我正在为防火墙配置文件编写解析器。 一般来说,我对Python和Python都是新手。你知道吗
问题是如果出现3个以上的参数,(xxxx,xxxx,xxxx),我该如何解析!=(xxxx,xxxx,xxxx,xxxx),如果每行不包含超过3个字符串,则所有规则都正常工作并正确解析所有内容,但我们可以看到[Firewall[F1]]在地址字段后包含“NAT”,并且无论我们如何更改规则都会被忽略。你知道吗
使用(def printTokens(s,loc,toks):#s=原始字符串,loc=位置,toks=匹配的令牌)
当使用第四个参数(“NAT”)和删除它时,请查看这两个输出。 提前谢谢!需要解析所有内容,包括实现规则的“NAT”。你知道吗
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']
['}']"""
如果要将任意数量的表达式与某个分隔符匹配,请使用PyParsing's delimitedList。默认情况下,它允许分隔符周围有空格;添加
combine=True
以不需要空格。你知道吗但是,如果您想在语法中允许可选项,您应该只添加一个可选项。对于接口规则,可以替换:
使用:
最后,您发布的代码的实际问题是使用
|
运算符,这是MatchFirst的简写形式。MatchFirst将按的顺序尝试给定的选项,并返回匹配的first的结果。如果改为使用Or,其缩写形式是^
运算符,则它将改为尝试所有选项,并返回与最长匹配的选项。你知道吗相关问题 更多 >
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