简约-一个简单的递归模式

Question

我希望能够解析简单的规则表达式，这些表达式可以在parsimonious中使用and和or等连接词连接在一起。

我已经尝试了一个非常基本的语法，它可以解析一个简单的表达式，但是一旦我开始引入连接词，它就失败了。

import parsimonious

grammar = parsimonious.grammar.Grammar( """
    rule = (rule)+
    rule = (fieldname function parameters) / (rule conjunction rule)
    fieldname = space? ("field1" / "field2" / "field3") space?
    function = space? ("equal to" / "not equal to") space?
    parameters = space? "()" space?
    conjunction = space? ("and" / "or") space?
    space = ~r"\s+"
    """)

测试一个简单的案例：

grammar.parse("field1 equal to ()")

成功解析(至少它看起来像是构建了一个节点树--我还没有深入了解它对内容的拆分有多好--但乍一看似乎还不错)

但是对于一个更复杂的情况：

grammar.parse("field1 equal to () and field2 not equal to ()")

它返回IncompleteParseError: Rule 'rule' matched in its entirety, but it didn't consume all the text. The non-matching portion of the text begins with 'and field2 not equal' (line 1, column 20).

我所阐述的语法试图允许任意连接的语句，但我肯定遗漏了一些东西。

我试着调整语法，以明确顶级类和低级类之间的区别：

grammar = parsimonious.grammar.Grammar( """
    rule = expr+
    expr = (fieldname function parameters) / (expr conjunction expr)
    fieldname = space? ("field1" / "field2" / "field3") space?
    function = space? ("equal to" / "not equal to") space?
    parameters = space? "()" space?
    conjunction = space? ("and" / "or") space?
    space = ~r"\s+"
    """)

现在，当运行两部分短语时：

grammar.parse("field1 equal to () and field2 not equal to ()")

我得到的是RecursionError: maximum recursion depth exceeded in comparison，而不是IncompleteParseError。

Answer 1

在expr的(expr conjunction expr)部分，你有一个左递归问题。因此，您需要将其分解为单独的规则，如下所示：

rule = expr+
expr = field_expr (conjunction expr)?
field_expr = fieldname function parameters
fieldname = space? ("field1" / "field2" / "field3") space?
function = space? ("not equal to" / "equal to") space?
parameters = space? "()" space?
conjunction = space? ("and" / "or") space?
space = ~r"\s+"