我使用Pyparsing库来计算简单的布尔查询,比如:
使用examples部分(simpleBool.py)中的simpleBool脚本,我在尝试验证表达式语法时遇到了一个障碍。下面这些表达式即使有明确的语法问题,也被认为是有效的:
有人知道如何用Pyparsing验证语法吗?在
以下是测试脚本,按要求:
#
# simpleBool.py
#
# Example of defining a boolean logic parser using
# the operatorGrammar helper method in pyparsing.
#
# In this example, parse actions associated with each
# operator expression will "compile" the expression
# into BoolXXX class instances, which can then
# later be evaluated for their boolean value.
#
# Copyright 2006, by Paul McGuire
# Updated 2013-Sep-14 - improved Python 2/3 cross-compatibility
#
from pyparsing import infixNotation, opAssoc, Keyword, Word, alphas
# define classes to be built at parse time, as each matching
# expression type is parsed
class BoolOperand(object):
def __init__(self,t):
self.label = t[0]
self.value = eval(t[0])
def __bool__(self):
return self.value
def __str__(self):
return self.label
__repr__ = __str__
__nonzero__ = __bool__
class BoolBinOp(object):
def __init__(self,t):
self.args = t[0][0::2]
def __str__(self):
sep = " %s " % self.reprsymbol
return "(" + sep.join(map(str,self.args)) + ")"
def __bool__(self):
return self.evalop(bool(a) for a in self.args)
__nonzero__ = __bool__
__repr__ = __str__
class BoolAnd(BoolBinOp):
reprsymbol = '&'
evalop = all
class BoolOr(BoolBinOp):
reprsymbol = '|'
evalop = any
class BoolNot(object):
def __init__(self,t):
self.arg = t[0][1]
def __bool__(self):
v = bool(self.arg)
return not v
def __str__(self):
return "~" + str(self.arg)
__repr__ = __str__
__nonzero__ = __bool__
TRUE = Keyword("True")
FALSE = Keyword("False")
boolOperand = TRUE | FALSE | Word(alphas,max=1)
boolOperand.setParseAction(BoolOperand)
# define expression, based on expression operand and
# list of operations in precedence order
boolExpr = infixNotation( boolOperand,
[
("not", 1, opAssoc.RIGHT, BoolNot),
("and", 2, opAssoc.LEFT, BoolAnd),
("or", 2, opAssoc.LEFT, BoolOr),
])
if __name__ == "__main__":
p = True
q = False
r = True
tests = [("p", True),
("q", False),
("p and q", False),
("p and not q", True),
("not not p", True),
("not(p and q)", True),
("q or not p and r", False),
("q or not p or not r", False),
("q or not (p and r)", False),
("p or q or r", True),
("p or q or r and False", True),
("(p or q or r) and False", False),
]
print("p =", p)
print("q =", q)
print("r =", r)
print()
for t,expected in tests:
res = boolExpr.parseString(t)[0]
success = "PASS" if bool(res) == expected else "FAIL"
print (t,'\n', res, '=', bool(res),'\n', success, '\n')
@PaulMcGuire回答:
改变boolExpr.parseString(t) [0]到boolExpr.parseString(t,parseAll=True)[0]。如果Pyparsing可以在字符串的前导部分找到一个有效的匹配项,那么Pyparsing将不会引发异常,即使在结尾处附加了垃圾。通过添加parseAll=True,可以告诉pyparsing整个字符串必须成功解析。在
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