend方法是追加一个列表的内容，如果用append方法，则结果会变成如下情况：
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]所以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}, 3Documentation: Returns the number of occurrences of the given value in list.Start / End / Elapsed: 20170422 19:30:33.807 / 20170422 19:30:33.807 / 00:00:00.00019:30:33.807 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，如果有需要，可以输入一个区间，比如指定起始位置和结束位置，比如如下代码
*** 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 / 00:00:00.00119:49:36.866 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
本来Python中比较两个字典是否相等只要用==就行了，这里为了能够精确的报错，遍历了所有的key和value来做比较。
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, 1Documentation: 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，同样是通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，这个方法就可以支持传入一个字典
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 / 00:00:00.00000:39:49.766 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确定
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 / 00:00:00.00119:45:59.200 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
源代码其实非常简单，传入一个字典和一个key，然后就返回这个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 ${new_lst}Documentation: Logs the given message with the given level.Start / End / Elapsed: 20170428 19:50:42.533 / 20170428 19:50:42.533 / 00:00:00.00019:50:42.533 INFO 1源代码
def get_from_list(self, list_, index): try: return list_[self._index_to_int(index)] except IndexError: self._index_error(list_, index)说明
执行方法就是获取列表和索引，然后返回列表的索引，值得一提的是，不论你传入的值是str类型还是int类型，都会被python代码转成int类型，所以源代码中${new_lst} get from list ${lst} 0这样写,执行结果也是一样的。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 / 00:00:00.00019:59:05.532 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中还可以传入两个参数，start表示列表的开始索引，end表示列表结束的索引，如果没有给，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 / 00:00:00.00120:08:08.700 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方法是非常强大的，可以做普通匹配，也可以做正则匹配，首先，传入的必须是字符串，否则报错，然后判定传入的条件是否带有正则，如果传入的条件是regexp=开头的，表示要使用正则匹配，否则就是普通查找。
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一样的底层方法，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 / 00:00:00.00020:20:00.780 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
源代码还是比较容易看懂的，如果开始和结束不传值，那么就是返回原来的列表，否则就返回切片的列表。
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 / 00:00:00.00120:27:02.737 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。如果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，如果传入的是负数，比如传入-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代码可以知道，传入的索引是int或者str类型都可以，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，可以参考以下代码
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 darwinType "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 / 00:00:00.00120:47:47.529 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.
第二句是说明如果传入的key如果不在这个字典里，那么这个操作就会被忽略。但是实际上我们可以看到代码执行时并不是这样的，第一个方法会把字典中与传入的keys不匹配的key全部拿出来，调用第二个方法来全部从字典里移除，也就是说传入的key如果在字典里不存在，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=400:00:00.001KEYWORD Collections . Keep In Dictionary ${dict}, eDocumentation: Keeps the given keys in the dictionary and removes all other.Start / End / Elapsed: 20170428 20:52:04.051 / 20170428 20:52:04.052 / 00:00:00.00120:52:04.052 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 / 00:00:00.00020:52:04.052 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}, 2Documentation: 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 / 00:00:00.00121:08:59.719 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，如果出现多次则报错。
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 / 00:00:00.00021:21:05.523 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，如果pop的键值在字典里不存在，那么框架是会报错处理的，但是如果给了一个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 / 00:00:00.00021:52:22.522 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，可以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 / 00:00:00.00022:13:57.288 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，需要注意的是，如果原始列表还需要使用的话，最好先保留一份，否则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库是一个非常简单的库，不过里面包含了大部分的字典和列表的操作，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。通过这些示例代码，基本上能够了解Robot Framework框架的使用方法，也能够对Robot Framework框架的编写方式和编写思想有了一定的认识。
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