Question

我正在为防火墙配置文件编写解析器。我是PyParsing和Python的新手。

问题是，如果发生超过3个参数，我如何解析，（xxxx，xxxx，xxxx）！=（xxxx，xxxx，xxxx，xxxx），所有规则都正常工作并且如果每行都包含no，则正确解析所有内容超过3个字符串，但我们可以看到[防火墙[F1]]在地址字段后面包含“NAT”，无论我们如何更改规则，都会被忽略。

使用（def printTokens（s，loc，toks）：＃s = orig字符串，loc = location，toks =匹配的标记）

使用第4个参数（“NAT”）时以及擦除时请查看2个输出。在此先感谢！需要解析包括“NAT”在内的所有规则。

from pyparsing import *

#===================================GRAMMER==================================
zone = Literal("zone")    
zoneid = Word(alphanums)
host = Literal("host")
hostid = Word(alphanums)
interface = Literal("interface")
interfaceid = Word(alphanums)
firewall = Literal("firewall")
firewallid = Word(alphanums)
router = Literal("router")
routerid = Word(alphanums)

fstop = Literal(".")
comma = Suppress(",") #Converter for ignoring the results of a parsed expression.
slash = Literal("/")
ocbracket = Literal("{")
ccbracket = Literal("}")
sobracket = Literal("[")
scbracket = Literal("]")
hyphen = Literal("-")
underline = Literal("_") 
word = Word(alphas)


#===================================IP-TYPE=================================

ip=Combine(Word(nums)+            
        fstop+ Word(nums) + 
        fstop+ Word(nums) + 
        fstop + Word(nums))

subnet = Combine(slash +Word(nums))

address = ip + Optional(subnet)


#===================================RULES===================================

#adword = address + word

zoneRule = zone + zoneid + address
hostRule = host + hostid + ocbracket
interfaceRule = interface + interfaceid + address 
interfaceRule2 = interface + interfaceid + address + word
firewallRule = firewall + firewallid + ocbracket
routerRule = router + routerid + ocbracket

endRule = ccbracket


rule = zoneRule | hostRule | interfaceRule | interfaceRule2 | firewallRule | routerRule | endRule 
rules = OneOrMore(rule)

#===================================DATA=====================================
details = """zone zone1 10.1.0.0/24                   
         zone backbone 10.254.0.0/24
         zone zone 10.2.0.0/24
         host ha {
             interface iha 10.1.0.1
         }
         host hb {
            interface ihb 10.2.0.1
         }
         firewall f1 {
            interface ifla 10.1.0.254 
            interface iflback 10.254.0.101 nat
         }
         router r2 {
            interface ir2back 10.254.0.102
         }
         router r3 {
            interface ir3b 10.2.0.103
         }"""

#==================================METHODS==================================

    def printTokens(s,loc,toks):   #s=orig string, loc=location, toks=matched tokens
    print (toks)

zoneRule.setParseAction(printTokens) 
hostRule.setParseAction(printTokens)
interfaceRule.setParseAction(printTokens)
interfaceRule2.setParseAction(printTokens) #takes in 4 instances where as 3 declared
firewallRule.setParseAction(printTokens)
routerRule.setParseAction(printTokens)
endRule.setParseAction(printTokens)

rules.parseString(details)


#================================OUTPUT RESULT WITH NAT=================================
"""
['zone', 'zone1', '10.1.0.0', '/24']
['zone', 'backbone', '10.254.0.0', '/24']
['zone', 'zone', '10.2.0.0', '/24']
['host', 'ha', '{']
['interface', 'iha', '10.1.0.1']        
['}']
['host', 'hb', '{']
['interface', 'ihb', '10.2.0.1']
['}']
['firewall', 'f1', '{']
['interface', 'ifla', '10.1.0.254']
['interface', 'iflback', '10.254.0.101']"""
#================================OUTPUT RESULT WITHOUT NAT=================================
"""['zone', 'zone1', '10.1.0.0', '/24']
['zone', 'backbone', '10.254.0.0', '/24']
['zone', 'zone', '10.2.0.0', '/24']
['host', 'ha', '{']
['interface', 'iha', '10.1.0.1']
['}']
['host', 'hb', '{']
['interface', 'ihb', '10.2.0.1']
['}']
['firewall', 'f1', '{']
['interface', 'ifla', '10.1.0.254']
['interface', 'iflback', '10.254.0.101']
['}']
['router', 'r2', '{']
['interface', 'ir2back', '10.254.0.102']
['}']
['router', 'r3', '{']
['interface', 'ir3b', '10.2.0.103']
['}']"""

Answer 1

如果要将任意数量的表达式与特定分隔符匹配，请使用PyParsing's delimitedList。默认情况下，它允许分隔符周围的空格;添加combine=True以不需要空格。

但是，如果要在语法中允许可选项，则只需添加一个可选项。对于您的接口规则，您可以替换：

interfaceRule = interface + interfaceid + address 
interfaceRule2 = interface + interfaceid + address + word

使用：

interfaceRule = interface + interfaceid + address + Optional(word)

最后，您发布的代码的实际问题是您使用的是|运算符，这是MatchFirst的简写形式。 MatchFirst将按顺序尝试给定选项，并返回 first 匹配的结果。如果您改为使用Or，那么简写形式为^运算符，那么它将尝试所有选项并返回带有最长的匹配。

PyParsing不同的字符串长度

1 个答案: